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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Ex. 22(A)

Ex. 22(B)

Test Yourself

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 $A B equals 3 x$ , length of $B C equals 4 x$

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 $B C equals square root of 2 x$

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

(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

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)

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

And

Now

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 $equals square root of 5 x$ , length of perpendicular = x

Since

Now

(i)

(ii)

Solution 4

Consider the diagram below:

Therefore if length of AB = x, length of $A C equals square root of 2 x$

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

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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(A)

Solution 1(a)

Solution 1(b)

Solution 1(c)

Solution 1(d)

Solution 1(e)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

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

Solution 1(a)

Solution 1(b)

Solution 1(c)

Solution 1(d)

Solution 1(e)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

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

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

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