law_of_sines.ppt

Use the Law of Sines to solve oblique triangles
(AAS or ASA).
Use the Law of Sines to solve oblique triangles
(SSA).
Find the areas of oblique triangles.
Use the Law of Sines to model and solve real-
life problems.
What You Should Learn

Plan for the day
 When to use law of sines and law of cosines
 Applying the law of sines
 Law of sines - Ambiguous Case

Introduction
In this section, we will solve
oblique triangles – triangles
that have no right angles.
As standard notation, the
angles of a triangle are labeled A, B, and C, and
their opposite sides are labeled a, b, and c.
To solve an oblique triangle, we need to know the
measure of at least one side and any two other
measures of the triangle—either two sides, two
angles, or one angle and one side.

Introduction
This breaks down into the following four cases:
1. Two angles and any side (AAS or ASA)
2. Two sides and an angle opposite one of them (SSA)
3. Three sides (SSS)
4. Two sides and their included angle (SAS)
The first two cases can be solved using the
Law of Sines, whereas the last two cases require the
Law of Cosines.

Introduction
The Law of Sines can also be written in the
reciprocal form: .

Given Two Angles and One Side – AAS
For the triangle below C = 102, B = 29, and
b = 28 feet. Find the remaining angle and
sides.

Example AAS - Solution
The third angle of the triangle is
A = 180 – B – C
= 180 – 29 – 102
= 49.
By the Law of Sines, you have
.

Example AAS – Solution
Using b = 28 produces
and
cont’d

Law of Sines
For non right triangles
Law of sines
Try this:
A
B
C
c
a
b
C
c
B
b
A
a
sin
sin
sin


mm
c
B
A o
o
45
,
67
,
43 





Let’s look at this: Example 1
Given a triangle, demonstrate using the Law of
Sines that it is a valid triangle (numbers are
rounded to the nearest tenth so they may be
up to a tenth off):
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
Is it valid??

And this: Example 2
Given a triangle, demonstrate using the Law of
Sines that it is a valid triangle (numbers are
rounded to the nearest tenth so they may be
up to a tenth off):
a = 5 A = 40o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
Is it valid??
Why does this work?

Looking at these two examples
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
a = 5 A = 40o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
In both cases a, b and A are the same (two sides and
an angle) but they produced two different triangles
Why??

Here is what happened
a = 5 A = 40o
b = 7 B =
c = C =
What is sin 64.1?
What is sin 115.9?
What is the relationship between these two angles?
Remember the sine of an angle in the first quadrant
(acute: 0o – 90o) and second quadrant,
(obtuse: 90o – 180o)are the same!




)
8999
(.
sin
5
40
sin
7
sin
sin
7
40
sin
5
1
o
o
B
B

The Ambiguous Case (SSA)
In our first example we saw that two angles and one
side determine a unique triangle.
However, if two sides and one opposite angle
are given, three possible situations can
occur:
(1) no such triangle exists,
(2) one such triangle exists, or
(3) two distinct triangles may satisfy the conditions.

Back to these examples
Given two sides and an angle across
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
a = 5 A = 40o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
The Ambiguous Case

Ambiguous Case
1. Is it Law of Sines or Law of Cosines
1. Law of Cosines – solve based upon one solution
2. Law of Sines – go to #2
2. Law of Sines - Is it the SSA case? (Two sides and angle opposite)
1. No – not ambiguous, solve based upon one solution
2. Yes – go to #3.
3. Is the side opposite the angle the shortest side?
2. Yes – go to #4
4. Is the angle obtuse?
1. No – go to #5
2. Yes – no solution
5. Calculate the height of the triangle
height = the side not opposite the angle x the sine of the angle
1. If the side opposite the angle is shorter than the height – no solution
2. If the side opposite the angle is equal to the height – one solution
3. If the side opposite the angle is longer than the height – two
solutions

How many solutions are there?
1. A = 30o a = 5 b = 3
2. B = 50o a = 6 b = 5
3. C = 80o b = 5 c = 6
4. A = 40o a = 4 b = 8
5. a = 5 b = 4 c = 6
6. B = 20o a = 10 c = 15
7. A = 25o a = 3 b = 6
8. C = 75o a = 5 b = 3
1
2
0
2
1
1
1
1

Example – Single-Solution Case—SSA
 For the triangle below, a = 22 inches, b = 12
inches, and A = 42. Find the remaining side
and angles.
One solution: a  b

Example – Solution SSA
By the Law of Sines, you have
Reciprocal form
Multiply each side by b.
Substitute for A, a, and b.
B is acute.
)
( 22
42
sin
12
sin 
B
)
(sin
sin
a
A
b
B 
a
A
b
B sin
sin

o
B 41
.
21


Example – Solution SSA
Now, you can determine that
C  180 – 42 – 21.41
= 116.59.
Then, the remaining side is
cont’d
)
59
.
116
sin(
)
42
sin(
22

c
C
A
a
c sin
sin

A
a
C
c
sin
sin

inches
40
.
29


What about more than one
solution?

How many solutions are there?
1. A = 30o a = 5 b = 3
2. B = 50o a = 6 b = 5
3. C = 80o b = 5 c = 6
4. A = 40o a = 4 b = 8
5. a = 5 b = 4 c = 6
6. B = 20o a = 10 c = 15
7. A = 25o a = 3 b = 6
8. C = 75o a = 5 b = 3
1
2
1
0
1
1
2
1
Solve #2, 7

Area of an Oblique Triangle
The procedure used to prove the Law of Sines leads to
a simple formula for the area of an oblique triangle.
Referring to the triangles below, that each triangle
has a height of h = b sin A.
A is acute. A is obtuse.

Area of a Triangle - SAS
SAS – you know two sides: b, c and
the angle between: A
Remember area of a triangle is
½ base ● height
Base = b
Height = c ● sin A
 Area = ½ bc(sinA)
A
B
C
c a
b
h
Looking at this from all three sides:
Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

Example – Finding the Area of a Triangular Lot
Find the area of a triangular lot having two sides of
lengths 90 meters and 52 meters and an included
angle of 102.
Solution:
Consider a = 90 meters, b = 52 meters, and the
included angle C = 102
Then, the area of the triangle is
Area = ½ ab sin C
= ½ (90)(52)(sin102)
 2289 square meters.

Homework 35
Section 6.1 p. 416:
3, 5, 19-24 all, 25-37 odd

Plan for the day
 Law of Cosines
 Finding the area of a triangle – Heron’s
Formula

Use the Law of Cosines to solve oblique
triangles (SSS or SAS).
Use the Law of Cosines to model and solve
real-life problems.
Use Heron’s Area Formula to find the area of a
triangle.
What You Should Learn

Introduction
Four cases.
1. Two angles and any side (AAS or ASA)
2. Two sides and an angle opposite one of them (SSA)
3. Three sides (SSS)
4. Two sides and their included angle (SAS)
The first two cases can be solved using the
Law of Sines, whereas the last two cases require the
Law of Cosines.

Law of Sines
For non right triangles
Law of sines
A
B
C
c
a
b
C
c
B
b
A
a
sin
sin
sin



Ambiguous Case
1. Is it Law or Sines or Law of Cosines
1. Law of Cosines – solve based upon one solution
2. Law of Sines – go to #2
2. Law of Sines - Is it the SSA case? (Two sides and angle opposite)
2. Yes – go to #3.
3. Is the side opposite the angle the shortest side?
2. Yes – go to #4
4. Is the angle obtuse?
1. No – go to #5
2. Yes – no solution
5. Calculate the height of the triangle
height = the side not opposite the angle x the sine of the angle
1. If the side opposite the angle is shorter than the height – no solution
2. If the side opposite the angle is equal to the height – one solution
3. If the side opposite the angle is longer than the height – two
solutions

Area of a Triangle - SAS
SAS – you know two sides: b, c and
the angle between: A
Remember area of a triangle is
½ base ● height
Base = b
Height = c ● sin A
 Area = ½ bc(sinA)
A
B
C
c a
b
h
Looking at this from all three sides:
Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc(sin A)

Law of Cosines: Introduction
Two cases remain in the list of conditions
needed to solve an oblique triangle – SSS
and SAS.
If you are given three sides (SSS), or two sides
and their included angle (SAS), none of the
ratios in the Law of Sines would be complete.
In such cases, you can use the Law of
Cosines.

Law of Cosines
Side, Angle, Side
A
B
C
c a
b
C
abCos
b
a
c
B
acCos
c
a
b
A
bcCos
c
b
a
2
2
2
2
2
2
2
2
2
2
2
2










Try these
1. B = 20o a = 10 c = 15
2. A = 25o b = 3 c = 6
3. C = 75o a = 5 b = 3

Law of Cosines
Side, Side, Side
A
B
C
c a
b

Law of Cosines
 SSS
ab
c
b
a
C
ac
b
c
a
B
bc
a
c
b
A
2
cos
2
cos
2
cos
2
2
2
2
2
2
2
2
2









Always solve for the angle across from the longest side first!

Why
It is wise to find the largest angle when you
have SSS. Knowing the cosine of an angle,
you can determine whether the angle is acute
or obtuse. That is,
cos  > 0 for 0 <  < 90
cos  < 0 for 90 <  < 180.
This avoids the ambiguous case!
Acute
Obtuse

Try these
1. a = 5 b = 4 c = 6
2. a = 20 b = 10 c = 28
3. a = 8 b = 5 c = 12

An Application of the Law of Cosines
The pitcher’s mound on a
women’s softball field is 43 feet
from home plate and the
distance between the bases is
60 feet (The pitcher’s mound is
not halfway between home
plate and second base.) How
far is the pitcher’s mound from
first base?

Solution
In triangle HPF, H = 45 (line HP bisects the
right angle at H), f = 43, and p = 60.
Using the Law of Cosines for this SAS case,
you have
h2 = f2 + p2 – 2fp cos H
= 432 + 602 – 2(43)(60) cos 45
 1800.3.
So, the approximate distance from the pitcher’s
mound to first base is
 42.43 feet.

Heron’s Area Formula
The Law of Cosines can be used to establish
the following formula for the area of a
triangle. This formula is called Heron’s Area
Formula after the Greek mathematician
Heron (c. 100 B.C.).

Area of a Triangle
Law of Cosines Case - SSS
)
)(
)(
( c
s
b
s
a
s
s
A 



A
B
C
c a
b
h
SSS – Given all three sides
Heron’s formula:
2
c
b
a
s
where




Try these
 Given the triangle with three sides of 6, 8, 10
find the area
 Given the triangle with three sides of 12, 15,
21 find the area

law_of_sines.ppt

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