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

The ICSE syllabus emphasises both academic excellence and the holistic development of the students. The board’s curriculum is designed to build a strong foundation and provide students with critical thinking skills and practical knowledge. But to prepare for various academic and professional pursuits, students need to have more than just the textbook. **Selina Solutions **provides comprehensive coverage and solutions for various subjects.

One of the most important chapters in **ICSE Class 9 Math** is Trigonometry. Trigonometry is a branch of mathematics that teaches about the relationships between the angles and sides of right-angled triangles using trigonometric ratios, which are mathematical functions such as Sine (sin), Cosine (cos), and Tangent (tan). The reciprocals of these ratios are cosecant (cosec), secant (sec) and cotangent (cot), and they provide alternative perspectives about right-angled triangles.

Selina Class 9 Mathematics Solutions extensively covers the concepts of trigonometric ratios and their reciprocals. The resource offers clear definitions and explanations in solutions to questions posed in the Selina textbook to help students grasp trigonometry concepts and apply them in real-life situations. With step-by-step solutions and illustrative representations of triangles, it helps in geometric interpretations of trigonometric ratios.

Students can use these Solutions to solve textbook exercises, practice problems and examples to reinforce their understanding and proficiency in the chapter. Doing so helps them develop a strong foundation in trigonometry, boosts their confidence and enhances their problem-solving skills. These skills have practical applications in fields such as engineering, architecture, physics, and navigation.

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

### Solution 1(a)

Correct option: (i)

^{ }

∴ If length of BC = 5x, length of AC = 13x

Now,

### Solution 1(b)

Correct option: (ii)
1^{}

^{ }

^{ }

∴ If length of BC = 3x, length of AB = 5x

Now,

So,

### Solution 1(c)

Correct option: (iv)
-1^{}

^{ }

∴ If length of AB = 5x, length of BC = 12x

Now,

So,

### Solution 1(d)

Correct option: (iii)
^{ }

In right-angled ΔADB,

Now,

### Solution 1(e)

Correct option: (i)
1^{}

^{ }

∴ If length of AB = 3x, length of AC = 5x

Now,

So,

### Solution 2

Given angle

(i)

(ii)

(iii)

(iv)

(v)

(vi)

### Solution 3

Given angle

(i)

(ii)

(iii)

(iv)

### Solution 4

Consider the diagram as

Given angle and

(i)

(ii)

(iii)

(iv)

(v)

(vi)

### Solution 5

Given angle and

(i)

(ii)

(iii)

(iv)

### Solution 6

Consider the diagram below:

Therefore if length of , length of

Since

Now

(i)

(ii)

### Solution 7

Given angle in the figure

Now

(i)

(ii)

(iii)

### Solution 8

Consider the diagram below:

Therefore if length of , length of

Since

Now

(i)

(ii)

### 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 , length of

Since

Now

Therefore

### Solution 11

Consider the diagram below:

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

Since

(i)

(ii)

(iii)

### Solution 12

Consider the diagram below:

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

Since

Now all other trigonometric ratios are

### Solution 13

Consider the diagram below:

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

Since

(i)

(ii)

(iii)

### Solution 14

Consider the diagram below:

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

Since

Now

Therefore

### Solution 15

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

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

### Solution 1(a)

Correct option: (i)

∴ If length of AB = x, length of AC = x

Now,

∴ If length of AC = x, length of BC = 2x

Now,

### Solution 1(b)

Correct option: (iii)
^{ }

In right-angled ΔADC,

In right-angled ΔACB,

Now,

### Solution 1(c)

Correct option: (i) ^{ }

Let the diagonals BD and AC of a rhombus intersect each other at O.

Given, BD = 12 cm, AC = 16 cm.

Diagonals of a rhombus bisect each other at right angles.

⇒ OB = OD = 6 cm, OA = OC = 8 cm and ∠AOB = 90^{o}

In right-angled ΔAOB,

Now,

### Solution 1(d)

Correct option: (iv)
6^{}

In right-angled ΔADB,

### Solution 1(e)

Correct option: (ii)
^{ }

∴ If length of AB = x, length of BC = x

Now,

So,

### Solution 2

Consider the given figure

(i)

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

(ii)

(iii)

Therefore

### Solution 3

Consider the given figure

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

Also

(i)

(ii)

(iii)

Therefore

### 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

(i)

(ii)

### Solution 5

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 6

Consider the figure below

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

### Solution 7

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 8

Consider the figure

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

Since

Now

Therefore

And

### Solution 9

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 10

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 11

Squaring both sides

### Solution 12

Squaring both sides

### Solution 13

Consider the diagram below:

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

Since

(i)

(ii)

(iii)

Therefore

(iv)

### Solution 14

Consider the diagram below:

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

Since

Now

Therefore

### Solution 15

Now

## Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] Exercise Test Yourself

### Solution 1

Consider the diagram below:

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

Since

Now

Therefore

### Solution 2

Consider the diagram below:

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

Since

Now

Therefore

### Solution 3

Consider the diagram below:

Therefore if length of hypotenuse , length of perpendicular = x

Since

Now

(i)

(ii)

### Solution 4

Consider the diagram below:

Therefore if length of AB = x, length of

Since

Now

Therefore

### Solution 5

Consider the diagram below:

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

Since

Now

Therefore

### Solution 6

Given angle and in the figure

Again

Now

(i)

(ii)

Therefore

(iii)

Therefore

### Solution 7

Since is mid-point of so

(i)

(ii)

### Solution 8

Consider the diagram below:

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

Since

(i)

(ii)

Therefore

### Solution 9

Consider the figure

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

Since

Now

Therefore

And

### Solution 10

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 11

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 12

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 13

Now

### Solution 14

Consider the figure

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

Since

Now

Therefore

### Solution 15

Consider the diagram

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

Since

Now

Therefore

Now

### Solution 16

Consider the figure

Therefore if length of , length of

Since

Now

(i)

(ii)

### Solution 17

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 18

Consider the figure

Therefore if length of , length of

Since

Now

Therefore

### Solution 19

Now

### Solution 20

Squaring both sides

### Solution 21

Now

(i)

(ii)

### Solution 22

Consider the given diagram as

Using Pythagorean Theorem

Now

Again using Pythagorean Theorem

Now

Therefore