Math 9 (module 7)

Mathematics
Learner’s Material
9
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Department of Education
Republic of the Philippines
Module 7:
Triangle Trigonometry

Math atics G a 9
Learner’s Material
First Edition, 2014
ISBN: 978-971-9601-71-5
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Published by the Department of Education
Secretary: Br. Armin A. Luistro FSC
Undersecretary: Dina S. Ocampo, PhD
Development eam of the Learner’s Material
Authors: Merden L. Bryant, Leonides E. Bulalayao, Melvin M. Callanta, Jerry D. Cruz, Richard F.
De Vera, Gilda T. Garcia, Sonia E. Javier, Roselle A. Lazaro, Bernadeth J. Mesterio, and
Rommel Hero A. Saladino
Consultants: Rosemarievic Villena-Diaz, PhD, Ian June L. Garces, PhD, Alex C. Gonzaga, PhD, and
Soledad A. Ulep, PhD
Editor: Debbie Marie B. Versoza, PhD
Reviewers: Alma D. Angeles, Elino S. Garcia, Guiliver Eduard L.Van Zandt, Arlene A. Pascasio, PhD,
and Debbie Marie B. Versoza, PhD
Book Designer: Leonardo C. Rosete, Visual Communication Department, UP College of Fine Arts
Management Team: Dir. Jocelyn DR. Andaya, Jose D. Tuguinayo Jr., Elizabeth G. Catao, Maribel S.
Perez, and Nicanor M. San Gabriel Jr.
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Table of Contents
Module 7. triangle trigonometry...................................................................................... 425
Module Map .................................................................................................................................. 426
Pre-Assessment............................................................................................................................ 427
Learning Goals.............................................................................................................................. 429
Lesson 1. The Six Trigonometric Ratios: Sine, Cosine, Tangent, Secant,
Cosecant, and Cotangent...................................................................................... 430
Lesson 2. Trigonometric Ratios of Special Angles............................................................ 447
Lesson 3. Angles of Elevation and Angles of Depression.............................................. 457
Lesson 4. Word Problems Involving Right Triangles ....................................................... 467
Lesson 5. Oblique Triangles ..................................................................................................... 477
Lesson 5.1. The Law of Sines and Its Applications........................................................... 480
Lesson 5.2. The Law of Cosines and Its Applications ...................................................... 497
Glossary of Terms......................................................................................................................... 506
References and Websites Links Used in this Module...................................................... 507

425
MODULE 7
TRIANGLE TRIGONOMETRY
I. INTRODUCTION AND FOCUS QUESTIONS
Have you ever wondered how towers and buildings were constructed? How do you determine
the distance traveled as well as the height of an airplane as it takes off? What about determining
the height of the mountain? We can do all these things even if we are not in the real place or sit-
uation. These are just some of many real-world applications of triangle trigonometry.
www.olx.com.ph www.nww2m.com
This module will help you understand how to use the concepts on triangle trigonometry in
solving different real-life problems involving right triangle.
II. LESSONS AND COVERAGE
In this module, you will find answers to questions above by studying the following lessons:
Lesson 1 – The Six Trigonometric Ratios: sine, cosine, tangent, secant,
cosecant, and cotangent
Lesson 2 – The Trigonometric Ratios of Special Angles
Lesson 3 – Angles of Elevation and Angles of Depression
Lesson 4 – Application: The Use of Trigonometric Ratios in Solving
Real-Life Problems involving Right Triangles
Lesson 5 – Oblique Triangles
5.1 – Law of sines and its applications
5.2 – Law of cosines and its applications

426
In these lessons, you should be able to:
Lesson 1
• illustrate the six trigonometric ratios;
• apply trigonometric ratios to solve right triangles given:
a. the length of the hypotenuse and the length of one leg
b. the length of the hypotenuse and one of the acute angles
c. the length of one leg and one of the acute angles
d. the length of both sides
Lesson 2
• determine trigonometric ratios involving special angles;
• compute the numerical values of trigonometric expressions
involving special angles;
Lesson 3
• illustrate angles of elevation and angles of depression;
• distinguish between angle of elevation and angle of depression;
• solve problems involving angles of elevation and depression;
Lesson 4
• use the trigonometric ratios in solving real-life problems involving
right triangles
Lesson 5
• illustrate the laws of sines and cosines
• solve problems involving oblique triangles
Module Map

427
III. PRE-ASSESSMENT
Let us find out how much you already know about this module. Answer the following ques-
tions as much as you can by writing on your answer sheet the letter that you think is the correct
answer. Take note of the items that you were not able to answer correctly and then let us find
out the correct answer as we go through this module.
1. With respect to the given angle, what is the ratio of the hypotenuse to the opposite side?
A. sine B. cosine C. tangent D. cosecant
2. Which of the following statements is correct?
A. x = 8 C. sin 60° =
y
4
B. sin30° =
1
x
D. cos 60° =
4
y
3. Given the figure on the right, which refers to the angle of depression?
A. ∠MKN C. ∠LKN
B. ∠MKL D. none of these
4. In the triangle PQR, what is the length of PQ?
A. 1 cm C. 9 cm
B. 5 cm D. 12 cm
5. In the triangle DEF, what is m ∠E to the nearest degree?
A. 16° C. 41°
B. 28° D. 62°
17
15
D
F
E
12 cm
P
Q R
13 cm
object
object
N
M
L
Keye
x
y60°
4

428
6. If p = 30 and q = 60, what is the measure of ∠R?
A. 40° C. 60°
B. 55° D. 65°
7. Find the value of side c to the nearest unit.
A. 20 C. 23
B. D. 24
8. What is the value of x?
A. 3 C. 6
B. 3 2 D. 6 2
9. The expression 2(sin 30°) – tan 45° is equal to
A. 0 B. 1 C. 2 D. –1
10. A kite held by 125 m of string makes an angle of elevation with the ground of 45°. About
how high is the kite above the ground?
A. 62.8 m C. 88.4 m
B. 75.1 m D. 113.6 m
11. From the top of a barn 7.62 m high, you see a cat on the ground. The angle of depression of
the cat is 40°. How many meters must the cat walk to reach the barn?
A. 9.08 m C. 9.81 m
B. 9.80 m D. 9.18m
12. ABCD is a parallelogram. If AB is 8 cm long, BC is 5 cm and their included angle measures
100°, how long is diagonal AC?
A. 12.95 cm C. 10.40 cm
B. 12.59 cm D. 10.14 cm
13. In right triangle PQR, PQ = 12 cm and QR = 5 cm. What is cos R?
A. 12
13
B. 5
13
C. 5
12
D. 12
5
x
45°
6
45°
40°
c
15
b
Q R
P
q
p
r

429
14. ∆XYZ is a non – right triangle. If XY measures 20 cm, XZ measures 15 cm and ∠Z measures
35° then what is the measure of ∠Y?
A. 25.84° B. 24.85° C. 25.48 D. 24.58°
15. A balloon is 50 m high. Its angle of elevation from observer A is 45° and from observer B it
is 30°. What is the maximum distance between the two observers? Express your answer to
the nearest meter.
A. 136 m B. 137 m C. 138 m D. 39 m
16. With the sun, a girl 1.4 m tall casts a 3.6 m shadow. Find the angle of elevation from the tip
of the shadow to the sun. Express your answer to the nearest degree.
A. 19° B. 20° C. 21° D. 22°
17. You are walking along a straight level path toward a mountain. At one point the angle of
elevation of the top of the mountain is 40°. As you walk 250 m closer, the angle of elevation
is 45°. How high is the mountain? Express your answer to the nearest tenth of a meter.
A. 1304 m B. 1340 m C. 1034 m D. 1043 m
18. Two points P and Q on the same side of a river are 12 m apart. A tree on the opposite side of
the river is directly opposite a point between P and Q. The lines of sight of a tree across the
river make angles of 78° and 57°, respectively, with the line joining P and Q. Find the dis-
tance from point Q to the tree. Express your answer to the nearest meter.
A. 14 m B. 15 m C. 16 m D. 17 m
19. A surveyor sights two signs and the angle between the two lines of sight measures 55°. If the
first and the second signs are 70 m and 50 m away, respectively, from the surveyor, find the
distance between the two signs. Express your answer to the nearest meter.
A. 81 m B. 82 m C. 83 m D. 84 m
20. Two ships leave the same port at the same time. One ship sails on a course of 110° at 32 mi/h.
The other sails on a course of 230° at 40 mi/h. Find the distance between them after 2 hours.
Express your answer to the nearest mile.
A. 124 mi B. 125 mi C. 126 mi D. 127 mi
IV. LEARNING GOALS
The learner demonstrates understanding of the basic concepts of trigonometry and is able to
apply the concepts of trigonometric ratios to formulate and solve real-life problems with pre-
cision and accuracy.

430
The Six Trigonometric Ratios: Sine, Cosine,
Tangent, Secant, Cosecant, and Cotangent
What to Know
Lesson 1 will help you recall the different concepts about triangles. This will guide you to
define and illustrate the six trigonometric ratios.
➤ Activity 1: Triangles of Different Sizes
This activity helps you recall the concepts of similar triangles.
Investigate the following triangles:
1. Draw three similar right triangles ABC, DEF, and GHI in different sizes in such a way that
m ∠C = m ∠F = m ∠I = 63°.
2. Measure the second acute angle in each of the triangles.
3. Use a ruler to measure the sides of the triangles to the nearest tenths in centimeters. Then
find each of the following ratios for all the three triangles:
a. the ratio of the leg opposite the 63° angle to the hypotenuse
b. the ratio of the leg adjacent to the 63° angle to the hypotenuse
Record your findings in the given table
Measures in ∆ABC in ∆DEF in ∆GHI
leg opposite the 63° angle
leg adjacent to the 63° angle
hypotenuse
leg opposite 63° angle
hypotenuse
leg adjacent to 63° angle
hypotenuse
leg opposite 63° angle
leg adjacent to 63° angle
4. Analyze the ratios of the sides across these triangles. What do you notice about:
a. the first ratio, leg opposite 63° angle to the hypotenuse?
b. the second ratio, leg adjacent to 63° angle to the hypotenuse?
c. the third ratio, leg opposite the 63° angle to the leg adjacent to 63° angle?
1

431
5. Using a scientific calculator, determine the value of sin 63°, cos 63°, and tan 63° then
compare them to the values obtained in step 4. What do you observe?
6. Complete each statement:
The ratio _____ is the approximate value of sine of 63°.
The ratio _____ is the approximate value of cosine of 63°.
The ratio _____ is the approximate value of tangent of 63°.
7. Complete the sentence:
In a right triangle having an acute angle,
a. the sine θ is the ratio between _________and ___________.
b. the cosine θ is the ratio between ___________ and ___________.
c. the tangent θ is the ratio between __________and ____________.
From Activity 1, you have discovered the different ratios derived from the sides of a right
triangle having an acute angle. The next activity will help you further develop the concept
you have determined.
➤ Activity 2: Measuring and Calculating
For each of the right-angled triangles in the worksheet do the following:
1. Measure the marked angles;
2. Label the opposite and adjacent sides with respect to the marked angle;
3. Use your calculator to work out the three ratios:
Opposite
Hypotenuse
,
Adjacent
Hypotenuse
,
Opposite
Adjacent
4. Record the measurements and calculations in this table.
Length of sides (in cm) Trigonometric ratios
Angle (°) Opposite Adjacent Hypotenuse
Adjacent
Hypotenuse
Opposite
Hypotenuse
Opposite
Adjacent
a)
b)
c)
d)
e)
f)

432
5. Examine the values of the three ratios. Could the ratios
Opposite
Hypotenuse
or
Adjacent
Hypotenuse
be greater than
1? Explain your answer.
6. What about the ratio
Opposite
Adjacent
? What greatest value could this ratio have? Why do you say so?
WORKSHEET
a. b.
c. d.
e. f.

433
To summarize what you have learned from the activities, study the concepts below about
the Six Trigonometric Ratios.
In a right triangle, we can define actually six trigonometric ratios. Consider the right triangle
ABC below. In this triangle we let θ represent -∠B. Then the leg denoted by a is the side adjacent
to θ, and the leg denoted by b is the side opposite to θ.
We will use the convention that angles are symbolized by capital letters, while the side
opposite each angle will carry the same letter symbol, in lowercase.
sin of θ = sin θ =
opposite
hypotenuse
cosecant of θ = csc θ =
hypotenuse
opposite
cosine of θ = cos θ =
adjacent
hypotenuse
secant of θ = sec θ =
hypotenuse
adjacent
tangent of θ = tan θ =
opposite
adjacent
cotangent of θ = cot θ =
adjacent
opposite
SOH – CAH – TOA is a mnemonic used
for remembering the equations.
Notice that the three new ratios at the
right are reciprocals of the ratios on the left.
Applying algebra shows the connection
between these functions.
csc θ =
hypotenuse
opposite
=
1
opposite
hypotenuse
=
1
sin θ
sec θ =
hypotenuse
adjacent
=
1
adjacent
hypotenuse
=
1
cos θ
cot θ =
adjacent
opposite
=
1
opposite
adjacent
=
1
tan θ
A
C B
θ
a
b c
hypotenuse
adjacent
opposite

434
Showing a formula for the Missing Parts of a Right Triangle
Example 1. Determine the equation or formula to find a missing part of the triangle.
a. Solve for s in the figure above. b. Solve for p in the figure above.
Solution: ∠P is an acute angle, t is the Solution: ∠P is an acute angle, t is the
hypotenuse, s is the side adjacent to ∠P. hypotenuse, and p is the opposite side of ∠P.
Use CAH, that is Use SOH, that is
cos θ =
adjacent
hypotenuse
cos P =
s
t
cos 53° =
s
t
s = 12 cos 53°
sin θ =
opposite
hypotenuse
sin P =
p
t
sin 53° =
p
12
p = 12 sin 53°
c. Solve for a in figure 2.
Solution: ∠B is an acute angle, b is the opposite side, and a is the side adjacent
to ∠B. Use TOA, that is
tan θ =
opposite
adjacent
=
b
a
tan B =
10.6
a
tan 67° =
10.6
a
a tan 67° = 10.6
a =
10.6
tan 67°
T
p
S
P
s
t = 12
53°
B
C A
ca
b = 10.6
67°

435
d. Solve for c in figure 2.
Solution: ∠B is an acute angle, b is the opposite side and c is the hypotenuse of the given
acute angle. Use SOH, that is
sin c =
opposite
hypotenuse
c sin 67° = 10.6
sin B =
b
c
c =
10.6
sin 67°
sin 67° =
10.6
c
Using the Calculator to Find Trigonometric Ratios
As you saw in previous sections, the values of the trigonometric ratios for any particular angle
are constant, regardless of the length of the sides. These values can also be found using a calcu-
lator. Also, you can use a calculator to find an angle when you are given a trigonometric ratio.
a. Finding a ratio given the angle
Example: To find the value of sin 38°, ensure that your calculator is operating in degrees.
Solution:
Press sin 38 = 0.615661475
The calculator should give sin 38° = 0.616, correct to three decimal places.
b. Finding an angle given the ratio
In finding the size of the angle to the nearest minute, given the value of the trigonometric
ratio, just follow the steps in the examples below.
Example: sin θ = 0.725, find θ to the nearest minute
Solution:
Press 2ndF sin 0.725 = 46.46884783
To convert this to degrees/minutes/seconds mode,
Press 2ndF D°M’S
The calculator gives you 46° 28’ (nearest minute)

436
c. Degrees and minutes
So far, all angles have been in whole degrees. However, one degree can be divided equally
into 60 minutes. Further, one minute can be divided equally into 60 seconds. Angle meas-
urement can also be expressed in degree/minutes form. We can use a calculator to convert a
degree measure from decimal form to degree, minute, and second form.
Example: Write 54.46° in degree and minutes, giving an answer correct to the nearest minute.
Solution:
Press 54.46° 2ndF D°M’S
The calculator gives 54°27’36”, or 54°28’. (nearest minute)
Try This
1. Use your calculator to find the value of the following, correct to two decimal places.
a. cos 85° d. cos 65°
b. sin 7° e. tan 23°
c. tan 35°
2. Using the degrees/minutes/seconds button on your calculator, write each of the following
in degrees and minutes, give answers to the nearest minute.
a. 17.8° d. 108.33°
b. 48.52° e. 35.24°
c. 63.7°
3. Find the size of the angle θ (to the nearest degree) where θ is acute.
a. sin θ = 0.529 d. sin θ = 0.256
b. cos θ = 0.493 e. tan θ = 0.725
c. tan θ = 1.8
4. Find m ∠θ, to the nearest minute, given that θ is acute.
a. sin θ = 0.9 d. cos θ = 0.501
b. cos θ = 0.013 e. tan θ = 2.3
c. tan θ = 0.958

437
A. Solving a right triangle given the measure of the two parts; the length of the
hypotenuse and the length of one leg
Solving a right triangle means finding the measure of the remaining parts.
Example:
Triangle BCA is right-angled at C. If c = 23 and b = 17, find ∠A, ∠B and a. Express your
answers up to two decimal places.
Solution: Sketch a figure:
a. Side b is the adjacent side of ∠A; c is the hypotenuse
of right triangle BCA. Use CAH, that is

cos θ =
adjacent
hypotenuse
cos A =
b
c
cos A =
17
23
cos A = 0.7391
A = cos–1
0.7391( )
We can use our scientific calculator to find an angle whose cosine value is 0.7391.
Using a scientific calculator, A = 42.340°
b. Since in part (a), it was already found that ∠A = 42.34°,
then ∠B = 90° – 42.34°
∠B = 47.66°.
c. Using the Pythagorean theorem:
a2
+ b2
= c2
a2
+ (17)2
= (23)2
a2
+ 289 = 529
a2
= 529 -289
a2
= 240
a = 240
a = 15.49
B. Solving a Right Triangle Given the Length of the Hypotenuse and the Measure of
One Acute Angle
Example:
Triangle BCA is right-angled at C if c = 27 and ∠A = 58°, find ∠B, b, and a.
B
AC
a
c = 23
b = 17

438
Solution:
a. To find B, since B and ∠A are complementary angles, then
∠B + ∠A = 90°
∠B = 90° – 58°
∠B = 32°
b. To find b, since b is the adjacent side of ∠A and c is the hypotenuse of right ∆BCA, then
use CAH.
cos θ =
adjacent
hypotenuse
cos A =
b
c
cos 58° =
b
27
b = 27 cos 58°
b = 27 0.5299( )
b = 14.31
c. To find a, since a is the opposite side of ∠A and c is the hypotenuse of right ∆BCA, then
use SOH.
sin θ =
opposite
hypotenuse
sin A =
a
c
sin 58° =
a
27
a = 27 sin 58°
a = 27 0.8480( )
a = 22.9
C. Solving a Right Triangle Given the Length of One Leg and the Measure of One
Acute Angle
Example:
Triangle ACB is right-angled at C. If ∠A = 63° and a = 11 cm, find ∠B, b, and c.
B
C Ab
a
c = 27
58°

439
Solution:
a. To find ∠B, take note that ∠B and ∠A
are complementary angles. Then,
∠B + ∠A = 90°
∠B = 90° – 63°
∠B = 27°
b. To find b, since b is the adjacent side c. To find c, since c is the
and a is the opposite side of ∠A, then hypotenuse and a is the opposite
use TOA. side of ∠A, then use SOH.
tan θ =
opposite
adjacent
tan A =
a
b
tan 63° =
11
b
b tan 63° = 11
b 1.9626( ) = 11
b =
11
1.9626
b = 5.60 cm
sin θ =
opposite
hypotenuse
sin A =
b
c
sin 63° =
11
c
c sin 63° = 11
c 0.8910( ) = 11
c =
11
0.8910
c = 12.35 cm
D. Solving a Right Triangle Given the Length of the Two Legs
Example:
Triangle ACB is right-angled at C. If a = 18.5 cm and b = 14.2 cm, find c, ∠A, and ∠B.
Solution:
a. To find c, use Pythagorean theorem:
c2
= a2
+ b2
c2
= (18.5)2
+ (14.2)2
c2
= 342.25 + 201.64
c2
= 543.89
c = 543.89
c = 23.32
A
C B
cb
a = 11
63°
A
C B
c
b = 14.2 cm
a = 18.5

440
b. To find ∠A, since a and b are c. Based on the fact that ∠A and ∠B are
opposite and adjacent side of ∠A complementary, the measure of angle
respectively, then use TOA. ∠B is 90° – 52° = 38°
tan θ =
opposite
adjacent
tan A =
a
b
tan A =
18.5
14.2
tan A = 1.3028
A = tan–1
1.3028( )
A = 52°
We can use our scientific calculator to find an angle whose tangent is 1.3028.
A = 52°
You have just learned the definition of the six trigonometric ratios. Make sure that you will
be able to use these in the succeeding activities.
What to PROCESS
In this stage, you will solve problems involving the measures of angles or sides of a right
triangle using the trigonometric ratios.
➤ Activity 3: Surfing the Trigonometry
Play the traditional Bingo Game as a class (activity with the teacher).
1. Make your own BINGO card. Draw a 5 x 5 grid, and number the squares in random fashion
from 1 to 24 (assuming a “Free Space”) or 1 to 25.
2. There must be a“Bingo caller”who will pick a random number from 1 to 24 by using a spin-
ner, picking numbers from a bag, or generating a random number from a graphing calculator.
3. When he/she announces the number, your teacher will present a question. You solve the
problem and mark the box on your BINGO card that corresponds to the question number.
4. BINGO is achieved in the normal manner – horizontally , diagonally, or vertically (or four
corners, if you wish).
5. Raise your hands if you have BINGO.
6. When somebody wins, everyone (including the winner) continues playing until all problems
have been solved, or until there are 5 or 6 winners.
7. Your teacher will collect your card after the game.

441
BINGO BOARD
Question:
1. Did you enjoy the activity? Why?
2. What have you learned from the activity?
3. What mathematical concept did you apply to answer the problems?
4. How will you use the concepts you have learned in your daily life?
5. Is it fun to study math? Why?
Now, you have learned the definition of the six trigonometric ratios and you were able to
use them in finding the missing sides or angles of an acute angle in the triangle. The next
activity will help you master the concepts.

442
➤ Activity 4: Try Me!
1. Group yourselves by 6 and choose your leader.
2. Activity sheets will be given.
3. You have 15 minutes to answer the given activity and discuss among yourselves.
4. The leader will go around each group to discuss or share outputs.
5. A volunteer will share what he/she has learned in the activity.
(In this exercise, all angle measures are in degrees, and the lengths of the sides are in centimeters)
Try this out!
Using the figure below, write
expression that gives the required
unknown value.
c
A C
B
a
b
1. If A = 15° and c = 37, find a.
2. If A = 76° and a = 13, find b.
3. If A = 49° and a = 10, find c.
4. If a = 21.2 and A = 71°,
find b.
5. If a = 13 and B = 16°, find c.
6. If A = 19° and a = 11, find c.
7. If c = 16 and a = 7, find b.
8. If b = 10 and c = 20, find a.
9. If a = 7 and b = 12, find A.
10. If a = 8 and c = 12, find B.
Use the given figure to solve the
remaining parts of right triangle
ACB.
B
C A
c
b
a
Try this out!
1. b = 17 cm and c = 23 cm
2. c = 16 and a = 7
3. b = 10 ad c = 20
4. b = 6 and c = 13
5. c = 13 and a = 12

443
Questions:
1. How did you find your answers?
2. What are the mathematical concepts that you have learned in the activity?
3. Do you think you can apply these concepts in your daily life? How? Why?
You skills in finding the missing sides and angles of a right triangle were developed through
the previous activities. The next one will help you reflect or understand and apply the six
trigonometric ratios.
What to REFLECT and UNDERSTAND
In this section, your goal is to check your understanding on how the six trigonometric ratios
are used.
Try this out! Try this out!
Try this out!
Sketch a figure and solve each right
triangle ABC with right angle at C,
given that:
1. A = 15° and c = 37
2. B = 64° and c = 19.2
3. A = 15° and c = 25
4. A = 45° and c = 16
5. B = 56° and c = 16
given that:
1. A = 76° and a = 13
2. A = 22° and b = 22
3. B = 30° and b = 11
4. B = 18° and a = 18
5. A = 77° and b = 42
given that:
1. a = 15.8, b = 21
2. a = 7, b = 12
3. a = 2, b = 7
4. a = 3, b = 3
5. a = 250, b = 250

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➤ Activity 5: Use! List! Explain!
Study the lesson. Then complete the following.
1. Use the figure below to answer the following questions.
a. What is the length of side t?
b. What is the measure of ∠B? ∠C?
c. How did you find the measure of the angles?
2. List the trigonometric ratio or ratios of B that involve the measure of the hypotenuse.
3. Explain how to determine which trigonometric ratio to use when solving for an unknown
part of a right triangle.
In the preceding activity, the discussion was about your knowledge on the six trigonometric
ratios. What are the new things that you have realized? Do you think you can use them in
your daily life? How and why?
The next activity will help you answer these questions.
What to TRANSFER
In this section you will also learn how to make and use a clinometer, an improvised
instrument used to measure the height of a certain object. Your goal now is to apply the
knowledge that you have learned about the six trigonometric ratios.
➤ Activity 6: The Clinometer
Objective: To make a clinometer and use it to measure the height of an object.
Materials required: Stiff card, small pipe or drinking straw, thread, a weight
(a metal washer is ideal)
Pre-requisite knowledge: Properties of right triangles
Procedure:
(A) To make a clinometer:
1. Attach a protractor on a cardboard and fix a viewing tube (straw or pipe) along the diameter.
2. Punch a hole (o) at the center of the semicircle.
3. Suspend a weight {w} from a small nail fixed to the center.
4. Ensure that the weight at the end of the string hangs below the protractor.
5. Mark degrees (in sexagecimal scale with 00 at the lowest and 10 to 900 proceeding both
clockwise and counterclockwise). [Fig 1].
C
TB
t
15
8

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(B) To determine the height of an object:
6. First measure the distance of the object from you. Let the distance be d.
7. Look through the straw or pipe at the top of the object. Make sure you can clearly see the
top of the object.
8. Hold the clinometer steady and let your partner record the angle the string makes on the
scale of the clinometer. Let this angle be θ.
9. Determine the height of the object.
10. Write your observation. You may change the distance of the object (by either moving the
object or by changing their position) and note how the angle of elevation varies.
Questions:
1. What have you realized after doing the activity?
2. How did you find the height of the object?
3. What other methods or devices can you suggest to solve the same type of problem?

446
➤ Activity 7: Trigonome – Tree
1. Look for a tree in your community then find
its height using a trigonometric ratio.
2. Draw an illustration.
3. Write a paragraph explaining how the
height of the tree was computed.
4. Determine the other trigonometric ratios.
5. What learning have you discovered in doing
such activity? Would you be able to use this
in your life? How and why?
In this section, your task was to measure the height of a tree using the trigonometric ratio.
How did you find the performance task? How did the task help you realize the real-life appli-
cation of the concepts?
SUMMARY/SYNTHESIS/GENERALIZATION
This lesson was about the six trigonometric ratios. Various activities were provided to help
you illustrate and define the six trigonometric ratios. You also learned how to use them
in finding the missing sides and angles of a right triangle and applied them to real-life
situations. Your knowledge in this lesson will help you understand the next topic, which is,
the trigonometric ratios of special angles.

447
Trigonometric Ratios of Special Angles
What to KNOW
In this lesson you will use the concepts you have learned in the previous lessons to evaluate
the trigonometric ratios of special angles. There are two triangles, the isosceles and equilateral
triangles that are frequently used in mathematics to generate exact values for the trigonometric
ratios. Consider the succeeding activities to develop mastery of this topic.
➤ Activity 1: Special Triangles & Exact Values
Form two groups to do the activity below. Share your output in class.
Case 1:
1. On a piece of paper, draw a square.
2. Set the length of each side to 1.
3. Cut the square along a diagonal. What triangles are formed?
4. Indicate the right angle in the triangle.
5. Determine the exact length of the hypotenuse using the Pythagorean theorem.
6. Label the hypotenuse and all other sides and angles.
7. Write the primary trigonometric ratios for 45° (SOH – CAH – TOA)
8. Rewrite the primary trigonometric ratios with rationalized denominators.
Case 2:
1. On a piece of paper, draw an equilateral triangle.
2. Set the length of each side to 2.
3. What are the measures of the angles of the triangle?
4. Label each angle.
5. Divide the triangle into two identical triangles using an altitude.
6. The base of the triangle is now divided into equal portions of length
7. Consider only one of the new triangles.
8. Indicate the right angle in the new triangle.
9. What is the third angle in the new triangle?
10. Determine the exact height (altitude) using the Pythagorean theorem.
11. Label the altitude (and all other sides and angles) with their measures.
12. Write the primary trigonometric ratios for 30° and 60° in the triangle.
13. Rewrite the primary trigonometric ratios with rationalized denominators.
2

a. b. c.
C
B A2
2 3
2 4
F E
D L
3
N M
3
3 2
leg
45°
45°
leg
hypotenuse
(leg times 2 )
shorter leg
(opposite the 30° angle)
60°
longer leg (opposite the 60° angle and
its length is shorter leg times 3 )
30°
hypotenuse
(2 times the shorter leg)
22
448
How did you find the activity? What did you discover from the activity? Do you think this
will be useful as you proceed to the next activity? You will see why.
➤ Activity 2: Compare My Size
Do the following activities and answer the questions that follow.
Use a protractor to find the measures of the angles of each triangle.
1. ∠A = _______ 1. ∠D = _______ 1. ∠L = _______
2. ∠B = _______ 2. ∠E = _______ 2. ∠M = _______
3. ∠C = _______ 3. ∠F = _______ 3. ∠N = _______
Questions:
1. What have you noticed about the lengths of the sides of each triangle?
2. What have you observed about the measures of the angles of each triangle?
3. What do you call these triangles?
4. Write the mathematical concepts that you learned from the activity.
In this activity, you have learned about some special angles. To evaluate the
trigonometric ratios of these special acute angles, we can use geometric methods. These
special acute angles are 30°, 45°, and 60°.
Key Concepts
In Geometry, the following sides of special right triangles are related as follows:

a. b.

m = 2 · 8 = 8 2 r = 2 · 3 2 = 6

c.
45° 45°
m
8
45°
45°
3 3
r
t
60° 30°
s
9
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45° – 45° – 90° Right Triangle Theorem 30° – 60° – 90° Right Triangle Theorem
In a 45° – 45° – 90° triangle, In a 30° – 60° – 90° triangle;
✔ the legs are congruent; ✔ the length of the hypotenuse is twice the
✔ the length of the hypotenuse is 2 length of the shorter leg
times the length of a leg ✔ the length of the longer leg is 3 times the
length of the shorter leg
hypotenuse = 2 leg hypotenuse = 2 shorter leg
longer leg = 3 shorter leg
Example 1: Find the length of the indicated side.

To practice the concepts that you have learned in the previous activity, take the next one to
fully develop these ideas.

9 = t 3 the longer leg = 3 shorter leg( )
t =
9
3

3
3
 =
9 3
3
= 3 3 solve for t( )
s = 2t hypotenuse = 2 shorter leg( )
s = 2  3 3 = 6 3 substitute 3 3 to t( )

1. 2. 3.
4. 5. 6.
7. 8. 9.
10. 11. 12.
13. 14. 15.
45°
45°
8
r
t
45°
i g
2
45°
600
n
45°
15 2
o
45°
m
3
45°
e
5
5
t
a
30°
10
60°
t
30°
2 3
60°
r
c
30°
40
r
i
9 3
60°
o
f
2 3
30°60°
x
s60°
i
o
12
30°45°
p
s7 2
60° 45°
4 3
e c
a e
i
l
30° 60°
n
a
10
g l
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➤ Activity 3: Practice Makes Perfect
Find the value of each variable used in the figures. If your answer is not an integer, express it in
simplest radical form.

Let a be the leg of a 45° – 45° – 90°
Triangle.
sin 45° = ______ sec 45° = ______
cos 45° = ______ csc 45° = ______
tan 45° = ______ cot 45° = ______
45°
45°
a
aa 2
Let a be the shorter leg of a 30° – 60° – 90°
Triangle.
sin 30° = ______ sec 30° = ______
cos 30° = ______ csc 30° = ______
tan 30° = ______ cot 30° = ______
30°
60°2a
a
a 3
451
Questions:
1. How did you answer the activity?
2. What mathematical concepts did you apply to find the answer?
3. What generalizations can you make after performing the activity?
Did you enjoy the activity? Has the activity helped you perform well in terms of the
concepts you used? If you still have questions regarding the ideas you have discovered, then
the next section will help you.
What to Process
In this section, you will study how to find the trigonometric ratios of special angles. The
succeeding activities will deepen your understanding of the concepts you learned from the
previous lesson.
➤ Activity 4: What Makes You Special?
1. Given the angles of the triangles below, find the values of the six trigonometric ratios. Then
answer the questions that follow.

Let a be the shorter leg of a 30° – 60° – 90°
triangle.
sin 60° = ______ sec 60° = ______
cos 60° = ______ csc 60° = ______
tan 60° = ______ cot 60° = ______
30°
60°
2a
a 3
a
452
2. Complete the table below that summarizes the values of the trigonometric ratios of the angles
30°, 45° and 60°.
TRIGONOMETRIC RATIOS OF THE ANGLES
θ sin cos tan csc sec cot
30°
45°
60°
Questions:
1. How did you find the values?
2. What did you discover about the values you obtained?
3. What do you think makes these angles special? Why?
Now that you know the values of the trigonometric ratios of special angles of a right
triangle, you are now ready to apply these concepts by doing the next activity.

453
What to REFLECT and understand
In this section you will learn the process of determining the values of expressions
involving trigonometric ratios of special angles even without the use of a calculator.
Remember that the two special right triangles are the 30° – 60° – 90° triangle and the isosceles
(45° – 45° – 90°) triangle.
Using the table of values in the previous activity, consider the following examples.
Example 1:
Determine the exact value of the expression sec 30° + cot 60°.
Solution: sec 30° + cot 60° =
2 3
3
+
3
3
=
3 3
3
= 3
Example 2:
Find the exact value of sin2
30° + cos2
45°.
Solution: sin2
30° + cos2
45° = sin 30°( )2
+ cos 45°( )2
=
1
2
⎛
⎝⎜
⎞
⎠⎟
2
+
2
2
⎛
⎝⎜
⎞
⎠⎟
2
=
1
4
+
2
4
=
3
4
Example 3:
Determine the value of angle x when sec x = 2
Solution: Referring to the table, sec 60° = 2. Therefore, x = 60°
Example 4:
If x = 45°, show that sec2
x + tan x = 3
Solution: Substituting x with 45°, we have

sec2
45° + tan 45°
= sec 45°( )2
+ tan 45°
= 2( )
2
+ 1
= 2 + 1
= 3

454
TRY THIS
• Find the exact value of (sec 30°) (cos 30°) – (tan 60°) (cot 60°)
• Evaluate:
5 sin2
30° + cos2
45° + 4 tan2
60°
2 sin 30° cos 45° + tan 45°
THINK ABOUT THIS
• Are all 30° – 60° – 90° triangles similar to each other? Explain.
• Suppose you know the length of one side of a 30° – 60° – 90° triangle, how can you
determine the lengths of the remaining sides without measuring them directly?
The next activity helps you deepen your understanding about the trigonometric ratios of
special angles. You will also need your knowledge of radicals.
➤ Activity 5: You Complete Me!
Look for a partner. Use the choices in the box to complete the equations below. Calculator is
not allowed.
1. cos 30° + ________ = 9. 3 csc 60° – ________ =
2. sin 30° – cos 60° = ________ 10. sin 45° (tan 45° – cos 60°) = ________
3. (sin 45°)2
= ________ 11. (sin 60°) (cos 30°) – (sin 30°) (cos 60°)= ____
4. tan 60° (________) = 1 12. (sin 30°) (cot 30°) (________) = 1
5. (sin 30°) (tan 45°) + (tan 30°) (sin 60°) = ___ 13. 6 cos 45° + 3 sec 45° = ________
6. (sin 60°) (cos 30°) + tan 45° = ________ 14. 3 (____) + 2 cos 60° – 2 tan 45° =
7. 2 sin 30° + ________ = 2 15. 2 (cos 30°)2
+ 3 (sin 30°)2
= ________
8. 3 (________) + 2 cos 45° =
Questions:
1. How did you answer the activity?
2. What mathematical concepts did you recall and use to find the exact values?
3. Do you think these concepts are important? Why?
4. How will you use these concepts in real-life situation?
sin 30°
1
2
6 2
9
4
sin 60°
1
tan 30°
1
2
sec 45°
tan 45°
2
4
csc 60°
7
4
cot 30°
0

5 3
60°30°
5
10
30°
45° 8m
p
p
q
r
455
Were you able to answer the activity correctly? In this section you have learned fully the les-
son on trigonometric ratios of special angles. You are now ready to use these concepts in
real-life situations.
What to Transfer
In this section, the objective is to apply your understanding of the lesson to real-life situations.
You will be given a task to demonstrate your learning.
➤ Activity 6: Application
You will be grouped into 5 to do the following tasks. Each group must present their output.
CASE 1: (Error analysis)
Agnes drew the triangle at the right.
Gina said that the lengths couldn’t be
correct. With which student do you
agree? Explain your answer.
CASE 2: (Formulating own problem)
Write a real-life problem that you can solve using a 30° – 60° – 90° triangle with a 12-meter
hypotenuse. Illustrate the problem with your complete solution.
CASE 3: (House repair)
After heavy winds brought by “Hanging Habagat” damaged a house, workers placed an
8-m brace against its side at a 45° angle. Then, at the same spot on the ground, they placed
a second longer brace to make a 30° angle with the side of the house. Show the complete
solution for each question below.
a. How long is the longer brace? Round your answer to the nearest tenth
of a meter.
b. About how much higher on the house does the longer brace reach than
the shorter brace?

d
456
CASE 4: (Geometry in 3 dimensions)
Find the length of d , the diagonal of a cube with
the following given sides.
a. 5 units
b. 7 units
c. sec x = 2 units
CASE 5:
As a hot-air balloon began to rise, the ground crew drove 1.2 mi to an observation station.
The initial observation from that instant is the angle between the ground and the line of
sight to the balloon is 30°. Approximately how high was the balloon at that point? (Assume
that the wind velocity was low and that the balloon rose vertically for the first few minutes.)
a. Draw the situation.
b. Show your solution
How did you find the activity? Did you learn from your group? From the lessons, what
mathematical concepts can you use in your daily life? When and how are you going to use
these concepts?
This lesson was about the trigonometric ratios of special angles. You have illustrated the
special acute angles 30°, 45°, and 60°.You have also studied how to evaluate the trigonometric
ratios of these special angles.You were given the chance to do some activities that apply what
you learned in this lesson. Your understanding of this lesson and other previously learned
concepts will help you facilitate your learning in the next lesson: the angle of elevation and
angle of depression.

457
Angles of Elevation and Angles
of Depression
What to KNOW
Suppose you are on top of a mountain looking down at a certain village, how will you directly
measure the height of the mountain? An airplane is flying a certain height above the ground.
Is it possible to find directly the distance along the ground from the airplane to an airport
using a ruler? The trigonometric ratios as you have learned in the previous lesson will help
you answer these questions. Perform the succeeding activities to apply these concepts in
solving real-life problems.
➤ Activity 1: Look Up! Look Down!
Follow the steps below and answer the questions that follow.
• Use a tape measure to measure the distance between your eyes and feet.
• Move around the room and find an object that is at the exact height as your eyes and label.
• Go outside the room and make an illustration of:
a. Tall objects/structures
1. 2. 3. 4.
b. Short objects/structures
1. 2. 3. 4.
3

EYE
Line of sight above observer
Angle of elevation
Horizontal line of sight
Angle of depression
Line of sight below observer
458
Questions:
1. How did you find the activity?
2. Describe the illustration or picture you have created from the activity.
3. What mathematical concepts did you learn from the activity? When you look up to tall objects
is there an angle formed? What about when you look down?
4. Do you think you can directly measure the height, the distance of the object you have listed
in the activity?
Solving real-life problems involving right triangles requires knowledge of some significant
terms, such as line of sight, angle of elevation, and angle of depression.
Definitions:
Line of sight is an imaginary line that connects the eye of an observer to the object
being observed.
The angle of elevation is the angle from the horizontal to the line of sight of the
observer to the object above.
The angle of depression is the angle from the horizontal to the line of sight of the
observer to the object below.
What to process
After learning the concepts on angle of elevation and angle of depression, you are now ready to
use these in solving such problems. In this section, you will be provided with an opportunity
to explore and solve problems in daily life through the following activities.

459
➤ Activity 2: Identify Me!
In the following figures, identify the segment that represents the line of sight, and identify the
angles (if any) that represent the angle of elevation or angle of depression.
Figure
Angle
of Elevation
Angle
of Depression
Line
of Sight

15.24
41°
x
460
Questions:
1. How did you identify the line of sight, angle of elevation and angle of depression?
2. What ideas have you learned from this activity?
3. Do you think you can use these ideas in your daily life?
4. Give and illustrate at least two situations in your life involving angle of elevation and angle
of depression.
You have just learned how to identify and describe line of sight, angle of elevation, and angle
of depression. The next activity will help you solve these problems.
The study of trigonometric ratios originated from geometric problems involving triangles.
To recall, solving a triangle means finding the lengths of the sides and measures of the angles of
the triangle. Trigonometric ratios may be used to solve problems involving angles of elevation
and depression.
Example 1:
A tower is 15.24 m high. At a certain distance away from the tower, an observer determines
that the angle of elevation to the top of it is 41°. How far is the observer from the base of
the tower?
Solutions:
A trigonometric ratio often helps us set up an
equation, which can then be solved for the missing
measurement. If two legs of the triangle are part of the
problem, then it is a tangent ratio. If the hypotenuse is
part of the problem, then it is either a sine or cosine ratio.
tan θ =
opposite
adjacent
tan 41° =
x
15.24
tan 41° = 15.24
x tan 41°
tan 41°
=
15.24
tan 41°

461
Example 2:
An airplane is flying at a height of 4 kilometers above the ground. The distance along the
ground from the airplane to the airport is 6 kilometers. What is the angle of depression from
the airplane to the airport?
Solution:
TRY THIS OUT!
You may ask your teacher to guide you in answering these problems.
1. A hiker is 400 meters away from the base of the radio tower. The angle of elevation to the
top of the tower is 46°. How high is the tower?
2. An observer in a lighthouse 48.8 m above sea level saw two vessels moving directly towards
the lighthouse. He observed that the angles of depression are 42° and 35°. Find the distance
between the two vessels, assuming that they are coming from the same side of the tower.
tan θ =
opposite
adjacent
tan θ =
4
6
tan θ = 0.6667
= tan–1
0.6667( )
θ = 33.69°

462
➤ Activity 3: Draw My Problem!
Objective:
To develop your accuracy in illustrating the pictures presented by the information in the given
word problems.
Materials:
Ruler, protractor, and drawing materials
Directions:
1. Look for a partner.
2. Draw the pictures presented by the information in the problems given. (These problems
don’t ask for an answer.)
3. Assume that buildings, ladders, etc. are all on level ground.
4. Clear, neat, and accurate illustrations are necessary.
1. The angle of elevation of the top of
the building from a point 30 meters
away from the building is 65°.
2. The angle of depression of a boy from a
point on a lighthouse 30.5 meters above
the surface of the water is 3°.
3. If an airplane that is cruising at an
altitude of 9 km wants to land at
NAIA, it must begin its descent so
that the angle of depression to the
airport is 7°.
4. A bird sits on top of a 5-meter lamppost.
The angle of depression from the bird
to the feet of an observer standing away
from the lamppost is 35°.
5. Two points on the same side of
a tree are 19.8 meters apart. The
angles of elevation of the top of the
tree are 21° from one point and 16°
from the other point.
Questions:
1. How did you find the activity?
2. Did you encounter any difficulty in illustrating the problems?
3. What mathematical concepts did you apply to have an accurate illustration?
4. Based on the drawings you have, how will you define angle of elevation and angle of depression
using your own words?You have just learned and practiced how to draw pictures presented by the
information in the given problems.Using this skill you can now easily illustrate and solve similar
problems involving angles of elevation and depression.

463
You have just learned and practiced how to draw pictures presented by the information in
the given problems. Using this skill you can now easily illustrate and solve similar problems
involving angles of elevation and depression.
What to Reflect and UNDERSTAND
In this section, you will illustrate and solve problems involving the angle of elevation and
angle of depression.You will also use the concepts that you acquired in the previous activity.
➤ Activity 4: Illustrate and Solve!
Look for a partner, then illustrate and solve the following problems. Use the template below.
Draw the diagram.
What is/are given?
What is to be determined?
Formula used
Solution
Problems:
1. A 12-meter high post casts a 19-meter shadow. Find the angle of elevation to the sun.
2. The angle of elevation from a boat to the top of a 92-meter hill is 12°. How far is the boat
from the base of the hill?

464
3. From the top of the control tower 250 m tall, an airplane is sighted on the ground below. If
the airplane is 170 m from the base of the tower, find the angle of depression of the airplane
from the top of the control tower.
4. From the top of a cliff 280 meters high, the angle of depression of a boat is 25°. How far from
the base of the cliff is the boat?
5. From an airplane at an altitude of 1200 m, the angle of depression to a rock on the ground
measures 28°. Find the horizontal distance from the plane to the rock.
Questions:
1. What have you noticed about the given problems?
2. How did you illustrate the information presented in the problem?
3. How did you solve them and what mathematical concepts did you apply?
4. How are these concepts important in your daily life?
5. Give one example in which you have experienced using these concepts in your life.
You have just learned how to solve word problems involving the angle of elevation and the
angle of depression.You also applied the different concepts of the trigonometric ratios. What
about real-life experience on how to use these mathematical concepts? Now, you can do this
by taking the next activity. The next activity will make use of your created clinometer.You will
be able to apply sine, cosine, and tangent ratios to find angles of elevation and depression.
You will be able to measure lengths and use measurements to determine angles measures.
What to transfer
In this section, your objective is to apply your understanding of the lesson regarding the angle
of elevation and depression to real-life situations. You will be given a task to demonstrate
this learning.

465
➤ Activity 5: What Can I Learn from My SHADOW?
Objectives:
• To apply sine, cosine, and tangent ratios to find angles of elevation and depression.
• To measure lengths and use measurements to determine angle measures.
Materials:
Tape measure, clinometer
Directions: You should be in groups of four members to do the activity.
Procedure:
1. Measure the height of a member in the group.
2. Measure the length of his/her shadow.
3. Using what you know about trigonometric ratios, determine the angle of elevation from the
ground to the sun. (Sketch a picture of the situation that will help in your computation.)
4. Measure the shadow of the object.
5. Using the angle of elevation and the measure of the shadow, use what you know about trig-
onometric ratios to determine the height of the object. (Sketch a picture of the situation that
will help you in your computation.)
Questions:
1. What have you learned in the activity?
2. Can the sine or cosine of an angle ever be greater than 1? If so, why?
3. Write a problem that applies angles of elevation and depression; show an illustration with
complete solution.
In this lesson you have studied how to illustrate and solve problems involving angle of
elevation and angle of depression.You were given an opportunity to do activities that helped
you identify and define angle of elevation and angle of depression through illustrations. Also
you have learned that in solving problems you need to draw a detailed diagram to help you
visualize them. These concepts that you have just learned will help you get through to the
next lesson.

466
Activity sheet:
What can I learn from my SHADOW?
Name: _________________________________ Date:______________
Group Members: ________________________________________________
Objective: Students will apply trigonometric ratios and other things they know about right
triangles to determine the height of an object outdoors.
Trigonometric Ratios:
You have used right triangles to determine some important relationships that you have listed.
Today, you and your group members will use these ratios to determine the height of an object
outside. Follow the steps below:
1. Pick one person in the group and measure the height.
Write the name of a person you are measuring.
2. Measure the length of that person’s shadow.
3. Using the appropriate trigonometric ratio, find the angle of elevation.
Sketch the picture.
4. Find the length of the shadow of the object your group has chosen.
5. Using the angle of elevation and the shadow length, find the height of the object.
6. Sketch a picture of the object, its shadow, and the angle of elevation.

467
Word Problems Involving
Right Triangles
What to KNOW
The concepts of trigonometric ratios which were discussed in lessons 1, 2, and 3 are essential
in solving word problems involving right triangles. Many real-life problems on right triangles
can be anchored on these concepts. Without knowledge of these concepts, you would be in
a dilemma learning the new lesson.
Before you proceed to the new lesson, let us recall these concepts through activity 1 and
activity 2.
➤ Activity 1: The Perfect Match
Using figures 1 and 2, match each trigonometric concept found in Column A with the correct
ratio found in Column B.
4
Column A Column B
1. tan A a.
2. cos A b.
3. sin P c.
4. cot P d.
5. sec P e.
b
c
p
q
q
r
r
p
a
b
B
C A
ca
b
Fig. 1 Fig. 2
P
R
Qr
q
p

468
Questions:
1. Did you find the activity helpful? Why?
2. What guarantees that your answers are correct?
3. How did you get the correct ratios?
The previous activity assessed your knowledge on the concept of trigonometric ratios. Now
that you have recalled the different trigonometric ratios, you may proceed to the next activity
involving the concepts of angles of elevation and depression.
➤ Activity 2: Who Am I?
In each of the following illustrations, identify whether ∠A is an angle of elevation or an angle
of depression.

469
Questions:
1. What did you realize in the activity?
2. Did the activity help you remember the concept of angles of elevation and depression? How?
3. How do you differentiate an angle of elevation from an angle of depression?
In the last activity, you have recalled the concepts of angles of elevation and depression.
These concepts will be used in the next activity which will require you to illustrate a
real-life application.
➤ Activity 3: Elevation or Depression
Consider the situation below.
A boy who is on the 2nd floor of their house watches his dog lying on the ground. The angle
between his eye level and his line of sight is 32°.
a. Which angle is identified in the problem: angle of elevation or angle of depression? Justify
your answer.
b If the boy is 3 meters above the ground, approximately how far is the dog from the house?
c. If the dog is 7 meters from the house, approximately how high is the boy above the
ground?
Questions:
1. Can you answer each problem? If so, how?
2. How will you illustrate the given situation in question b? in question c?
3. Use your illustrations to find –
a) the distance between the dog and the house.
b) the distance between the boy and the ground.
What to Process
Your goal in this section is to apply the concepts of angle of elevation and angle of depression
in solving real-life problems.

470
➤ Activity 4: Find Me!
Work in pairs and find what is asked. (If answers are not exact, round off to the nearest tenth.)
1. Find the mesure of ∠A 2. Find the altitude.
3. Find the distance between the car and the building.

4. Find the height of the flagpole. 5. Find the distance of the bird above the
ground.

471
Questions:
1. Does the activity help you reinforce your knowledge of angles of elevation and
depression? How?
2. Given the same situation in problem 3, if you were asked to find the distance between the
car and the observer, how will you go about it?
Pictures depicting real-life situations were given in the previous activity and your knowledge of
the concept of angles of elevation and depression was assessed. These concepts are important
for you to work on the next activity.
➤ Activity 5: Let's Work–It–Out
Group yourselves into 5 or as instructed by your teacher. Let a representative from each group
pick one problem. Members of each group should work together to solve the problem they have
picked. Present and explain your solution.
1. A ladder 8 meters long leans against the wall of a building. If the foot of the ladder
makes an angle of 68° with the ground, how far is the base of the ladder from the wall?
2. Adrian is flying a kite. He is holding the end of the string at a distance of 1.5 m above
the ground. If the string is 20 m long and makes an angle of 40° with the horizontal,
how high is the kite above the ground?
3. A man, 1.5 m tall, is on top of a building. He observes a car on the road at an angle
of 75°. If the building is 30 m high, how far is the car from the building?
4. A four-meter ladder leans against a wall. If the foot of the ladder makes an angle of
80° with the ground, how high up the wall does the ladder reach?
5. An airplane took off from an airport and traveled at a constant rate and angle of
elevation. When the airplane reached an altitude of 500 m, its horizontal distance
from the airport was found to be 235 m. What was the angle when the airplane rose
from the ground?
Questions:
1. What are your thoughts and feelings about the activity? Why?
2. How will you assess yourself in terms of the given activity? (Are you good at it or do you still
find difficulties in solving application problems on right triangles?)
3. Why do you assess yourself as such?

C Bx
A
40°
1000m
472
What to Reflect and understand
Your goal in this section is to look closer into real-life problems involving right triangles.
Two activities are presented to test your understanding.
➤ Activity 6: Where Have I Gone Wrong?
1. Read carefully the problems given below.
2. Analyze the suggested solutions and find out what is wrong with them.
3. Write your solutions in the 2nd column of the table given below.
4. On the 3rd column of the table, write an explanation why your solution is correct.
Problem No. 1: An airplane is flying at constant altitude of 100 m above the ground. At that
instant, it was recorded that the angle of depression of the airport is 40°. Find the horizontal
distance between the airplane and the airport.
Suggested Solution My Solution My Explanation
Problem 1:
Let A represent the airplane
and B the airport.
tan 40° =
opposite
adjacent
tan 40° =
x
1000
x = 1000 tan 40°( )
= 1000 0.8391( )
x = 839.1
∴ Tℎe horizontal distance
between the airplane and the
airport is 839.1 m.

G A
M
2 m 1 m
473
Problem No. 2: PAGASA announces that a typhoon is going to enter the Philippine Area of
Responsibility. Strong winds and heavy rainfall are expected over Bulacan and nearby provinces
in Central Luzon.
Adrian lives in Bulacan. He noticed that one of the lampposts installed in their garden is about
to collapse. As a precautionary measure, he attached a 2-meter wire to the lamppost to support
it. One end of the wire is attached one meter from the base of the lamppost and the other end
is attached to the base of a nearby tree. Determine the angle the wire makes with the ground.
Suggested Solution My Solution My Explanation
Problem 2:
sin G =
adjacent
hypotenuse
By Pythagorean Theorem,
GA = GM2
– MA2
GA = 22
– 12
GA = 3
sin G =
3
2
G = sin–1 3
2
⎛
⎝⎜
⎞
⎠⎟
G = sin–1
0.8660( )
∴m ∠ G = 60°
Questions:
1. Did you encounter difficulties in doing this activity? Why?
2. How did you overcome these difficulties?
Being able to analyze what is wrong with a given solution and present the right solution to
the problem will give you an edge in planning solutions to the problems that you encounter
in life. Activity 6 provided you with an opportunity to improve your thinking skills which
you will need in doing the next activity.

474
➤ Activity 7: Problem Solving
Read each of the following problems carefully and solve.
1. Obiwan is standing at a distance of 15 m from the base of a tree. From where he is standing,
he can see the top of the tree. If the tree is 15 m high and Obiwan is 1 m tall, what is the
angle of elevation of the top of the tree?
2. A bamboo pole is leaning against a tree. If the height of the tree is 12.2 meters and the angle
made by the pole and the ground is 40°, what is the length of the pole?
3. The angle of elevation of the sun is 27° at the same time that a flagpole casts a shadow 12 m
long. How high is the flagpole?
4. From the top of a lighthouse 29.5 m high, the angle of depression of an observer to a boat
is 28.5°. How far is the boat from the lighthouse?
5. According to a lighting specialist for an art gallery, for best illumination of a piece of art,
it is recommended that a ceiling mounted light be 1.8 m from the piece of art and that the
angle of depression of the art piece be 38°. How far from the wall should the light be placed
so that the recommendations of the specialist are met? Give your answer to the nearest tenth
of a meter.
Questions:
1. Can you solve problem 1 without computing? How?
2. How did you solve problem 2? Can you illustrate what the problem states?
3. In problem 3, if the angle of elevation of the sun is increased, what happens to the length of
the shadow of the flagpole? If the angle of elevation of the sun is decreased, what happens
to the length of the shadow? Support your answer.
4. How did you answer problem 4? problem 5?
What to transfer
Your goals are:
1. to apply the concepts you have previously learned in solving problems involving right
triangles through practical work approach;
2. to formulate a problem on right triangles using practical situations in your home or in
school and present its solution.

475
➤ Activity 8: Go the Distance (an outdoor activity)
1. Group yourselves per instructions of your teacher.
2. Each group must have the following tools:
a. improvised clinometer
b. meter stick
c. chalk or marking pen
d. scientific calculator
3. Locate your station by measuring the horizontal distance (assigned by the teacher to the
group prior to this activity) from the flagpole.
4. Once you have located your station, determine the angle of elevation of the top of the flagpole.
5. Sketch a figure to model the situation.
6. Using a scientific calculator and the model you made, find the height of the flagpole.
7. Compare the height you got from those of the other groups. Did you get the same height
as the other groups did? If not, what do you think caused the differences in your answers?
Why?
8. The next time you do a similar activity, what should you keep in mind to ensure the accuracy
of your answer?
In Activity 8, you have experienced learning outside the classroom setup. Through this activity,
you have learned that in realizing the right solutions, accuracy plays a very important role not
only in measurements but also in computations. This is just what you need to do Activity 9.

476
➤ Activity 9: Formulating Real–Life Problems
Note: This is a group activity. Refer to the groupings in Activity 8.
A. Warm-up Exercise:
Together with your group mates, formulate the problem based on Activity 8. Make sure to
use the distance assigned to your group.
B. Write a problem that involves right triangle using situations found at home or in school and
provide solutions to this problem.
Questions:
1. How do you feel about the activity?
2. Were you able to write a problem that is required in part B of the activity? If so, how did you
come up with it?
Note: Suggested Scoring Rubrics for Activity 9 (Formulating Real-Life Problems) are found
in the Teaching Guide.
Lesson 4 is all about solving word problems involving right triangles. The different
trigonometric ratios, angle of elevation, and angle of depression were recalled in this lesson
since these are vital concepts which you must understand to successfully solve right triangles.
Activities that will help you remember the difference between angles of elevation and angles
of depression were included in this lesson. Opportunities for cooperative learning where you
can develop team work and application problems using real-life situations that will help you
understand how the concept of trigonometry can be applied in the real world are likewise
provided through some of the activities in this lesson. The concept of right triangles discussed
in this lesson can serve as springboard for the next lesson which is solving oblique triangles.

acute triangle obtuse triangle
477
Oblique Triangles
Introduction
The triangles we see around us are not all right triangles. Look at the pictures below. Can you
see the triangular patterns? Do the triangles you see in the pictures contain a right angle?
5
The triangles that you see in the pictures are known as oblique triangles.
➤ Activity 1:
Look around the classroom.
1. Can you see things in the shape of oblique triangles? List down at least five things you see
around that may not be considered as right triangles. These are oblique triangles.
2. Take a second look at the things you have listed and find out what common characteristic
these triangles have.
3. If you were asked to classify the things in your list into two, how would you do it? What is
your basis for classifying them as such?
4. Can you now define an oblique triangle? Give your definition of an oblique triangle based
on what you have observed.
5. On the basis of your observation, how would you classify oblique triangles?
The following are the classifications of oblique triangles.

40°
35°
45°
45°
30°
65°
60°
40°
30°
45°80°
1. 2.
3. 4.
5. 6.
478
An oblique triangle is a triangle which does not contain any right angle.
Oblique triangles may be classified into two—acute and obtuse.
An acute triangle is a triangle whose angles are all less than 90°.
An obtuse triangle is a triangle in which one of the angles is more than 90°.
Lesson 5 is divided into two sections, namely Lesson 5.1 and Lesson 5.2. Lesson 5.1 deals with
The Law of Sines and Its Applications while Lesson 5.2 deals with The Law of Cosines and Its
Applications. These two laws are essential in solving oblique triangles since the trigonometric
ratios involving parts of a right triangle are not applicable in these types of triangles.
Oral Exercises:
Find the missing angle/s and identify whether the triangle is acute, obtuse, or neither.

7. 8.
9. 10.
90° 25°
20°135°
60°
28°
65°35°
479
Preliminary Exercises:
A. Use a scientific calculator to find the values of each of the following:
1. sin 20°, sin 160°
2. sin 50°, sin 130°
3. sin 125°, sin 55°
4. sin 162°, sin 18°
5. sin 89°, sin 91°
B. Observe the given angles and the values you have obtained. What do you notice? Based on
your observations, what can you conclude?

40°
20°
5
25°
3
6.5
70° 45°
6
32° 120°34°
6
8
85°
5.2
4.5 10
480
The Law of Sines
and Its Applications
What to know
➤ Activity 1:
Look at the triangles below.

triangle 1 triangle 2 triangle 3
triangle 4 triangle 5 triangle 6
1. What do you notice about each of the given triangles?
2. If you were asked to group the triangles into two, how will you group them? What is your
basis for grouping them that way?
3. Examine closely the triangles in Group 1. Can you solve the missing parts of the triangles?
How about those in Group 2? If you can solve the missing parts of these two groups of
triangles using the previous concepts you have learned, show how. If not, proceed to the
next activity.
➤ Activity 2: Exploring Possibilities
Consider the triangle below.
5.1

B
C
A
a b
c
B
A C
c
b
a = 6
42° 70°
481
1. Draw an altitude of the triangle from vertex C. Label the altitude as h.
2. Using trigonometric ratios, form an equation that will solve –
a) sin A b) sin B
3. Manipulate the equations obtained in No. 3 to find expressions that will solve for h. Can you
equate these expressions? Why?
4. If we divide both sides of the equation by the product of a and b, what is the resulting equation?
5. Next, draw an altitude from vertex A and label the altitude as h.
6. Using trigonometric ratios, form an equation that will solve –
a) sin B b) sin C
7. Manipulate the equations obtained in no. 6 to find expressions that will solve for h. Can you
equate these expressions? Why?
8. If we divide both sides of the equation by the product of b and c, what is the resulting equation?
9. Observe the equations obtained in nos. 4 and 8. Based on your observation, what can be said
about
sin A
a
,
sin B
b
, and
sin C
c
.
The Law of Sines is easy to follow and very useful in solving oblique triangles when
you know the following information:
• two angles and one side (SAA Case & ASA Case)
• two sides and an angle opposite one of these sides (SSA Case)
The Law of Sines is described by the relation,
Law of Sines

In any ABC,
sin A
a
=
sin B
b
=
sin C
c
Illustrative Example 1: SAA Case
Find the missing parts of ABC on the right.
Given: two angles and one side
∠A = 42°
∠C = 70°
a = 6
Solutions:
Since, side b and ∠B are not yet given, we can use the formula sin A
a
=
sin C
c
.

B
A C
c = 8.43
b = 8.31
a = 6
42° 70°
68°
482
sin A
a
=
sin C
c
Formula to use to solve for c
sin 42°
6
=
sin 70°
c
Substitute the given values
c sin 42° = 6 sin 70° Cross multiply
0.6691 c = 6 0.9397( ) Compute for the values of sin 42° and
sin 70° using a scientific calculator
0.6691 c = 5.6382 Simplify the resulting equation
c =
5.6382
0.6691
Solve for c
c = 8.43
To solve for side b, the formula to be used is
sin A
a
=
sin B
b
.. Notice that -∠B
is unknown. But you have learned in your Grade 8 Mathematics that in any triangle,
the sum of the measures of the three angles is 180°. Using this concept,
∠A + ∠B + ∠C = 180°, we have
42° + ∠B + 70° = 180°
112° + ∠B = 180°
∠B = 180° – 112°
∠B = 68°
Thus, we can now solve for side b.
Following the steps used earlier in solving for c, we now have

Thus, the triangle with the measures
of its parts is shown at the right.
sin A
a
=
sin B
b
0.6691b = 5.5632
sin 42°
6
=
sin 68°
b
b sin 42° = 6 sin 68°
0.6691b = 6 0.9272( )
b =
5.5632
0.6691
b = 8.31

60° 40°
c = 7
A B
C
ab
483
Illustrative Example 2: ASA Case
Determine the measures of the missing parts
of ABC on the right.
Given: two angles and the included side
Solutions:
Since the measures of the two angles of the triangle are known, the measure of the third angle can
be determined using the concept that the sum of the measures of the angles of a triangle is 180°.

To solve for side a, we can use the formula
sin A
a
=
sin C
c
.
sin A
a
=
sin C
c
Formula to use to solve for a
sin 60°
a
=
sin 80°
7
a sin 80° = 7 sin 60° Cross multiply
0.9848 a = 7 0.8660( ) Compute for the values of sin 80° and
0.9848 a = 6.062 Simplify the resulting equation
a =
6.062
0.9848
Solve for a
a = 6.16
sin A
a
=
sin C
c
Formula to use to solve for a
sin 60°
a
=
sin 80°
7
a sin 80° = 7 sin 60° Cross multiply
0.9848 a = 7 0.8660( ) Compute for the values of sin 80° and
0.9848 a = 6.062 Simplify the resulting equation
a =
6.062
0.9848
Solve for a
a = 6.16
To solve for side b, use the formula
sin B
b
=
sin C
c
and follow the steps earlier.

∠A = 60°
∠B = 40°
c = 7
∠A + ∠B + ∠C = 180°
60° + 40° + ∠C = 180°
100° + ∠C = 180°
∠C = 180° – 100°
∠C = 80°
sin B
b
=
sin C
c
sin 40°
b
=
sin 80°
7
b sin 80° = 7 sin 40°
0.9848 b = 7 0.6428( )
0.9848 b = 4.4996
b =
4.4996
0.9848
b = 4.57

C
A Bc
a = b sin A
b
C
A Bc
a
b
a
b sin A
b
C
A Bc
a
b
C
A Bc
a b
C
A c
a
b
C
A
a
60° 40°
c = 7
A B
C
a = 6.16
b = 4.57
80°
C
C
A
A
B
B B
c
ab
aa
b
b sin A
484
Shown below is the triangle with its complete parts.
The SSA CASE
When the measures of two sides and an angle opposite one of them are given, you may be able
to use the Law of Sines to find the measure of another angle of the triangle. These may, however,
lead to the following possibilities:
THE SSA POSSIBILITIES
If ∠A is an acute angle and a ≥ b, then
there is exactly one solution
If ∠A is an acute angle, a < b, and
a > b sin A, then there are two solutions.
(ambiguous case)
a = b sin A, then there is exactly one
solution.
a < b sin A, then there are no solutions.
If ∠A is an obtuse or a right angle and
a > b, then there is exactly one solution.

If ∠A is an obtuse or a right angle and
a ≤ b, then there is exactly no solution.

b CA
c = 19
a = 10
B
120°
485
Illustrative Example 3:
Solve for the missing parts of ABC on the right.
Given: two sides and an angle opposite one of these sides
Solutions:
∠C is an obtuse angle and c > a, thus there is exactly one solution.
Since a, c, and ∠C are known, we can use the formula
sin A
a
=
sin C
c
.
sin A
a
=
sin C
c
Formula to use to solve for A
sin A
10
=
sin 120°
19
19 sin A( ) = 10 sin 120°( ) Cross multiply
19 sin A( )= 10 0.8660( ) Compute for the values of sin 120°
using a scientific calculator
19 sin A( )= 8.66 Simplify the resulting equation
sin A =
8.66
19
= 0.4558 Solve for A
A = 27.12°
sin A
a
=
sin C
c
Formula to use to solve for A
sin A
10
=
sin 120°
19
19 sin A( ) = 10 sin 120°( ) Cross multiply
19 sin A( )= 10 0.8660( ) Compute for the values of sin 120°
using a scientific calculator
19 sin A( )= 8.66 Simplify the resulting equation
sin A =
8.66
19
= 0.4558 Solve for A
A = 27.12°
Using the concept that the sum of the angles of a triangle is 180°, we have
a = 10
c = 19
∠C = 120°
∠A + ∠B + ∠C = 180°
27.12° + ∠B + 120° = 180°
∠B + 147.12° = 180°
∠B = 180° – 147.12°
∠B = 32.88°

b = 11.91
CA
c = 19
a = 10
B = 32.88°
120°27.12°
486
Following the steps used earlier in solving for c, we can now solve for b.

Shown below is the triangle with its complete parts.
sin B
b
=
sin C
c
sin 32.88°
b
=
sin 120°
19
b sin 120° = 19 sin 32.88°
b 0.8660( ) = 19 0.5429( )
0.8660 b = 10.3151
b =
10.3151
0.8660
= 11.91

b = 15
C
A
B
c = 20
a
b = 15
27°
b = 15
C
A
a
27°
c = 20
B
b = 15
C
A
B
c = 20
a
27°
487
Solve ABC given that b = 15, c = 20, and ∠B = 27°.
b ? c sin B 15 ? 20 (0.4540)
15 ? 20 sin 27° 15 > 9.08
Since ∠B is an acute angle and b > c sin B, therefore this is an ambiguous
case and there are two solutions.
Solution 1: Solution 2:
Solving for ∠C, Solving for ∠C,

Note:
If ∠A and ∠B are supplementary angles,
then sin A = sin B.
Solving for ∠A, Solving for ∠A,
∠A + ∠B + ∠C = 180°
∠A + 27° + 34.58° = 180°
∠A + 61.58° = 180°
∠A = 180° – 61.58°
∠A = 118.42°
∠A + ∠B + ∠C = 180°
∠A + 27° + 145.42° = 180°
∠A + 172.42° = 180°
∠A = 180° – 172.42°
∠A = 7.58°
sin B
b
=
sin C
c
sin 27°
16
=
sin C
20
16 sin C = 20 sin 27°
sin C =
20 sin 27°
16
sin C =
20 0.4540( )
16
sin C = 0.5675
∠C = 34.58°
∠C = 180° – 34.58°
∠ C = 145.42°

b = 15
C
A
a = 29.05
27°
c = 20
B
b = 15
C
A
B
c = 20
a = 4.36
27° 145.42°
7.58°
118.42°
34.58°
1. 2. 3.
4. 5.
41°
36°
10
x
X
x
x
X
125°
20
5
95°
25°
5.5
50°
100°
25
30°
30
37
488
Solving for a, Solving for a,

The triangles with their complete parts are shown below.
Exercises:
A. State an equation that will solve x.
B. For each triangle given in Exercise A, solve for the unknown parts. Round off your answers
to the nearest tenth.
sin A
a
=
sin B
b
sin 118.42°
a
=
sin 27°
15
0.8795
a
=
0.4540
15
0.4540 a = 15 0.8795( )
0.4540 a = 13.19
a =
13.19
0.4540
a = 29.05
sin A
a
=
sin B
b
sin 7.58°
a
=
sin 27°
15
0.1319
a
=
0.4540
15
0.4540 a = 15 0.1319( )
0.4540 a = 1.9785
a =
1.9785
0.4540
a = 4.36

a.) b.)
a.) b.)
C
A
50°
B 45°
24
A
B
C
105°
12
40°
B
C
A
43
27
95°
17
27°
C
A
B
13
12
11
10
9
8
7
6
5
4
3
2
1
489
More Exercises:
1. Find the length of AB. Round off to the nearest tenth.
2. Find m ∠A. Round off your answer to the nearest tenth.
What to Process
Your goal here is to apply the Law of Sines in solving oblique triangles.
➤ Activity 3: Pair Me Up
What is the title of this picture?
Solve the triangles on the next page. If answers are not whole numbers, round off to the nearest
tenth. Then, write the corresponding letter on the blanks below to decode a message.

_____ _____ _____ _____ _____ _____ _____ _____ _____
14.1 15° 24.2 14.1 21.9 60° 19.6 24.2 14.7
_____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____
105° 120° 15° 15° 14 20 30° 55° 150° 45° 19.3 105°
_____ _____ _____ _____ _____ _____ _____ _____ _____ _____ _____
105° 14 38.6 120° 25° 14 105° 65 150° 19.3 5.4

T = ?
c = ?
W = ?
L = ?
G = ?
H = ?
E = ?
a = ?
o = ?
i = ?
n = ?
m = ?
k = ? f = ?
s = ?
r = ?
45° 30° 30° 120°
10
14
15°
20
60° 75°
16
45°
15°
18
490
Questions:
1. How did you fare in Activity 3?
2. Were you able to solve each triangle and the whole puzzle? If not, why?
3. The next time you do an activity, what measures will you take so you could finish
the activity?
In Activity 3 you were given illustrations of triangles in which you were asked to find the
missing parts and it gave you an opportunity to practice some more in solving oblique
triangles using the Law of Sines. In Activity 4 you are to draw the triangles based on given
conditions before you solve the triangle.
➤ Activity 4: Picture Me
Draw your own triangles given the indicated measures and solve for the other parts using the
Law of Sines. Round off all decimals to the nearest tenth.
1. ∠A = 73°, a = 18, b = 11
∠B = _____ ∠C=_____ c =_____
2. ∠A = 26°, ∠C = 35°, b = 13
∠B = _____ a =_____ c =_____

40°
30° O (ship)
E
S
N
W
A
N
B
30 km
491
Questions:
1. In each problem, which part of the triangle did you solve first? Why?
2. Do the diagrams you made really help you in solving the given triangles? How?
The previous activity assessed your ability to draw triangles given some conditions. Being
able to translate problems in the form of drawings or diagrams helps in analyzing what tool
should be used to solve it. In the next activity, you will see just how important diagrams are
in helping you understand the given problem.
➤ Activity 5:
Read the problem and analyze the partial solution presented below. Then, answer the last question
in the problem on your own.
Problem: Two towns, A and B, located along the coast of the Pacific Ocean are 30 km apart on
a north-south line. From a ship, the line of sight of town A is W30°N, while that of B is S40°W.
a) How far is the ship from town A? b) How far is the ship from town B?
Since line AB is along the
north-south direction,
a.) ∠B = 40°. Why?
b.) ∠AOB = 80°. Why?
c.) ∠A = 60°. Why?

40°
O (ship)
A
B
30 km
60°
80°
A
B
C
D
ac
b
h
A
B
C
D
a
c
b
h
A
B
C
Da
c
b
h
Fig. 1 Fig. 2 Fig. 3
492
Solving for AO,
The distance of the ship from town A
is 19.58 km.
Now, it is your turn to do the rest of
the problem.
Questions:
1. What are your thoughts about the activity?
2. If the partial solution was not provided, can you show the entire solution? If not, why?
3. Are the given sketches helpful? How did they help you?
What to reflect and understand
Your goal in this lesson is to reflect on how the Law of Sines can be applied to solve real-life
problems involving oblique triangles.
➤ Activity 6: Exploring Possibilities
Consider the different positions of ABC as it is rotated around a point.
In each position, the horizontal side is considered the base and the altitude is drawn from
the opposite vertex.
1. In each figure, explore how the altitude (h) of each triangle can be expressed using the sine
ratio. List as many expressions as possible.
2. Recall the formula for area of a triangle,

A
=
1
2
bh. Replace h by each of the expressions you
got in no. 1 and replace b by each of the bases in figures 1 to 3. Write down the formulas you
got on the board.
3. Analyze the resulting formulas.
sin B
AO
=
sin O
AB
AO =
AB sin B( )
sin O
=
30 sin 40°( )
sin 80°
AO =
30 0.6428( )
0.9848
= 19.58
∴

M
K L
150 m
52°
83°
493
4. Test the formulas using the triangles below.
Triangle 1 Triangle 2
5. Once you have tested the formulas, check your answers against the area you will get using
the original formula,

A =
1
2
bh . Are the formulas you have tested accurate?
The area of a triangle is one-half the product of any two sides and the sine of the included angle.
In symbols,
➤ Activity 7: Help Me Find a Solution
Work in pairs and solve the problems below. Read each problem carefully and discuss with your
partner how you will go about solving the problem.
1. A newly-wed couple wants to purchase a piece of land which is triangular in shape. If the
two sides of the lot measure 25 m and 30 m and their included angle is 60°, what is its area?
If the lot costs Php2,000 per square meter, how much would it cost the couple to buy the
lot?
2. A triangular parcel of land with points M, K, L was to be fenced. No data for the lengths of
sides MK and KL are available as shown in the figure below. How much fencing material is
to be used in the lot? Determine the area of the lot.
3. A pilot approaching a 10 000-foot runway finds that the angles of depression of the ends of
the runway are 12° and 15°. How far is the plane from the nearer end of the runway?

A =
1
2
ab sin C or A =
1
2
bc sin A or A =
1
2
ac sin B

494
4. A man measured the angle of elevation of the top of a tower to be 70°. When he walked
30 m further, the angle of elevation of the top of the tower was 35°. Find the distance from
the top of the tower to the second observation point.
5. From the top of a building 24 m high, the angle of elevation of a weather balloon is 54°, and
from the bottom of the building the angle is 62°. How high above the ground is the balloon?
Questions:
1. Does having a partner in solving the problems somehow make you comfortable?
2. Which among the given problems is the easiest for you and your partner to solve? Why?
3. In which problem did you encounter the most difficulty? Why?
4. Explain briefly how you solved each problem.
Activity 7 gave you a chance to work with a partner in solving real-life problems using the
Law of Sines. This also gave you an opportunity to discuss and clarify any misconceptions
on the topic. Now, let’s see if you can now solve real-life problems on your own by moving
on to Activity 8.
➤ Activity 8: All Alone but Still Going Strong
Read the given problems carefully and solve them. (Work individually.)
1. To find the distance between two points A and B that lie on opposite banks of a
river, a surveyor lays off a line segment AC of length 250 m along one bank and
determines that the measures of ∠BAC and ∠ACB are 64° and 43°, respectively.
Find the distance from A to B.
2. As you walk through a level path toward a mountain, the measure of the angle of
elevation to the peak of a mountain from one point is 33°. From another point
1000 ft closer, the angle of elevation is 35°. How high is the mountain?
3. Maggie observes a car and a tree from a window. The angle of depression of the
car is 45° and that of the tree is 30°. If the distance between the car and the tree is
100 m, find Maggie’s distance from the tree.
Question:
1. How does it feel solving the problems by yourself?
2. Were you able to solve all the problems? If yes, how did you do it? If not, why? What will you
do next time to make sure that you finish the task?
What to transfer
Your goal at this point is to create a short story using the concept you have learned.

495
➤ Activity 9:
1. Create a short story based on the scene pictured below.
2. Incorporate the concepts you have learned in the story by creating at least 2 problems. Present
the solutions to the problem.
3. Present your work to the class.
Questions:
1. How does it feel solving the problems by yourself?
2. Were you able to solve all the problems? If yes, how did you do it? If not, why? What will you
do next time to make sure that you finish the task?
➤ Activity 10: Did You Grasp the Idea?
Goal of task related to understanding: Your task is to survey the area of a triangular parcel
of land using the concept of sine.
Role: You are a surveyor.
Audience: The target audience are the prospective buyers of the land.
Situation: Three prospective buyers of the triangular parcel of land are interested in buying
the land. All of the prospective buyers want the land so much because each side of the land
is along a road which makes the land very accessible. The only thing that keeps them from

496
buying the land is that they do not know the area of the land. Prospective buyer no. 1 wants
the area to be within the range of 500 m2
to 750 m2
, prospective buyer no. 2 wants the area
to be within the range of 750 m2
to 850 m2
, and prospective buyer no. 3 wants the area to be
exactly 850 m2
. To make sure that the area of the land is what they really wanted, they agreed
to hire you as surveyor and share the expenses for your fee.
Product or Performance: Submit a sketch of the land drawn to scale and computations
indicating the area of the triangle.
Standards of Performance: Your work must meet the standards in the following rubric.
RUBRIC FOR PERFORMANCE
CRITERIA 4 3 2 1
Mathematical
Concept Used
Has shown full
understanding
of the concept as
evidenced in the
work presented.
Also used other
concepts leading
to the same
solution
Has shown full
understanding
of the concept as
evidenced in the
work presented
Has shown
some degree of
understanding
of the concept as
evidenced in the
work presented
Has no knowledge
of the concept as
evidenced in the
work presented
Accuracy of
Computations
The computations
were presented
logically and
done accurately.
Alternative
solution/s leading
to the same
answer has been
presented.
The computations
were presented
logically and done
accurately.
The computations
were presented
logically but
there were
minimal errors in
computations.
The computations
were incoherent
and erroneous.
Presentation
Sketch was
drawn precisely
according to
the scale used.
Presentation of
the sketch is clear
and easy to follow
and has add-on
features (such as
flowers and trees)
making the sketch
attractive.
Sketch was
drawn precisely
according to
the scale used.
Presentation of
the sketch is clear
and easy to follow.
Sketch was
drawn precisely
according to
the scale used.
Presentation of
the sketch is fairly
easy to follow.
Sketch was not
drawn to scale and
is hard to follow.

AB
C
c
a b
A B
C
c = 13.22
b
a = 8
° 63.39°
°
497
The Law of Cosines
and Its Applications
What to KNOW
Oblique triangles can also be solved using the Law of Cosines. This law states the following:
The square of the length of one side is equal to the sum of the squares of the other two sides
minus the product of twice the two sides and the cosine of the angle between them.
In symbols,
Law of Cosines
In any ABC,
a2
= b2
+ c2
– 2bc (cos A)
b2
= a2
+ c2
– 2ac (cos B)
c2
= a2
+ b2
– 2ab (cos C)
The Law of Cosines can be used in the following situations:
• two sides and the included angle are known
• three sides are known
Let us use ABC to illustrate the Law of Cosines.
Given : two sides and the included angle
Solutions:
To solve for c,
c2
= a2
+ b2
– 2ab (cos C) c2
= 208 – 33.3312
c2
= 82
+ b2
– 2 (8)(12)(cos 80°) c2
= 174.6688
c2
= 64 + 144 – 192 (0.1736) c = 13.22
5.2
∠C = 180°
a = 8
b = 12

A
C
c = 13.22
b = 12
a = 8
36.61° 63.39°
80°
A
B
C
c = 20
b = 10
a = 15
498
To determine the measure of ∠A,
a2
= b2
+ c2
– 2bc (cos A) (317.28)(cos A) = 254.6688
82
= 122
+ 13.222
– 2 (12)(13.22)(cos A) cos A =
254.6688
317.28
= 0.8027
64 = 144 + 174.6688 – (347.28)(cos A)
64 = 318.6688 – (317.28)(cos A) A = 36.61°
(317.28)(cos A) = 318.6688 – 64
Since the measure of ∠C is given and the measure of ∠A is now known, the measure of ∠B
can be computed using the equation∠A + ∠B + ∠C = 180°.

∠B = 180° – 116.61°
∠B = 63.39°
The triangle with its complete parts is shown below.
Determine the missing parts of ABC.
Given: three sides
a = 15
b = 10
c = 20
Solution:
a2
= b2
+ c2
– 2bc (cos A); b2
= a2
+ c2
– 2ac (cos B)
c2
= a2
+ b2
– 2ab (cos C)
Let's solve for the measure of ∠A .
a2
= b2
+ c2
– 2bc (cos A)
152
= 102
+ 202
– 2(10)(20)(cos A)
225 = 100 + 400 – 400 (cos A)
∠A + ∠B + ∠C = 180°
36.61° + ∠B + 80° = 180°
116.61 + ∠B = 180°

A = 46.57°
B = 28.96°C
c = 20
b = 10
a = 15
104.47°
499
–275 = – 400 cos A( )
–275
–400
=
–400 cos A( )
–400
0.6875 = cos A
A = 46.57°
Using the formula b2
= a2
+ c2
– 2ac (cos B) and following the steps used above, let’s find the
measure of ∠B .

Since two angles are already known, substitute their values in the equation
∠A + ∠B + ∠C = 180° to solve for ∠C
46.57° + 28.96° + ∠C = 180°
75.53° + ∠C = 180°
∠C = 180° – 75.53°
∠C = 104.47°
Below is the triangle with its complete parts.
What to Process
Your goal here is to apply the Law of Cosines in solving oblique triangles.
b2
= a2
+ c2
– 2ac cos B( )
102
= 152
+ 202
– 2 15( ) 20( ) cos B( )
100 = 225 + 400 – 600 cos B( )
–525 = – 600 cos B( )
–525
–600
=
–600 cos B( )
–600
0.8750 = cos B
B = 28.96°

1. 2.
3. 4.
5.
108° 120°
B
B
B
B
B
A
A
A
A A
C
C
C
C
C
3
9
4 4
37
8
8 5
6
6
5
130°
500
➤ Activity 1: Practice Makes Perfect
Solve each triangle. (If answers are not exact, round off to the nearest hundredth.)
Questions:
1. Were you able to solve the given triangles? How?
2. In each of the given triangles, which part did you solve first and why?
3. Explain how you solved each of the given triangles.
In Activity 1 you were given illustrations of triangles and you were asked to find the missing
parts. The next activity, however, will give you an opportunity to practice drawing triangles
based on specific conditions. This activity would help prepare you in dealing with application
problems.

501
➤ Activity 2: Sketch Me If You Can
Draw the triangle that best describes each of the situations given below. Label the given parts.
1. In ∆PQR, PQ is 7 cm long, QR is 5 cm, and the angle between them measures 52°.
2. ∆XYZ is an isosceles triangle. Its vertex angle Y measures 35° and each leg is 8 cm
long.
3. In ∆ABC, AB is 15 cm long, BC is 4 cm long and makes an angle measuring 138°
with AB.
4. In ∆ABC, all sides measure 10 cm.
5. ∆PQR is isosceles. ∠Q is the vertex angle and it measures 55° and one leg is 7 cm
long.
Questions:
1. How do you find Activity 2?
2. What should you keep in mind if you are to draw the diagrams yourself? Why?
Being able to illustrate a problem through diagrams will help you connect with the problem
which will make solving easier. The previous activity provided you this skill. In the next
activity, illustrations based on real-life situations are already provided. Your task is to apply
the Law of Cosines to solve the problems.
➤ Activity 3: Find Me Please
Below are sketches depicting real-life situations. Your task is to find the missing part indicated
in each sketch.

502
Questions:
1. How did you solve the problem in situation 1? in situation 2?
2. Did the illustrations help you in solving the given triangles? How did they help you?
What to reflect and UNDERSTAND
Your goal in this lesson is to reflect on how the Law of Cosines can be applied to solve real-
life problems involving oblique triangles.
➤ Activity 4:
For each of the given situational problems:
1. sketch the required triangle to solve the problem
2. use the Law of Cosines to solve it
1. Peter has three sticks measuring 19 cm, 23 cm, and 27 cm. He lays them down to
form a triangle. Find the measure of the angle formed by the 19-cm and the 23-cm
sides to the nearest degree.
2. An aircraft tracking station determines the distance from common point O to each
aircraft and the angle between the aircrafts. If ∠O between the two aircrafts is 49°
and the distances from point O to the two aircrafts are 50 km and 72 km, find the
distance between the two aircrafts. (Round off the answer to the nearest tenth.)
3. A triangular parcel of land has sides 50 ft., 40 ft., and 35 ft. What are the measures
of the angles between the sides? Express answers to the nearest degree.
Questions:
1. What methods/strategies did you use to solve problem no. 1? Show us how you did it.
2. In problem no. 2, how did you find the distance between the two airplanes? Explain how you
solved the distance.
3. In problem no. 3, is it possible to solve for the three angles of the triangle given the lengths
of the three sides? How?
What to Transfer
Your goal at this stage is to formulate a problem using a real-life situation.

503
➤ Activity 5: What Is Your Problem?
1. Create a problem situation based on what you see in the illustration below.
2. Present a solution to the problem you made.
Questions:
1. What is the problem you made all about?
2. Were you able to incorporate the Law of Cosines in the problem? How?
3. Present and explain your solution to the class.
Lesson 5 deals with solving word problems involving oblique triangles. The lesson is divided
into two—Lesson 5.1 and Lesson 5.2. Lesson 5.1 is about the Law of Sines and Its Applications
while Lesson 5.2 focuses on the Law of Cosines and Its Applications. These laws are essential
in the solution of oblique triangles. Different activities regarding these concepts are given
to ensure that students can apply these concepts in solving problem situations in real life.
Through the given activities, students will have opportunities to develop their critical thinking
skills which play a vital role in solving problems we encounter in the real world.

504
Before we end this Module, let us summarize the concepts you have learned by doing a
final activity.
➤ Activity 6: Laws to Validate
Solve each problem inside the box. Then, look for the answer on the decoder and write the cor-
responding letter in the box directly above the answer to complete the key ideas.

505
decoder
Triangle trigonometry can be used to solve problems involving angles of
and
8 60° 8 96.41° opp
hyp
33. 59° 15.69 ft 39.8° 72.97°
adj
opp
8 opp
adj
21.35 ft. 8 157.9 ft. 157.9 ft. 15.69 ft. 39.8° 72.97°
However, many interesting problems involve non-right triangles. The laws needed to solve
these types of problems are the
60° opp
hyp
27.18 157.9 ft. 39.8° 55.3°
157.9 ft. 15.69 72.97° 8 157.9 ft. opp
hyp
72.97° adj
opp
25.07 39.8° 157.9 ft. 15.69 72.97° 8 157.9 ft.

506
Glossary of Terms
acute triangle – a triangle whose angles are all less than 90°
adjacent side – the side next to the reference angle in a right triangle
angle – the figure formed by two rays, called the sides of the angle, sharing a common endpoint
called the vertex.
angle of depression – the angle from the horizontal line to the line of sight of the observer to
the object below.
angle of elevation – is the angle from the horizontal to the line of sight of the observer to the
object above it
clinometer – a device used to measure angles of elevation or depression
cosecant (csc) – In a right-angled triangle, the cosecant of an angle is the length of the hypotenuse
divided by the length of the opposite side.
cosine (cos) – In a right-angled triangle, the cosine of an angle is the length of the adjacent side
divided by the length of the hypotenuse.
cotangent (cot) – In a right-angled triangle, the cotangent of an angle is the length of the adjacent
side divided by the length of the opposite side.
hypotenuse – the longest side of a right triangle. The hypotenuse is always opposite the 90°
angle in a right triangle.
Law of Cosines – the square of any side of a triangle is equal to the sum of the squares of the
other two sides minus the product of these sides and the cosine of their included angle
Law of Sines – the sine of an angle of a triangle divided by its opposite side is equal to the sine
of any other angle divided by its opposite side
line of sight – an imaginary line that connects the eye of an observer to the object being observed
oblique triangle – a triangle which does not contain any right angle
obtuse triangle – a triangle in which one of the angles is more than 90°
opposite side – the side across the reference angle in a right triangle
ratio – a relationship between two numbers of the same kind
right triangle – a triangle in which one angle is a right angle (that is, a 90-degree angle). The
relation between the sides and angles of a right triangle is the basis for trigonometry.
secant (sec) – In a right-angled triangle, the secant of an angle is the length of the hypotenuse
divided by the length of the adjacent side.
sine (sin) – In a right-angled triangle, the sine of an angle is the length of the opposite side
divided by the length of the hypotenuse.
special angles – angles with a reference angle of 30°, 45°, or 60°
tangent (tan) – In a right-angled triangle, the tangent of an angle is the length of the opposite
side divided by the length of the adjacent side.

507
triangle – one of the basic shapes in geometry: a polygon with three vertices and three sides,
which are line segments
trigonometric ratio – a ratio that describes a relationship between a side and an angle of a triangle
Trigonometry – a branch of mathematics that studies triangles and the relationship between
the lengths of their sides and the angles between those sides
References
Deauna, M & Lamayo, F. (1998). Basic Trigonometry for Secondary Schools. SIBS Publishing
House. 927 Quezon Avenue, Quezon City
Bass, L., et al. (2008). Prentice Hall Mathematics Texas Geometry, Pearson Education, Inc.
Publishing as Pearson Prentice Hall, Boson, Massachusetts 02116
Lynch, C. & Olmstead, E. (1998). South-Western MathMatters an Integrated Approach, Book 3.
South-Western Educational Publishing. Cincinnati, Ohio
Boyd, et al. (1998). Glencoe Algebra 1 Integration Applications Connections. The McGraw-Hill
Companies, Inc. Glencoe/McGraw-Hill, 936 Eastwind Drive Westerville, OH 43081-3329
Lapinid, M., and Buzon, O. (2007). Advanced Algebra, Trigonometry and Statistics: Patterns and
Practicalities, by Don Bosco Press, Inc., Makati City, Philippines.
Deauna, M. and Lamayo, F. (1990). Integrated Mathematics IV, Phoenix Publishing House, Inc.,
Quezon City, Philippines.
Hayden, J. and Hall, B. (1995). Prentice Hall Trigonometry, Anvil Publishing, Inc., Quezon City,
Philippines.
Basic Education Curriculum (2002). – Mathematics IV
Ramos Jr., A., and Gureng, P. (2001). Foundations and Applications – Advanced Algebra and
Trigonometry IV, Rex Book Store, Inc., Manila, Philippines.
BEAM Module 13: Laws on Triangle Trigonometry Made Simple (Student Activity 5: Laws to
Validate), pp. 34 – 37
*DepEd Materials: BEAM Module 1 Triangle Trigonometry

508
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http://ceemrr.com/Geometry2/GeneralDefinitions/GeneralDefinitions7.html
http://www.toolingu.com/definition-800210-12780-trigonometric-ratio.html
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www.olx.com.ph
www.nww2m.com
http://www.classroomclipart.com
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Values.pdf

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