Class 10 Trigonometric Ratios

Topic to be covered

`☼` Introduction.
`☼` Trigonometric Ratios.


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?

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`

`(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

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

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

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

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

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`