The Selina Solutions for the questions given in Chapter 22, Trigonometrical Ratios, of the Class 9 Selina textbooks are available here. In this chapter students learn about the topic of Trigonometrical Ratios of a triangle, dealing with the relations of sides and angles of right-angled triangles. Students can easily score full marks in the exams by solving all the questions present in the Selina textbook.

The Class 9 Selina Solutions Maths is very easy to understand. These solutions cover all the exercise questions included in the book and are according to the syllabus prescribed by the ICSE or CISCE. Here, the PDF of the Class 9 Maths Chapter 22 Selina Solutions is available, which can be downloaded as well as viewed online. Students can also avail the Selina Solutions and download it for free to practice offline as well.

**Download PDF of Selina Solutions for Class 9 Maths Chapter 22:-**Download Here

**Exercise 22(A)**

**1. From the following figure, find the values of:**

**(i) sin A**

**(ii) cos A**

**(iii) cot A**

**(iv) sec C**

**(v) cosec C**

**(vi) tan C**

**Solution:**

Given, âˆ ABC = 90^{o}

AC^{2} = AB^{2} + BC^{2} (AC is hypotenuse)

AC^{2} = 3^{2} + 4^{2}

= 9 + 16

= 25

Taking square root on both sides, we get

AC = 5 cm

(i) sin A = perpendicular/hypotenuse

= BC/AC

= 4/5

(ii) cos A = base/hypotenuse

= AB/AC

= 3/5

(iii) cot A = base/perpendicular

= AB/BC

= 3/4

(iv) sec C = hypotenuse/base

= AC/BC

= 5/4

(v) cosec C = hypotenuse/perpendicular

= AC/AB

= 5/3

(vi) tan C = perpendicular/base

= AB/BC

= 3/4

**2. Form the following figure, find the values of:**

**(i) cos B**

**(ii) tan C**

**(iii) sin ^{2}B + cos^{2}B**

**(iv) sin B. cos C + cos B. sin C**

**Solution:**

Given, âˆ BAC = 90^{o}

BC^{2} = AB^{2} + AC^{2} (As BC is the hypotenuse)

17^{2} = 8^{2} + AC^{2}

AC^{2} = 289 â€“ 64

= 225

Taking square root on both sides, we get

AC = 15 cm

(i) cos B = base/hypotenuse

= AB/BC

= 8/17

(ii) tan C = perpendicular/base

= AB/AC

= 8/15

(iii) sin B = perpendicular/hypotenuse

= AC/BC

= 15/17

cos B = base/hypotenuse

= AB/BC

= 8/17

Now,

sin^{2} B + cos^{2} B = (15/17)^{2} + (8/17)^{2}

= (225 + 64)/289

= 289/289

= 1

(iv) sin B = perpendicular/hypotenuse

= AC/BC

= 15/17

cos B = base/hypotenuse

= AB/BC

= 8/17

sin C = perpendicular/hypotenuse

= AB/BC

= 8/17

cos C = base/hypotenuse

= AC/BC

= 15/17

Now,

sin B. cos C + cos B. sin C = 15/17 x 15/17 + 8/17 x 8/17

= (225 + 64)/289

= 289/289

= 1

**3.** **From the following figure, find the values of:**

**(i) cos AÂ **

**(ii) cosec A**

**(iii) tan ^{2}A – sec^{2}AÂ **

**(iv) sin C**

**(v) sec CÂ **

**(vi) cot ^{2}Â C â€“ 1/sin^{2} C**

**Solution:**

Considering the given diagram, we have

âˆ ADB = 90^{o}Â andÂ âˆ BDC = 90^{o}

So, by Pythagoras theorem

AB^{2} = AD^{2} + BD^{2} (As AB is the hypotenuse in âˆ†ABD)

= 3^{2} + 4^{2}

= 9 + 16

= 25

Taking square root on both sides, we get

AB = 5

Also,

BC^{2} = BD^{2} + DC^{2} (As BC is the hypotenuse in âˆ†BDC)

DC^{2} = BC^{2} – BD^{2}

= 12^{2} – 4^{2}

= 144 â€“ 16

= 128

Taking square root on both sides, we get

DC = 8âˆš2

Now,

(i) cos A = base/hypotenuse

= AD/AB

= 3/5

(ii) cosec A = hypotenuse/perpendicular

= AB/BD

= 5/4

(iii) tan A = perpendicular/base

= BD/AD

= 4/3

sec A = hypotenuse/base

= AB/AD

= 5/3

tan^{2} A â€“ sec^{2} A = (4/3)^{2} â€“ (5/3)^{2}

= 16/9 â€“ 25/9

= -9/9

= -1

(iv) sin C = perpendicular/hypotenuse

= BD/BC

= 4/12

= 1/3

(v) sec C = hypotenuse/base

= BC/DC

= 12/8âˆš2

= 3/2âˆš2

= 3âˆš2/4

(vi) cot C = base/perpendicular

= DC/BD

= 8âˆš2/4

= 2âˆš2

sin C = perpendicular/hypotenuse

= BD/BC

= 4/12

= 1/3

Now,

cot^{2} C â€“ 1/sin^{2} C = (2âˆš2)^{2} â€“ 1/(1/3)^{2}

= 8 â€“ 1/(1/9)

= 8 â€“ 9

= -1

**4. From the following figure, find the values of:**

**(i) sin BÂ **

**(ii) tan C**

**(iii) sec ^{2}Â B – tan^{2}BÂ **

**(iv) sin ^{2}C + cos^{2}C**

**Solution:**

From the figure, we have

âˆ ADB = 90^{o} and âˆ ADC = 90^{o}

So, by Pythagoras theorem

AB^{2} = AD^{2} + BD^{2} (As AB is the hypotenuse in âˆ†ABD)

13^{2} = AD^{2} + 5^{2}

AD^{2} = 13^{2} – 5^{2}

= 169 â€“ 25

= 144

Taking square root on both sides, we get

AD = 12

Also,

AC^{2} = AD^{2} + DC^{2} (As AC is the hypotenuse in âˆ†ADC)

AC^{2} = 12^{2} + 16^{2}

= 144 + 256

= 400

Taking square root on both sides, we get

AC = 20

Now,

(i) sin B = perpendicular/hypotenuse

= AD/AB

= 12/13

(ii) tan C = perpendicular/base

= 12/16

= Â¾

(iii) sec B = hypotenuse/base

= AB/BD

= 13/5

tan B = perpendicular/base

= AD/BD

= 12/5

Hence,

sec^{2} B â€“ tan^{2} B = (13/5)^{2} â€“ (12/5)^{2}

= (169 – 144)/25

= 25/25

= 1

(iv) sin C = perpendicular/hypotenuse

= AD/AC

= 12/20

= 3/5

cos C = base/hypotenuse

= DC/AC

= 16/20

= 4/5

Hence,

sin^{2} C + cos^{2} C = (3/5)^{2} + (4/5)^{2}

= (9 + 16)/25

= 25/25

= 1

**5.** **Given: sin A =Â 3/5, find:**

**(i) tan A**

**(ii) cos A**

**Solution:**

Letâ€™s consider the diagram below:

Given, sin A = 3/5

â‡’ perpendicular/hypotenuse = 3/5

BC/AC = 3/5

Hence,

If the length of BC is 3x, the length of AC is 5x

We have,

AB^{2} + BC^{2} = AC^{2} [By Pythagoras Theorem]

AB^{2} + (3x)^{2} = (5x)^{2}

AB^{2} = 25x^{2} â€“ 9x^{2}

= 16x^{2}

Taking square root on both sides, we get

AB = 4x, which is the base

Now,

(i) tan A = perpendicular/base

= 3x/4x

= 3/4

(ii) cos A = base/hypotenuse

= 4x/5x

= 4/5

**6. From the following figure, find the values of:**

**(i) sin A**

**(ii) sec A**

**(iii) cos ^{2}Â A + sin^{2}A**

**Solution:**

From the given figure, we have

âˆ ABC = 90^{o} and AC is the hypotenuse âˆ†ABC

So, by Pythagoras Theorem

AC^{2} = AB^{2} + BC^{2}

= a^{2} + a^{2 }

= 2a^{2}

Taking square root on both sides, we get

AC = âˆš2a

Now,

(i) sin A = perpendicular/hypotenuse

= BC/AB

= a/âˆš2a

= 1/âˆš2

(ii) sec A = hypotenuse/base

= AC/AB

= âˆš2a/a

= âˆš2

(iii) sin A = perpendicular/hypotenuse

= BC/AC

= a/âˆš2a

= 1/âˆš2

cos A = base/hypotenuse

= AB/AC

= a/âˆš2a

= 1/âˆš2

Hence,

cos^{2} A + sin^{2} A = (1/âˆš2)^{2} + (1/âˆš2)^{2}

= Â½ + Â½

= 1

Â

**7. Given: cos A = 5/13Â **

**Evaluate: (i) (sin A â€“ cot A)/2tan AÂ (ii) cot A + 1/cos A**

**Solution:**

Letâ€™s consider the following diagram:

Given, cos A = 5/13

â‡’ base/hypotenuse = 5/13

AB/AC = 5/13

Hence,

If length of AB = 5x, the length of AC = 13x

So, by Pythagoras Theorem

AB^{2} + BC^{2} = AC^{2}

(5x)^{2} + BC^{2} = (13x)^{2}

BC^{2} = 169x^{2} â€“ 25x^{2}

= 144x^{2}

Taking square root on both sides, we get

BC = 12x, which is the perpendicular

Now,

tan A = perpendicular/base

= 12x/5x

= 12/5

sin A = perpendicular/base

= 12x/13x

= 12/13

cot A = base/perpendicular

= 5x/12x

= 5/12

(i) (sin A â€“ cot A)/2tan A = [(12/13) â€“ (5/12)]/ 2(12/5)

= 79/156 x 5/24

= 395/3744

(ii) cot A + 1/cos A = 5/12 + 1/(5/13)

= 5/12 + 13/5

= 181/60

**8. Given: sec A =Â 29/21, evaluate: sin A â€“Â 1/tan A**

**Solution:**

Letâ€™s consider the diagram below:

Given, sec A = 29/21

â‡’ hypotenuse/base = 29/21

AC/AB = 29/21

Hence,

If length of AB = 21x, the length of AC = 29x

So, by Pythagoras Theorem

AB^{2} + BC^{2} = AC^{2}

(21x)^{2} + BC^{2} = (29x)^{2}

BC^{2} = 841x^{2} â€“ 441x^{2}

= 400x^{2}

Taking square root on both sides, we get

BC = 20x, which is the perpendicular

Now,

sin A = perpendicular/hypotenuse

= 20x/29x

= 20/29

tan A = perpendicular/base

= 20x/21x

= 20/21

Therefore,

sin A â€“ 1/tan A = 20/29 â€“ 1/(20/21)

= 20/29 â€“ 21/20

= – 209/580

**9. Given: tan A =Â 4/3, find: cosec A/(cot A â€“ sec A)**

**Solution:**

Letâ€™s consider the diagram below:

Given, tan A = 4/3

â‡’ perpendicular/base = 4/3

BC/AB = 4/3

Hence,

If length of AB = 3x, the length of BC = 4x

So, by Pythagoras Theorem

AB^{2} + BC^{2} = AC^{2}

(3x)^{2} + (4x)^{2} = AC^{2}

AC^{2} = 9x^{2} + 16x^{2}

= 25x^{2}

Taking square root on both sides, we get

AC = 5x, which is the hypotenuse

Now,

sec A = hypotenuse/base

= AC/AB

= 5x/3x

= 5/3

cot A = base/perpendicular

= AB/BC

= 3x/4x

= Â¾

cosec A = hypotenuse/perpendicular

= AC/BC

= 5x/4x

= 5/x

Therefore,

cosec A/(cot A â€“ sec A) = (5/4)/(3/4 â€“ 5/3)

= (5/4)/(-11/12)

= – 60/44

= – 15/11

**10. Given: 4 cot A = 3, find;**

**(i) sin A**

**(ii) sec A**

**(iii) cosec ^{2}Â A – cot^{2}A.**

**Solution:**

Letâ€™s consider the diagram below:

Given, 4 cot A = 3

cot A = 3/4

â‡’ base/perpendicular = 4/3

AB/BC = 3/4

Hence,

If length of AB = 3x, the length of BC = 4x

So, by Pythagoras Theorem

AB^{2} + BC^{2} = AC^{2}

(3x)^{2} + (4x)^{2} = AC^{2}

AC^{2} = 9x^{2} + 16x^{2}

= 25x^{2}

Taking square root on both sides, we get

AC = 5x, which is the hypotenuse

Now,

(i) sin A = perpendicular/hypotenuse

= 4x/5x

= 4/5

(ii) sec A = hypotenuse/base

= AC/AB

= 5x/3x

= 5/3

(iii) cosec A = hypotenuse/perpendicular

= AC/BC

= 5x/4x

= 5/4

cot A = 3/4

Hence,

cosec^{2} A â€“ cot^{2} A = (5/4)^{2} â€“ (3/4)^{2}

= (25 – 9)/16

= 16/16

= 1

**Exercise 22(B)**

**1. From the following figure, find:**

**(i) yÂ **

**(ii) sin x ^{o}**

**(iii) (sec x ^{o}Â – tan x^{o}) (sec x^{o}Â + tan x^{o})**

**Solution:**

In the given figure,

(i) As itâ€™s a right-angled triangle, so using Pythagorean Theorem

2^{2} = y^{2 }+ 1^{2}

y^{2} = 2^{2} – 1^{2}

= 4 â€“ 1

= 3

Taking square root on both sides, we get

y = âˆš3

(ii) sin x^{o} = perpendicular/hypotenuse

= âˆš3/2

(iii) tan x^{o} = perpendicular/base

= âˆš3

sec x^{o} = hypotenuse/base

= 2

Therefore,

(sec x^{o} â€“ tan x^{o}) (sec x^{o} + tan x^{o}) = (2 – âˆš3)(2 + âˆš3)

= 4 – âˆš3

= 1

**2. Use the given figure to find:**

**(i) sin x ^{o}Â **

**(ii) cos y ^{o}**

**(iii) 3 tan x ^{o}Â – 2 sin y^{o}Â + 4 cos y^{o}**

**Solution:**

Letâ€™s consider the given figure,

As the triangle is a right-angled triangle, so using Pythagorean Theorem

AD^{2} = 8^{2} + 6^{2}

= 64 + 36

= 100

Taking square root on both sides, we get

AD = 10

Also, by Pythagorean Theorem

BC^{2} = AC^{2} â€“ AB^{2}

= 17^{2} – 8^{2}

= 289 â€“ 64

= 225

Taking square root on both sides, we get

BC = 15

Now,

(i) sin x^{o} = perpendicular/hypotenuse

= 8/17

(ii) cos y^{o} = base/hypotenuse

= 6/10

= 3/5

(iii) sin y^{o} = perpendicular/base

= AB/AD

= 8/10

= 4/5

And,

cos y^{o} = 6/10

= 3/5

So,

tan x^{o} = perpendicular/base

= AB/BC

= 8/15

Therefore,

3 tan x^{o} â€“ 2 sin y^{o} + 4 cos y^{o}

= 3(8/15) â€“ 2(4/5) + 4(3/5)

= 8/5 â€“ 8/5 + 12/5

= 12/5

**3. In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC. Find:**

**(i) cosÂ âˆ DBCÂ (ii) cotÂ âˆ DBA**

**Solution:**

Letâ€™s consider the given figure,

As the triangle is a right-angled triangle, so using Pythagorean Theorem

AC^{2} = 5^{2} + 12^{2}

= 25 + 144

= 169

Taking square root on both sides, we get

AC = 13

InÂ âˆ†CBDÂ and âˆ†CBA,

âˆ CÂ is common to both the triangles

âˆ CDB = âˆ CBA = 90^{o}

Hence, âˆ CBD = âˆ CAB

Thus, âˆ†CBDÂ and âˆ†CBAÂ are similar triangles according to AAA criterion

So, we have

AC/BC = AB/BD

13/5 = 12/BD

BD = 60/13

Now,

(i) cos âˆ DBC = base/hypotenuse

= BD/BC

= (60/13)/5

= 12/13

(ii) cot âˆ DBA = base/perpendicular

= BD/AB

= (60/13)/12

= 5/13

**4. In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm. Find:**

**(i) tanÂ âˆ DBC**

**(ii) sinÂ âˆ DBA**

**Solution:**

Considering the given figure, we have

A right-angled triangle ABC, so by using Pythagorean Theorem we have

AC^{2} = BC^{2} + AB^{2}

= 4^{2} + 3^{2}

= 16 + 9

= 25

Taking square root on both sides, we get

AC = 5

In âˆ†CBDÂ and âˆ†CAB, we have

âˆ BCD = âˆ ACB (Common)

âˆ CDB = âˆ CBA = 90^{o}

Hence, âˆ†CBDÂ ~ âˆ†CABÂ by AA similarity criterion

So,

AC/BC = AB/BD

5/3 = 4/BD

BD = 12/5

Now, using Pythagorean Theorem in âˆ†BDC

DC^{2} = BC^{2} â€“ BD^{2}

= 3^{2} â€“ (12/5)^{2}

= 9 â€“ 144/25

= (225 â€“ 144)/25

= 81/25

Taking square root on both sides, we get

DC = 9/5

Therefore,

AD = AC â€“ DC

= 5 â€“ 9/5

= 16/5

Now,

(i)Â tan âˆ DBC = perpendicular/ base

= DC/BD

= (9/5)/(12/5)

= 3/4

Â

(ii) sin âˆ DBA = AD/AB

= (16/5)/4

= 4/5

**5. In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cosÂ âˆ ABC.**

**Solution:**

Letâ€™s consider the figure below:

In the isoscelesÂ âˆ†ABC, we have

AB = AC = 15 cm

BC = 18 cm

Now, the perpendicular drawn from angle A to its opposite BC divides its into two equal parts

i.e., BD = DC = 9cm

Hence,

cos **âˆ **ABC = base/hypotenuse

= BD/AB

= 9/15

= 3/5

**6. In the figure given below, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. Find:**

**(i) sin BÂ **

**(ii) tan C**

**(iii) sin ^{2}Â B + cos^{2}BÂ **

**(iv) tan C – cot B**

**Solution:**

Letâ€™s consider the figure below:

In the isoscelesÂ âˆ†ABC, we have

AB = AC = 5 cm

BC = 8 cm

Now, the perpendicular drawn from angle A to its opposite BC divides its into two equal parts

i.e., BD = DC = 4cm

As, âˆ ADB = 90^{o }in âˆ†ABD, we have

AB^{2} = AD^{2} + BD^{2}

AD^{2 }= AB^{2} – BD^{2}

= 5^{2} – 4^{2}

= 25 â€“ 16

= 9

Taking square root on both sides, we get

AD = 3

Now,

(i) sin B = AD/AB

= 3/5

(ii) tan C = AD/DC

= 3/4

(iii) sin B = AD/AB

= 3/5

cos B = BD/AB

= 4/5

Hence,

sin^{2} B + cos^{2} B = (3/5)^{2} + (4/5)^{2}

= 9/25 + 16/25

= 25/25

= 1

(iv) tan C = AD/DC

= Â¾

cot B = BD/AD

= 4/3

Hence,

tan C â€“ cot B = Â¾ – 4/3

= (9 – 16)/12

= -7/12

**7. In triangle ABC;Â âˆ ABC = 90 ^{o},Â âˆ CAB = x^{o}, tan x^{o}Â =Â Â¾ and BC = 15 cm. Find the measures of AB and AC.**

**Solution:**

Letâ€™s consider the figure below:

Given, tan x^{o}Â = Â¾

â‡’ perpendicular/base = Â¾

BC/AB = Â¾

Hence,

If length of base AB = 4x, the length of perpendicular BC = 3x

So, by Pythagoras Theorem

BC^{2} + AB^{2} = AC^{2}

(3x)^{2} + (4x)^{2} = AC^{2}

AC^{2} = 9x^{2} + 16x^{2}

= 25x^{2}

Taking square root on both sides, we get

AC = 5x, which is the hypotenuse

Now, we have

BC = 15

â‡’ 3x = 15

x = 5

Therefore, AB = 4x = 4(5) = 20 cm

And, AC = 5x = 5 Ã— 5 = 25 cm

**8. Using the measurements given in the following figure:**

**(i) Find the value of sinÂ Ã˜ and tan Î¸**

**(ii) Write an expression for AD in terms ofÂ Î¸**

**Solution:**

Letâ€™s consider the figure below:

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

Now, in right angled âˆ†BCD using Pythagorean Theorem, we have

BD^{2} = BC^{2} + CD^{2} [As AB is the hypotenuse]

CD^{2} = BD^{2} – BC^{2}

CD^{2} = 13^{2} – 12^{2}

= 169 â€“ 144

= 25

Taking square root on both sides, we get

CD = 5

As BCDE is a rectangle,

ED = 12 cm, EB = 5 cm and AE = (14 â€“ 5) cm = 9Â cm

Now,

(i) sin Ã˜ = CD/BD

= 5/13

tan Î¸ = ED/AE

= 12/9

= 4/3

(ii) sec Î¸ = AD/AE

= AD/9

AD = 9 sec Î¸

Or

cosec Î¸ = AD/ED

= AD/12

AD = 12cosec Î¸

**9. In the given figure:**

**BC = 15 cm and sin B = 4/5**

** (i) Calculate the measure of AB and AC.**

**(ii) Now, if tanÂ âˆ ADC = 1; calculate the measures of CD and AD.**

**Also, show that: tan ^{2}B â€“ 1/cos^{2} BÂ = -1**

**Solution: **

Given, BC = 15 cm and sin B = 4/5

â‡’ Perpendicular/hypotenuse = AC/AB

= 4/5

Hence, if the length of perpendicular is 4x, the length of hypotenuse will be 5x

In right triangle ABC, we have

BC^{2} + AC^{2} = AB^{2} [By Pythagoras Theorem]

BC^{2} = AB^{2} – AC^{2}

^{ } = (5x)^{2} â€“ (4x)^{2}

= 25x^{2} â€“ 16x^{2}

= 9x^{2}

Taking square root on both sides, we get

BC = 3x

Now, as BC = 15 (given)

3x = 15

x = 15/3

x = 5

(i) AC = 4x

= 4(5)

= 20 cm

And,

AB = 5x

= 5(5)

= 25 cm

(ii) Given,

tan âˆ ADC = 1

perpendicular/base = AC/CD

= 1/1

Hence,

If length of perpendicular is x, then the length of hypotenuse will be x

And, we have

AC^{2} + CD^{2} = AD^{2} [Using Pythagoras Theorem]

x^{2} + x^{2} = AD^{2}

AD^{2} = 2x^{2}

Taking square root on both sides, we get

AD = âˆš2x

Now,

AC = 20 â‡’x = 20

So,

AD = âˆš2x = âˆš2(20) = 20âˆš2 cm

And,

CD = 20 cm

Hence,

tan B = AC/BC

= 20/15

= 4/3

cos B = BC/AB

= 15/25

= 3/5

Thus,

tan^{2} B â€“ 1/cos^{2} B = (4/3)^{2} â€“ 1/(3/5)^{2}

= 16/9 â€“ 1/(9/25)

= 16/9 â€“ 25/9

= – 9/9

= -1

**10. If sin A + cosec A = 2;**

**Find the value of sin ^{2}Â A + cosec^{2}Â A.**

**Solution:**

Given, sin A + cosec A = 2

On squaring on both sides, we have

(sin A + cosec A)^{2} = 2^{2}

sin^{2} A + cosec^{2} A + 2sinA. cosec A = 4

sin^{2} A + cosec^{2} A + 2 = 4 [As sin A. cosec A = sin A x 1/sin A = 1]

sin^{2} A + cosec^{2} A = 4 â€“ 2 = 2

Hence, the value of (sin^{2} A + cosec^{2} A) is 2

## Selina Solutions for Class 9 Maths Chapter 22- Trigonometrical Ratios

Chapter 22, Trigonometrical Ratios, consists of 2 exercises and the solutions given here contain answers to all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

22.1 Introduction

22.2 Concept of perpendicular, base and hypotenuse in a right triangle

22.3 Notation of angles

22.4 Trigonometrical Ratios

22.5 Reciprocal Relations

## Selina Solutions for Class 9 Maths Chapter 22- Trigonometrical Ratios

The ratio between the lengths of a pair of two sides of a right-angled triangle is called a Trigonometrical Ratio. The three sides of a right-angled triangle give six trigonometrical ratios; namely sine, cosine, tangent, cotangent, secant, and cosecant. Here, in this chapter, students are taught about the different Trigonometrical Ratios. Read and learn Chapter 22 of Selina textbook to get to know more about Trigonometrical Ratios. Learn the Selina Solutions for Class 9 effectively to attain excellent results in the examination.