Mathematics-Inroduction to Trignometry Class 10 | Smart eTeach

INTRODUCTION TO TRIGONOMETRY 173
8
There is perhaps nothing which so occupies the
middle position of mathematics as trigonometry.
– J.F. Herbart (1890)
8.1 Introduction
You have already studied about triangles, and in particular, right triangles, in your
earlier classes. Let us take some examples from our surroundings where right triangles
can be imagined to be formed. For instance :
1. Suppose the students of a school are
visiting Qutub Minar. Now, if a student
is looking at the top of the Minar, a right
triangle can be imagined to be made,
as shown in Fig 8.1. Can the student
find out the height of the Minar, without
actually measuring it?
2. Suppose a girl is sitting on the balcony
of her house located on the bank of a
river. She is looking down at a flower
pot placed on a stair of a temple situated
nearby on the other bank of the river.
A right triangle is imagined to be made
in this situation as shown in Fig.8.2. If
you know the height at which the
person is sitting, can you find the width
of the river?.
INTRODUCTION TO
TRIGONOMETRY
Fig. 8.1
Fig. 8.2

174 MATHEMATICS
3. Suppose a hot air balloon is flying in
the air. A girl happens to spot the
balloon in the sky and runs to her
mother to tell her about it. Her mother
rushes out of the house to look at the
balloon.Now when the girl had spotted
the balloon intially it was at point A.
When both the mother and daughter
came out to see it, it had already
travelled to another point B. Can you
find the altitude of B from the ground?
In all the situations given above, the distances or heights can be found by using
some mathematical techniques, which come under a branch of mathematics called
‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’
(meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact,
trigonometry is the study of relationships between the sides and angles of a triangle.
The earliest known work on trigonometry was recorded in Egypt and Babylon. Early
astronomers used it to find out the distances of the stars and planets from the Earth.
Even today, most of the technologically advanced methods used in Engineering and
Physical Sciences are based on trigonometrical concepts.
In this chapter, we will study some ratios of the sides of a right triangle with
respect to its acute angles, called trigonometric ratios of the angle. We will restrict
our discussion to acute angles only. However, these ratios can be extended to other
angles also. We will also define the trigonometric ratios for angles of measure 0° and
90°. We will calculate trigonometric ratios for some specific angles and establish
some identities involving these ratios, called trigonometric identities.
8.2 Trigonometric Ratios
In Section 8.1, you have seen some right triangles
imagined to be formed in different situations.
Let us take a right triangle ABC as shown
in Fig. 8.4.
Here, ∠ CAB (or, in brief, angle A) is an
acute angle. Note the position of the side BC
with respect to angle A. It faces ∠ A. We call it
the side opposite to angle A. AC is the
hypotenuse of the right triangle and the sideAB
is a part of ∠ A. So, we call it the side
adjacent to angle A.
Fig. 8.4
Fig. 8.3

Note that the position of sides change
when you consider angle C in place of A
(see Fig. 8.5).
You have studied the concept of ‘ratio’ in
your earlier classes. We now define certain ratios
involving the sides of a right triangle, and call
them trigonometric ratios.
The trigonometric ratios of the angle A
in right triangle ABC (see Fig. 8.4) are defined
as follows :
sine of ∠ A =
side opposite to angle A BC
hypotenuse AC
=
cosine of ∠ A =
side adjacent to angle A AB
hypotenuse AC
=
tangent of ∠ A =
side opposite to angle A BC
side adjacent to angle A AB
=
cosecant of ∠ A =
1 hypotenuse AC
sine of A side opposite to angle A BC
= =
∠
secant of ∠ A =
1 hypotenuse AC
cosine of A side adjacent to angle A AB
= =
∠
cotangent of ∠ A =
1 side adjacent to angle A AB
tangent of A side opposite to angle A BC
= =
∠
The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A
and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively,
the reciprocals of the ratios sin A, cos A and tan A.
Also, observe that tan A =
BC
BC sin AAC
ABAB cos A
AC
= = and cot A =
cosA
sin A
.
So, the trigonometric ratios of an acute angle in a right triangle express the
relationship between the angle and the length of its sides.
Why don’t you try to define the trigonometric ratios for angle C in the right
triangle? (See Fig. 8.5)
Fig. 8.5

176 MATHEMATICS
The first use of the idea of ‘sine’ in the way we use
it today was in the work Aryabhatiyam byAryabhata,
in A.D. 500. Aryabhata used the word ardha-jya
for the half-chord, which was shortened to jya or
jiva in due course. When the Aryabhatiyam was
translated into Arabic, the word jiva was retained as
it is. The word jiva was translated into sinus, which
means curve, when theArabic version was translated
into Latin. Soon the word sinus, also used as sine,
became common in mathematical texts throughout
Europe.An English Professor of astronomy Edmund
Gunter (1581–1626), first used the abbreviated
notation ‘sin’.
The origin of the terms ‘cosine’and ‘tangent’was much later. The cosine function
arose from the need to compute the sine of the complementary angle. Aryabhatta
called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the
English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.
Remark : Note that the symbol sin A is used as an
abbreviation for ‘the sine of the angle A’. sinA is not
the product of ‘sin’ and A. ‘sin’ separated from A
has no meaning. Similarly, cosAis not the product of
‘cos’ and A. Similar interpretations follow for other
trigonometric ratios also.
Now, if we take a point P on the hypotenuse
AC or a point Q onAC extended, of the right triangle
ABC and draw PM perpendicular to AB and QN
perpendicular to AB extended (see Fig. 8.6), how
will the trigonometric ratios of ∠ A in ∆ PAM differ
from those of ∠ A in ∆ CAB or from those of ∠ A in
∆ QAN?
To answer this, first look at these triangles. Is ∆ PAM similar to ∆ CAB? From
Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the
triangles PAM and CAB are similar. Therefore, by the property of similar triangles,
the corresponding sides of the triangles are proportional.
So, we have
AM
AB
=
AP MP
AC BC
= ⋅
Aryabhata
C.E. 476 – 550
Fig. 8.6

From this, we find
MP
AP
=
BC
sin A
AC
= .
Similarly,
AM AB
AP AC
= = cos A,
MP BC
tan A
AM AB
= = and so on.
This shows that the trigonometric ratios of angle A in ∆ PAM not differ from
those of angle A in ∆ CAB.
In the same way, you should check that the value of sin A (and also of other
trigonometric ratios) remains the same in ∆ QAN also.
From our observations, it is now clear that the values of the trigonometric
ratios of an angle do not vary with the lengths of the sides of the triangle, if
the angle remains the same.
Note : For the sake of convenience, we may write sin2
A, cos2
A, etc., in place of
(sin A)2
, (cos A)2
, etc., respectively. But cosec A = (sin A)–1
≠ sin–1
A (it is called sine
inverse A). sin–1
Ahas a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well. Sometimes, the
Greek letter θ (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle. If we know any one
of the ratios, can we obtain the other ratios? Let us see.
If in a right triangle ABC, sin A =
1
,
3
then this means that
BC 1
AC 3
= , i.e., the
lengths of the sides BC andAC of the triangle
ABC are in the ratio 1 : 3 (see Fig. 8.7). So if
BC is equal to k, then AC will be 3k, where
k is any positive number. To determine other
trigonometric ratios for the angle A, we need to find the length of the third side
AB. Do you remember the Pythagoras theorem? Let us use it to determine the
required length AB.
AB2
= AC2
– BC2
= (3k)2
– (k)2
= 8k2
= (2 2 k)2
Therefore, AB = 2 2 k±
So, we get AB = 2 2 k (Why is AB not – 2 2 k ?)
Now, cos A =
AB 2 2 2 2
AC 3 3
k
k
= =
Similarly, you can obtain the other trigonometric ratios of the angleA.
Fig. 8.7

178 MATHEMATICS
Remark : Since the hypotenuse is the longest side in a right triangle, the value of
sin A or cos A is always less than 1 (or, in particular, equal to 1).
Let us consider some examples.
Example 1 : Given tan A =
4
3
, find the other
trigonometric ratios of the angle A.
Solution : Let us first draw a right ∆ ABC
(see Fig 8.8).
Now, we know that tan A =
BC 4
AB 3
= .
Therefore, if BC = 4k, then AB = 3k, where k is a
positive number.
Now, by using the Pythagoras Theorem, we have
AC2
= AB2
+ BC2
= (4k)2
+ (3k)2
= 25k2
So, AC = 5k
Now, we can write all the trigonometric ratios using their definitions.
sin A =
BC 4 4
AC 5 5
k
k
= =
cos A =
AB 3 3
AC 5 5
k
k
= =
Therefore, cot A =
1 3 1 5, cosec A =
tanA 4 sin A 4
= = and sec A =
1 5
cos A 3
= ⋅
Example 2 : If ∠ B and ∠ Q are
acute angles such that sin B = sin Q,
then prove that ∠ B = ∠ Q.
Solution : Let us consider two right
triangles ABC and PQR where
sin B = sin Q (see Fig. 8.9).
We have sin B =
AC
AB
and sin Q =
PR
PQ
Fig. 8.8
Fig. 8.9

Then
AC
AB
=
PR
PQ
Therefore,
AC
PR
=
AB
, say
PQ
k= (1)
Now, using Pythagoras theorem,
BC = 2 2
AB AC−
and QR = 2 2
PQ – PR
So,
BC
QR
=
2 2 2 2 2 2 2 2
2 2 2 2 2 2
AB AC PQ PR PQ PR
PQ PR PQ PR PQ PR
k k k
k
− − −
= = =
− − −
(2)
From (1) and (2), we have
AC
PR
=
AB BC
PQ QR
=
Then, by using Theorem 6.4, ∆ ACB ~ ∆ PRQ and therefore, ∠ B = ∠ Q.
Example 3 : Consider ∆ ACB, right-angled at C, in
which AB = 29 units, BC = 21 units and ∠ ABC = θ
(see Fig. 8.10). Determine the values of
(i) cos2
θ + sin2
θ,
(ii) cos2
θ – sin2
θ.
Solution : In ∆ ACB, we have
AC = 2 2
AB BC− = 2 2
(29) (21)−
= (29 21)(29 21) (8)(50) 400 20units− + = = =
So, sin θ =
AC 20 BC 21, cos =
AB 29 AB 29
= θ = ⋅
Now, (i) cos2
θ + sin2
θ =
2 2 2 2
2
20 21 20 21 400 441
1,
29 29 84129
+ +   
+ = = =   
   
and (ii) cos2
θ – sin2
θ =
2 2
2
21 20 (21 20)(21 20) 41
29 29 84129
+ −   
− = =   
   
.
Fig. 8.10

180 MATHEMATICS
Example 4 : In a right triangle ABC, right-angled at B,
if tan A = 1, then verify that
2 sin A cos A = 1.
Solution : In ∆ ABC, tan A =
BC
AB
= 1 (see Fig 8.11)
i.e., BC = AB
Let AB = BC = k, where k is a positive number.
Now, AC = 2 2
AB BC+
= 2 2
( ) ( ) 2k k k+ =
Therefore, sin A =
BC 1
AC 2
= and cos A =
AB 1
AC 2
=
So, 2 sin A cos A =
1 1
2 1,
2 2
  
=  
  
which is the required value.
Example 5 : In ∆ OPQ, right-angled at P,
OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
Determine the values of sin Q and cos Q.
Solution : In ∆ OPQ, we have
OQ2
= OP2
+ PQ2
i.e., (1 + PQ)2
= OP2
+ PQ2
(Why?)
i.e., 1 + PQ2
+ 2PQ = OP2
+ PQ2
i.e., 1 + 2PQ = 72
(Why?)
i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm
So, sin Q =
7
25
and cos Q =
24
25
⋅
Fig. 8.12
Fig. 8.11

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