# Class 9 SELINA Solutions Maths Chapter 22 - Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]

## Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] Exercise Ex. 22(B)

### Solution 1

Consider the given figure

(i)

Since the triangle is a right angled triangle, so using Pythagorean Theorem

(ii)

(iii)

Therefore

### Solution 2

Consider the given figure

Since the triangle is a right angled triangle, so using Pythagorean Theorem

Also

(i)

(ii)

(iii)

Therefore

### Solution 3

Consider the given figure

Since the triangle is a right angled triangle, so using Pythagorean Theorem

In and , the is common to both the triangles, so therefore .

Therefore and are similar triangles according to AAA Rule

So

(i)

(ii)

### Solution 4

Consider the given figure

Since the triangle is a right angled triangle, so using Pythagorean Theorem

In and , the is common to both the triangles, so therefore.

Therefore and are similar triangles according to AAA Rule

So

Now using Pythagorean Theorem

Therefore

(i)

(ii)

### Solution 5

Consider the figure below

In the isosceles , and the perpendicular drawn from angle to the side divides the side into two equal parts

### Solution 6

Consider the figure below

In the isosceles , and the perpendicular drawn from angle to the side divides the side into two equal parts

Since

(i)

(ii)

(iii)

Therefore

(iv)

Therefore

### Solution 7

Consider the figure

Therefore if length of base = 4x, length of perpendicular = 3x

Since

Now

Therefore

And

### Solution 8

Consider the figure

A perpendicular is drawn from D to the side AB at point E which makes BCDE is a rectangle.

Now in right angled triangle BCD using Pythagorean Theorem

Since BCDE is rectangle so ED 12 cm, EB = 5 and AE = 14 - 5 = 9

(i)

(ii)

Or

### Solution 9

Given

Therefore if length of perpendicular = 4x, length of hypotenuse = 5x

Since

Now

(i)

And

(ii)

Given

Therefore if length of perpendicular = x, length of hypotenuse = x

Since

Now

So

And

Now

So

### Solution 10

Squaring both sides

### Solution 11

Squaring both sides

### Solution 12

Consider the diagram below:

Therefore if length of BC = 3x, length of AB = 4x

Since

(i)

(ii)

(iii)

Therefore

(iv)

### Solution 13

Consider the diagram below:

Therefore if length of AB = 15x, length of AC = 17x

Since

Now

Therefore

### Solution 14

Now

### Solution 15

Since is mid-point of so

(i)

(ii)

### Solution 16

Consider the diagram below:

Therefore if length of AB = 4x, length of BC = 3x

Since

(i)

(ii)

Therefore

### Solution 17

Consider the figure

Therefore if length of base = 4x, length of perpendicular = 3x

Since

Now

Therefore

And

### Solution 18

Consider the figure

The diagonals of a rhombus bisects each other perpendicularly

Therefore if length of base = 3x, length of hypotenuse = 5x

Since

Now

Therefore

And

Since the sides of a rhombus are equal so the length of the side of the rhombus

The diagonals are

### Solution 19

Consider the figure below

In the isosceles , the perpendicular drawn from angle to the side divides the side into two equal parts

Since

(i)

(ii)

(iii)

Therefore

### Solution 20

Consider the figure below

Therefore if length of perpendicular = 4x, length of hypotenuse = 5x

Since

Now

Therefore

And

Again

Therefore if length of perpendicular = x, length of base = x

Since

Now

Therefore

And

### Solution 21

Now

### Solution 22

Consider the figure

Therefore if length of perpendicular = x, length of base = x

Since

Now

Therefore

### Solution 23

Consider the diagram

Therefore if length of base = 3x, length of perpendicular = 2x

Since

Now

Therefore

Now

### Solution 24

Consider the figure

Therefore if length of , length of

Since

Now

(i)

(ii)

### Solution 25

Consider the diagram below:

Therefore if length of , length of

Since

Consider the diagram below:

Therefore if length of , length of

Since

Now

Therefore

### Solution 26

Consider the given diagram as

Using Pythagorean Theorem

Now

Again using Pythagorean Theorem

Now

Therefore

### Solution 27

Consider the figure

Therefore if length of , length of

Since

Now

Therefore

### Solution 28

Now

### Solution 29

Squaring both sides

### Solution 30

Now

(i)

(ii)

## Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] Exercise Ex. 22(A)

### Solution 1

Given angle

(i)

(ii)

(iii)

(iv)

(v)

(vi)

### Solution 2

Given angle

(i)

(ii)

(iii)

(iv)

### Solution 3

Consider the diagram as

Given angle and

(i)

(ii)

(iii)

(iv)

(v)

(vi)

### Solution 4

Given angle and

(i)

(ii)

(iii)

(iv)

### Solution 5

Consider the diagram below:

Therefore if length of , length of

Since

Now

(i)

(ii)

### Solution 6

Given angle in the figure

Now

(i)

(ii)

(iii)

### Solution 7

Consider the diagram below:

Therefore if length of , length of

Since

Now

(i)

(ii)

### Solution 8

Consider the diagram below:

Therefore if length of , length of

Since

Now

Therefore

### Solution 9

Consider the diagram below:

Therefore if length of , length of

Since

Now

Therefore

### Solution 10

Consider the diagram below:

Therefore if length of AB = 3x, length of BC = 4x

Since

(i)

(ii)

(iii)

### Solution 11

Consider the diagram below:

Therefore if length of AB = 3x, length of AC = 5x

Since

Now all other trigonometric ratios are

### Solution 12

Consider the diagram below:

Therefore if length of AB = 12x, length of BC = 5x

Since

(i)

(ii)

(iii)

### Solution 13

Consider the diagram below:

Therefore if length of perpendicular = px, length of hypotenuse = qx

Since

Now

Therefore

### Solution 14

Consider the diagram below:

Therefore if length of AB = x, length of AC = 2x

Since

Consider the diagram below:

Therefore if length of AC = x, length of

Since

Now

Therefore

### Solution 15

Consider the diagram below:

Therefore if length of base = 12x, length of perpendicular = 5x

Since

Now

Therefore

### Solution 16

Consider the diagram below:

Therefore if length of base = 3x, length of perpendicular = 4x

Since

Now

Therefore

### Solution 17

Consider the diagram below:

Therefore if length of hypotenuse , length of perpendicular = x

Since

Now

(i)

(ii)

### Solution 18

Consider the diagram below:

Therefore if length of AB = x, length of

Since

Now

Therefore

### Solution 19

Consider the diagram below:

Therefore if length of base = x, length of perpendicular = x

Since

Now

Therefore

### Solution 20

Given angle and in the figure

Again

Now

(i)

(ii)

Therefore

(iii)

Therefore