Class 10 Trigonometric Ratios of Some Specific Angles

### topic to be covered

☼ Trigonometric Ratios of Some Specific Angles .

### Trigonometric Ratios of Some Specific Angles

From geometry, you are already familiar with the construction of angles of 30°, 45°, 60° and 90°.

In this section, we will find the values of the trigonometric ratios for these angles and, of course, for 0°.

Trigonometric Ratios of 45°
In Delta ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠ C = 45° (see Fig. 8.14).

So, BC = AB
Now, Suppose BC = AB = a.
Then by Pythagoras Theorem, AC^2 = AB^2 + BC^2 = a^2 + a^2 = 2a^2,

and, therefore AC = a sqrt2

sin45^0 = text(side opposite to angle 45°)/text(hypotenuse) = (BC)/(AC) = a/(a sqrt2) = 1/sqrt2

cos45^0 = text(side adjacent to angle 45°)/text(hypotenuse) = (AB)/(AC) = a/(a sqrt2) = 1/sqrt2

tan45^0 = text(side opposite to angle 45°)/text(side adjacent to angle 45°) = (BC)/(AB) = a/a = 1

also cosec45^0 = 1/(sin45^0) = sqrt2 , sec45^0 = 1/(cos45^0) = sqrt2 , cot45^0 = 1/(tan45^0) = 1.

Trigonometric Ratios of 30° and 60° :

Let us now calculate the trigonometric ratios of 30° and 60°. Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore,

Draw the perpendicular AD from A to the side BC (see Fig. 8.15).

Now Delta ABD ≅ Delta ACD
Therefore, BD = DC
and ∠ BAD = ∠ CAD (CPCT)
Now observe that:

Delta ABD is a right triangle, right- angled at D with ∠ BAD = 30° and ∠ ABD = 60°
(see Fig. 8.15).

As you know, for finding the trigonometric ratios, we need to know the lengths of the
sides of the triangle. So, let us suppose that AB = 2a.

Then, BD = 1/2 BC = a

and AD^2 = AB^2 – BD^2 = (2a)^2 – (a)^2 = 3a^2,

Therefore, AD = a sqrt3

Now, we have :

sin30^0 = (BD)/(AB) = a/(2a) = 1/2 , cos30^0 = (AD)/(AB) = ( a sqrt3)/(2a) = sqrt3/2

tan30^0 = (BD)/(AD) = a/(a sqrt3) = 1/sqrt3

Also cosec30^0 = 1/(sin30^0) = 2 , sec30^0 = 1/(cos30^0) = 2/sqrt3

cot 30^0 = 1/(sin30^0) = sqrt3

Similarly,  sin60^0 = (AD)/(AB) = (a sqrt3)/(2a) = sqrt3/2 , cos60^0 = 1/2 , tan 60^0 = sqrt3

cosec60^0 = 2/sqrt3 , sec60^0 = 2 and cot60^0 = 1/sqrt3

bbul"Trigonometric Ratios of 0° and 90°"

Let us see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC
(see Fig. 8.16), till it becomes zero.

As ∠ A gets smaller and smaller, the length of the side BC decreases.The point C gets closer to point B, and finally when ∠ A becomes very close to 0°, AC becomes almost the same as AB (see Fig. 8.17).

When ∠ A is very close to 0°, BC gets very close to 0 and so the value of
sinA = (BC)/(AC)  is very close to 0. Also, when ∠ A is very close to 0°, AC is nearly the
same as AB and so the value of cos A = (AB)/(AC) is very close to 1.

This helps us to see how we can define the values of sin A and cos A when A = 0°. We define : sin 0° = 0 and cos 0° = 1.
Using these, we have : tan0^0 = (sin 0^0)/(cos 0^0) = 0 , cot 0^0 = 1/(tan 0^0) which is not defined.

sec^0 = 1/( cos 0^0) = 1 and cosec 0^0 = 1/( sin 0^0) which is again not defined.

Now, let us see what happens to the trigonometric ratios of ∠ A, when it is made larger and larger in Delta ABC till it becomes 90°. As ∠ A gets larger and larger, ∠ C gets smaller and smaller.

Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when ∠ A is very close to 90°, ∠ C becomes very close to 0° and the side AC almost coincides with side BC (see Fig. 8.18).

When ∠ C is very close to 0°, ∠ A is very close to 90°, side AC is nearly the same as side BC, and so sin A is very close to 1. Also when ∠ A is very close to 90°, ∠ C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0.
So, we define : sin 90° = 1 and cos 90° = 0.

We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60°
and 90° in Table 8.1, for ready reference

bbul"Remark" : From the table above you can observe that as ∠ A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0. Let us illustrate the use of the values in the table above through some examples.

Q 3250101014

In Delta ABC, right-angled at B, AB = 5 cm and ∠ ACB = 30° (see Fig). Determine the lengths of the sides BC and AC.
Class 10 Chapter 8 Example 6
Solution:

To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore

(AB)/(BC) = tanC

5/(BC) = tan30^0 =1/sqrt3

which gives BC = 5 sqrt3 cm

To find the length of the side AC, we consider

sin30^0 = (AB)/(AC)  (Why?)

1/2 = 5/(AC)

AC = 10 cm

Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,

AC = sqrt(AB^2+BC^2) = sqrt(5^2+(5 sqrt3)^2) cm = 10 cm
Q 3260101015

In Delta PQR, right -angled at Q (see Fig), PQ = 3 cm and PR = 6 cm.
Determine ∠ QPR and ∠ PRQ.
Class 10 Chapter 8 Example 7
Solution:

Given PQ = 3 cm and PR = 6 cm

Therefore, (PQ)/(PR) = sinR
or sinR = 3/6 = 1/2

so ∠ PRQ = 30°

and therefore, ∠ QPR = 60°.  (Why?)

You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.
Q 3270101016

If sin (A – B) = 1/2 , cos (A+B) = 1/2 , 0^0 < A + B ≤ 90°, A > B, find A and B
Class 10 Chapter 8 Example 8
Solution:

sin (A – B) = 1/2 , cos (A+B) = 1/2 , therefore A - B = 30^0  ......(1)

Also, since cos (A + B) = 1/2 therefore, A + B = 60° .........(2)

Solving (1) and (2), we get : A = 45° and B = 15°.