Class 10 Trigonometric Ratios

### Topic to be covered

☼ Introduction.
☼ Trigonometric Ratios.

### Introduction

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?

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.

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.

### Trigonometric Ratios

As you've 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 side AB is a part of ∠ A. So, we call it the side adjacent to angle A.

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 color{orange}{∠ A = text( side opposite to angle A)/text(hypotenuse) = (BC)/(AC)}

cosine of ∠ A = text( side adjacent to angle A)/text(hypotenuse) = (AB)/(AC)

tangent of ∠ A = text(side opposite to angle A)/text(side adjacent to angle A) = (BC)/(AB)

cosecant of ∠ A = 1/text(sine of ∠ A) = text(hypotenuse)/text(side opposite to angle A) = (AC)/(AB)

secant of ∠ A = 1/text(cosine of ∠ A) = text(hypotenuse)/text(side adjacent to angle A) = (AC)/(AB)

cotangent of ∠ A = 1/text(tangent of ∠ A) = text(side adjacent to angle A)/text(side opposite to angle A)

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)/(AB) = ((BC)/(AC)/(AB)/(AC)) = (sinA)/(cosA)  and cotA = (cosA)/(sinA)

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)

Remark : Note that the symbol sin A is used as an abbreviation for ‘the sine of the angle A’. sin A is not the product of ‘sin’ and A. ‘sin’ separated from A
has no meaning. Similarly, cos A is 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 on AC 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 Delta PAM differ from those of ∠ A in Delta CAB or from those of ∠ A in D QAN?

To answer this, first look at these triangles. Is Delta PAM similar to Delta 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)/(AC) = (MP)/(BC)

From this, we find (MP)/(AP) = (BC)/(AC) = sinA
Similarly,

(AM)/(AP) = (AB)/(AC) = cos A , (MP)/(AM) = (BC)/(AB) = tanA and so on.

This shows that the trigonometric ratios of angle A in Delta PAM not differ from
those of angle A in Delta CAB.

In the same way, you should check that the value of sin A (and also of other trigonometric ratios) remains the same in Delta 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 sin^2 A, cos^2 A , etc., in place of (sin A)^2, (cos A)^2, etc., respectively. But cosecA = (sin A)^(–1) ≠ sin^(–1) A (it is called sine inverse A).

sin^(–1) A has 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)/(AC) = 1/3 i.e.

the lengths of the sides BC and AC 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.

Now, 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.

AB^2 = AC^2 – BC^2 = (3k)^2 – (k)^2 = 8k^2 = (2 sqrt2 k)^2

Therefore, AB = pm 2 sqrt2 k

So, we get AB = 2sqrt2 k (Why is AB not -2 sqrt2k)

Now cosA = (AB)/(AC) = (2 sqrt2 k)/(3k) = (2 sqrt2)/3

Similarly, you can obtain the other trigonometric ratios of the angle A.

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).
Q 3129080811

Given tanA = 4/3 find the other trigonometric ratios of the angle A.
Class 10 Chapter 8 Example 1
Solution:

Let us first draw a right Delta ABC (see Fig 8.8).

Now, we know that tan A = (BC)/(AB) = 4/3

Therefore, if BC = 4k, then AB = 3k, where k is a positive number.

Now, by using the Pythagoras Theorem, we have

AC^2 = AB^2 + BC^2 = (4k)^2 + (3k)^2 = 25k^2
So, AC = 5k
Now, we can write all the trigonometric ratios using their definitions.

sinA = (BC)/(AC) = (4k)/(5k) = 4/5

cosA = (AB)/(AC) = (3k)/(5k) = 3/5

Therefore, cot A = 1/(tanA) = 3/4 , cosecA = 1/(sinA) = 5/4 and secA = 1/(cosA) = 5/3
Q 3179891716

If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.
Class 10 Chapter 8 Example 2
Solution:

Let us consider two right triangles ABC and PQR where sin B = sin Q (see Fig. ).

We have sinB = (AC)/(AB)

and sinQ = (PR)/(PQ)

Then (AC)/(AB) = (PR)/(PQ)

Therefore, (AC)/(PR) = (AB)/(PQ) = k ....(1)

Now, using Pythagoras theorem,

BC = sqrt(AB^2-AC^2)

QR = sqrt(PQ^2-PR^2)

so (BC)/(QR) = sqrt(AB^2-AC^2)/sqrt(PQ^2-PR^2) = sqrt(k^2PQ^2-k^2PR^2)/sqrt(PQ^2-PR^2) = (k sqrt(PQ^2-PR^2))/sqrt(PQ^2-PR^2) = k .........(2)

From (1) and (2), we have

(AC)/(PR) = (AB)/(PQ) = (BC)/(QR)

Then, by using Theorem 6.4, Delta ACB ~ Delta PRQ and therefore, ∠ B = ∠ Q.
Q 3119091810

Consider Delta ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ ABC = θ (see Fig). Determine the values of
(i) cos^2 θ + sin^2 θ,
(ii) cos^2 θ – sin^2 θ.
Class 10 Chapter 8 Example 3
Solution:

In Delta ACB, we have

AC = sqrt(AB^2-BC^2) = sqrt((29)^2-(21)^2)

= sqrt((29-21)(29+21)) = sqrt((8)(50)) = sqrt(400) = 20 units

So sintheta = (AC)/(AB) = 20/29 , costheta = (BC)/(AB) = 21/29

Now (i) cos^2 theta + sin^2 theta = (20/29)^2 + ( 21/29)^2 = (20^2+21^2)/(29)^2 = (400+441)/(841) = 1

and (ii) cos^2theta - sin^2 theta = (21/29)^2 - (20/29)^2 = ((21+20)(21-20))/(29)^2 = 41/841
Q 3149091813

In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.
Class 10 Chapter 8 Example 4
Solution:

In Delta ABC, tan A = (BC)/(AB) = 1 (see Fig )

i.e., BC = AB
Let AB = BC = k, where k is a positive number.

Now AC = sqrt(AB^2+BC^2)

= sqrt(k^2+k^2) = k sqrt2

Therefore, sinA = (BC)/(AC) = 1/sqrt2 and cosA = (AB)/(AC) = 1/sqrt2

so 2 sinA cosA = 2 (1/sqrt2) (1/sqrt2) = 1 which is the required value.
Q 3159091814

In Delta OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Fig).
Determine the values of sin Q and cos Q.
Class 10 Chapter 8 Example 5
Solution:

In Delta OPQ, we have
OQ^2 = OP^2 + PQ^2
i.e., (1 + PQ)^2 = OP^2 + PQ^2 (Why?)
i.e., 1 + PQ^2 + 2PQ = OP^2 + PQ^2
i.e., 1 + 2PQ = 72 (Why?)
i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm

so sinQ = 7/25 and cosQ = 24/25