Concise Selina Solutions for Class 9 Maths Chapter 23 - Trigonometrical Ratios of Standard Angles

Selina Solutions are considered to be very useful when you are preparing for the ICSE Class 9 Maths exams. Here, we bring to you the Selina Solutions for Class 9 Maths, providing the detailed explanations to the questions present in the exercises of Chapter 23- Trigonometrical Ratios of Standard Angles. These answers have been devised by the subject matter experts as per the syllabus prescribed by the CISCE for the ICSE.

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Exercise 23A Page: 291

1. Find the value of:

(i) sin 30o cos 30o

(ii) tan 30o tan 60o

(iii) cos2 60o + sin2 30o

(iv) cosec2 60o – tan2 30o

(v) sin2 30o + cos2 30o + cot2 45o

(vi) cos2 60o + sec2 30o + tan2 45o.

Solution:

(i) Given sin 30o cos 30o

By substituting the values, we get

sin 30o cos 30o = ½ (√3/2)

= √3/4

(ii) Given tan 30o tan 60o

By substituting the values, we get

tan 30o tan 60o = 1/√3 (√3)

= 1

(iii) Given cos2 60o + sin2 30o

By substituting the values, we get

cos2 60o + sin2 30o = (½)2 +(½)2

= ¼ + ¼

= ½

(iv) Given cosec2 60o – tan2 30o

By substituting the values, we get

cosec2 60o – tan2 30o = (2/√3)2 – (1/√3)2

= 4/3 – 1/3

= 1

(v) Given sin2 30o + cos2 30o + cot2 45o

By substituting the values, we get

sin2 30o + cos2 30o + cot2 45o = (½)2 + (√3/2)2 + 12

= ¼ + ¾ + 1

= 2

(vi) Given cos2 60o + sec2 30o + tan2 45o

By substituting the values, we get

cos2 60o + sec2 30o + tan2 45o = (½)2 + (2/√3)2 + 12

= ¼ + 4/3 + 1

= 31/12

2. Find the value of:

(i) tan2 30o + tan2 45o + tan2 60o

(ii) Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 1

(iii) 3 sin2 30o + 2 tan2 60o – 5 cos2 45o.

Solution:

(i) Given tan2 30o + tan2 45o + tan2 60o

By substituting the values, we get

tan2 30o + tan2 45o + tan2 60o = (1//√3)2 + 12 + (/√3)2

= 1/3 + 1 + 3

= 13/3

= 4 1/3

(ii) Given

 
Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 2

By substituting the values, we get

= ½ + 2/1 – 5/2

= (1 + 4 – 5)/2

= 0

(iii) Given 3 sin2 30o + 2 tan2 60o – 5 cos2 45o.

By substituting the values, we get

3 sin2 30o + 2 tan2 60o – 5 cos2 45o. = 3 (½)2 + 2 (√3)2 + 5 (1/√3)2

= ¾ + 6 – 5/2

= (3 + 24 – 10)/4

= 4 ¼

3. Prove that:

(i) sin 60o cos 30o + cos 60o. sin 30o = 1

(ii) cos 30o. cos 60o – sin 30o. sin 60o = 0

(iii) cosec2 45o – cot2 45o = 1

(iv) cos2 30o – sin2 30o = cos 60o.

(v) Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 3

(vi) 3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.

Solution:

(i) Given sin 60o cos 30o + cos 60o. sin 30o

LHS = sin 60o cos 30o + cos 60o. sin 30o

Now we have to prove that RHS = 1

= (√3/2) (√3/2) + ½ ½

= ¾ + ¼

= 1

= RHS

(ii) Given cos 30o. cos 60o – sin 30o. sin 60o = 0

LHS = cos 30o. cos 60o – sin 30o. sin 60o

= (√3/2) ½ – ½ (√3/2)

= (√3/4) – (√3/4)

= 0

= RHS

(iii) Given cosec2 45o – cot2 45o = 1

LHS = cosec2 45o – cot2 45o = 1

= (√2)2 – 12

= 2 – 1

= 1

= RHS

(iv) Given cos2 30o – sin2 30o = cos 60o.

LHS = cos2 30o – sin2 30o = cos 60o.

= (√3/2)2 – (½)2

= ¾ – ¼

= ½

= cos 60o

= RHS

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 4

(vi) Given 3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.

LHS = 3 cosec2 60o – 2 cot2 30o + sec2 45o = 0.

= 3 (2/√3)2 – 2 (√3)2 + (√2)2

= 4 – 6 + 2

= 0

= RHS

4. Prove that

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 5

Solution:

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 6

(ii) RHS =

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 7

Now substituting the values, we get

= (1 – 1/3)/ (1 + 1/3)

= ½

Consider LHS

cos (2 × 30o)

= cos 60o

= ½

Therefore, LHS = RHS

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 8

= √3

LHS,

tan (2 × 30o)

= tan 60o

= √3

Therefore, LHS = RHS

5. ABC is an isosceles right-angled triangle. Assuming of AB = BC = x, find the value of each of the following trigonometric ratios:

(i) sin 45o

(ii) cos 45o

(iii) tan 45o

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 9

Solution:

Given that AB = BC = x

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 10

(i) sin 45o = AB/AC

= x/x√2

= 1/√2

(ii) cos 45o = BC/AC

= x/x√2

= 1/√2

(iii) tan 45o = AB/BC

= x/x

= 1

6. Prove that:

(i) sin 60o = 2 sin 30o cos 30o.

(ii) 4 (sin4 30o + cos4 60o) – 3 (cos2 45o – sin2 90o) = 2

Solution:

(i) LHS = sin 60o

= √3/2

RHS = 2 sin 30o cos 30o

= 2 (√3/2) (½)

= √3/2

Therefore LHS = RHS

(ii) LHS = 4 (sin4 30o + cos4 60o) – 3 (cos2 45o – sin2 90o)

Now by substituting the values we get

= 4[(½)4 + (½)4] – 3 [(1/√2)2 + 14]

= 4(1/16 + 1/16) – 3 (½ – 1)

= 8/16 + 3/2

= 2

LHS = RHS

7. (i) If sin x = cos x and x is acute, state the value of x.

(ii) If sec A = cosec A and 0o ≤ A ≤ 90o, state the value of A.

(iii) If tan θ= cot θ and 0o ≤ θ ≤ 90o, state the value of θ.

(iv) If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.

Solution:

(i) The angle, x is acute and hence we have, 0 < x

We know that

Cos2x + sin2 x = 1

Since cos x = sin x

Above equation will become

2 sin2 x = 1

Sin x = 1/√2

Therefore, x = 45o

(ii) sec A = cosec A

Cos A = sin A

Cos2 A = sin2 A

Cos2x + sin2 x = 1

Above equation will become

Cos2 A = 1 – cos2 A

2 cos2 A = 1

Cos A = 1/√2

A = 45o

(iii) tan θ = cot θ

tan θ = 1/tan θ

tan2 θ = 1

tan θ = 1

tan θ = tan 45o

θ = 45o

(iv) sin x = cos y = sin (90o – y)

If x and y are acute angles

x = 90o – y

which implies,

x + y = 90o

hence x and y are complementary angles.

8. (i) If sin x = cos y, then x + y = 45o; write true of false.

(ii) sec θ. Cot θ = cosec θ; write true or false.

(iii) For any angle θ, state the value of:

Sin2 θ + cos2 θ.

Solution:

(i) sin x = cos y = sin (π/2 – y)

If x and y acute angles,

x = (π/2 – y)

x + y = π/2

x + y = 45o is false

(ii) sec θ. Cot θ = 1/ cos θ. cos θ/ sin θ

= cosec θ

sec θ. Cot θ = cosec θ

is true.

(iii) Sin2 θ + cos2 θ = Sin2 θ + 1 – sin2 θ.

= 1

9. State for any acute angle θ whether:

(i) sin θ increases or decreases as θ increases:

(ii) cos θ increases or decreases as θ increases.

(iii) tan θ increases or decreases as θ decreases.

Solution:

(i) For acute angles, remember what sine means: opposite over hypotenuse. If we increase the angle, then the opposite side gets larger. That means “opposite/hypotenuse” gets larger or increases.

(ii)For acute angles, remember what cosine means: base over hypotenuse. If we increase the angle, then the hypotenuse side gets larger. That means “base/hypotenuse” gets smaller or decreases.

(iii)For acute angles, remember what tangent means: opposite over base. If we decrease the angle, then the opposite side gets smaller. That means “opposite /base” gets decreases.

10. If √3 = 1.732, find (correct to two decimal place) the value of each of the following:

(i) sin 60o 

(ii) 2/ tan 30o

Solution:

(i) sin 60o = √3 /2

= 1.732/2

= 0.87

(ii) 2/ tan 30o = 2/ (1/√3)

= 2√3 

= 2 (1.732)

= 3.46

Exercise 23b Page: 293

1. Given A = 60o and B = 30o, prove that:

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

(iii) cos (A – B) = cos A cos B + sin A sin B

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 11

Solution:

(i) Given A = 60o and B = 30o

LHS = sin (A + B)

= sin (60o + 30o)

= sin 90o

= 1

RHS = sin A cos B + cos A sin B

= sin 60o cos 30o + cos 60o sin 30o

= √3/2 (√3/2) + ½ ½

= ¾ + ¼

= 1

LHS = RHS

(ii) LHS = cos (A + B)

= cos (60o + 30o)

= cos 90o

= 0

RHS = cos A cos B – sin A sin B

RHS = cos 60o cos 30o – sin 60o sin 30o

= ½ (√3/2) – (√3/2) ½

= √¾ – √3/4

= 0

LHS = RHS

(iii) LHS = cos (A – B)

= cos (60o – 30o)

= cos 30o

= √3/2

RHS = cos A cos B + sin A sin B

RHS = cos 60o cos 30o + sin 60o sin 30o

= ½ (√3/2) + (√3/2) ½

= √¾ + √3/4

= √3/2

LHS = RHS

(iv) LHS = tan (A – B)

= tan (60o – 30o)

= tan 30o

= 1/√3

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 12

= (√3 – 1/√3)/ 1 + √3 (1/√3)

= 2/ 2 √3

= 1/√3

Therefore, LHS = RHS

2. If A =30o, then prove that:

(i) sin 2A = 2sin A cos A = Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 13

(ii) cos 2A = cos2A – sin2A

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 14

(iii) 2 cos2 A – 1 = 1 – 2 sin2A

(iv) sin 3A = 3 sin A – 4 sin3A.

Solution:

(i) Given A = 30o

Sin 2A = sin 2(30o)

= sin 60o

= √3/2

2 sin A cos A = 2 sin 30o cos 30o

= 2 (½) (√3/2)

= √3/2

Now,

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 15

(ii) cos 2A = cos 2 (300)

= cos 60o

= ½

Cos2 A – sin2 A = cos2 30o – sin2 30o

= ¾ – ¼

= ½

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 16

= 2/4

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 17
= ½

(iii) 2 cos2 A – 1 = 2 cos2 30o – 1

= 2 (¾) – 1

= 3/2 – 1

= ½

1 – 2 sin2 A = 1 – 2 sin2 30o

= 1 – 2 (¼)

= ½

2 cos2 A – 1 = 1 – sin2 A

(iv) sin 3A = sin 3 (30o)

= sin 90o

= 1

3 sin A – 4 sin3 A = 3 sin 30o – 4 sin3 30o

= 3 (½) – 4 (½)3

= 3/2 – ½

= 1

Therefore,

Sin 3A = 3 sin A – 4 sin3 A

3. If A = B = 45o, show that:

(i) sin (A – B) = sin A cos B – cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

Solution:

Given that A = B = 45o

(i) LHS = sin (A – B)

= sin (45o – 45o)

= sin 0o

= 0

RHS = sin A cos B – cos A sin B

= sin 45o cos 45o – cos 45o sin 45o

= 1/√2 (1/√2) – 1/√2 (1/√2)

= 0

Therefore, LHS = RHS

(ii) LHS = cos (A + B)

= cos (45o + 45o)

= cos 90o

= 0

RHS = cos A cos B – sin A sin B

= cos 45o cos 45o – sin 45o sin 45o

= 1/√2 (1/√2) – 1/√2 (1/√2)

= 0

Therefore, LHS = RHS

4. If A = 30o; show that:

(i) sin 3 A = 4 sin A sin (60o – A) sin (60o + A)

(ii) (sin A – cos A)2 = 1 – sin 2A

(iii) cos 2A = cos4 A – sin4 A

(iv) Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 18

(v) Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 19= 2 cos A.

(vi) 4 cos A cos (60o – A). cos (60o + A) = cos 3A

(vii) Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 20

Solution:

Given that A = 30o

(i) According to the question we have

LHS = sin 3A

= sin 3 (30o)

= sin 90o

= 1

RHS = 4 sin A sin (60o – A) sin (60o + A)

= 4 sin A sin (60o – 30o) sin (60o + 30o)

= 4 (½) (½) (1)

= 1

LHS = RHS

(ii) According to the question we have

LHS = (sin A – cos A)2

= (sin 30o – cos 30o)2

= (½ – √3/2)2

= ¼ + ¾ – √3/2

= 1 – √3/2

= (2 – √3)/2

RHS = 1 – sin 2A

= 1 – sin 2 (30o)

= 1 – sin 60o

= 1 – √3/2

= (2 – √3)/2

Therefore, LHS = RHS

(iii) According to the question we have

LHS = cos 2A

= cos 2 (30o)

= cos 60o

= ½

RHS = cos4 A – sin4 A

= cos4 30o – sin4 30o

= (√3/2)4 – (½)4

= 9/16 – 1/16

= ½

LHS = RHS

(iv) According to the question we have

LHS = (1 – cos 2A)/ sin 2A

= 1 – cos 2(30o)/ sin 2 (30o)

= 1 – ½/ (√3/2)

= 1/√3

RHS = tan A

= tan 30o

= 1/√3

LHS = RHS

(v) According to the question we have

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 21

= 2√3/2

= √3

RHS = 2 cos A

= 2 cos 30o

= 2 (√3/2)

= √3

(vi) According to the question we have

LHS = 4 cos A cos (60o – A) cos (60o + A)

= 4 cos A cos (60o – 30o) cos (60o + 30o)

= 4 cos 30o cos 30o cos 90o

= 4 (√3/2) (√3/2) 0

= 0

RHS = cos 3A

= cos 3 (30o)

= cos 90o

= 0

LHS = RHS

(vii) According to the question we have

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 22

= ¾ + 9/4

= 12/4

= 3

= RHS

Hence the proof

Exercise 23b Page: 297

1. Solve the following equations for A, if:

(i) 2 sin A = 1 

(ii) 2 cos 2 A = 1

(iii) sin 3 A = √3/2 

(iv) sec 2 A = 2

(v) √3 tan A = 1 

(vi) tan 3 A = 1

(vii) 2 sin 3 A = 1 

(viii) √3 cot 2 A = 1

Solution:

(i) Given 2 sin A = 1

Sin A = ½

Sin A = sin 30o

Therefore, A = 30o

(ii) Given 2 cos 2A = 1

Cos 2A = ½

Cos 2A = cos 60o

2A = 60o

A = 30o

(iii) Given sin 3A = √3/2

Sin 3A = sin 60o

3A = 60o

A = 20o

(iv) Given sec 2A = 2

Sec 2A = sec 60o

2A = 60o

A = 30o

(v) Given √3 tan A = 1

Tan A = 1/√3

Tan A = tan 30o

A = 30o

(vi) Given tan 3A = 1

Tan 3A = tan 45o

3A = 45o

A = 15o

(vii) Given 2 sin 3A = 1

Sin 3A = ½

Sin 3A = sin 30o

3A = 30o

A = 10o

(viii) Given √3 cot 3A = 1

Cot 2A = 1/√3

Cot 2A = cot 60o

2A = 60o

A = 30o

2. Calculate the value of A, if:

(i) (sin A – 1) (2 cos A – 1) = 0

(ii) (tan A – 1) (cosec 3A – 1) = 0

(iii) (sec 2A – 1) (cosec 3A – 1) = 0

(iv) cos 3A. (2 sin 2A – 1) = 0

(v) (cosec 2A – 2) (cot 3A – 1) = 0

Solution:

(i) Given (sin A – 1) (2 cos A – 1) = 0

It can be written as

(sin A – 1) = 0 and (2 cos A – 1) = 0

Sin A = 1 and cos A = ½

Sin A = sin 90o and cos A = cos 60o

A = 90o and A = 60o

(ii) Given (tan A – 1) (cosec 3A – 1) = 0

It can be written as

(tan A – 1) = 0 and (cosec 3A – 1) = 0

tan A = 1 and cosec 3A = 1

tan A = tan 45o and cosec 3A = cosec 90o

A = 45o and A = 30o

(iii) Given (sec 2A – 1) (cosec 3A – 1) = 0

It can be written as

(sec 2A – 1) = 0 and (cosec 3A – 1) = 0

sec 2A = 1 and cosec 3A = 1

sec 2A = sec 60o and cosec 3A = cosec 90o

A = 0o and A = 30o

(iv) Given cos 3A (2 sin 2A – 1) = 0

It can be written as

Cos 3A = 0 and 2 sin 2A – 1 = 0

Cos 3A = cos 90o and sin 2A = ½

3A = 90o and sin 2A = sin 30o

A = 30o and 2A = 30o which implies A = 15o

(v) Given (cosec 2A – 2) (cot 3A – 1) = 0

It can be written as

(cosec 2A – 2) = 0 and (cot 3A – 1) = 0

cosec 2A = 2 and cot 3A = 1

cosec 2A = cosec 30o and cot 3A = cot 45o

2A = 30o and 3A = 45o

A = 15o and A = 15o

3. If 2 sin xo – 1 = 0 and xo is an acute angle; find:

(i) sin xo 

(ii) xo 

(iii) cos xo and tan xo.

Solution:

(i) Given 2 sin xo – 1 = 0

2 sin xo = 1

sin xo = ½

(ii) we have sin xo = ½

sin xo = sin 30o

xo = 30o

(iii) we have xo = 30o

Cos xo = cos 30o = √3/2

Tan xo = tan 30o = 1/√3

4. If 4 cos2 xo – 1 = 0 and 0 Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 23a xo Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 23b 90o, find:

(i) xo 

(ii) sin2 xo + cos2 xo

(iii) Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 23

Solution:

(i) Given 4 cos2 xo – 1 = 0

4 cos2 xo = 0

cos2 xo = (½)2

cos xo = ½

cos xo = cos 60o

xo = 60o

(ii) sin2 xo + cos2 x0 = sin2 60o + cos2 60o

= (√3/2)2 + (½)2

= ¾ + ¼

= 1

(iii) Given

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 24

= 1/ (½)2 – (√3)2

= 4 – 3

= 1

5. If 4 sin2 θ – 1= 0 and angle θ is less than 90o, find the value of θ and hence the value of cos2 θ + tan2 θ.

Solution:

Given 4 sin2 θ – 1= 0

sin2 θ = ¼

sin θ = ½

sin θ = sin 30o

θ = 30o

cos2 θ + tan2 θ = cos2 30o + tan2 30o

= (√3/2)2 + (1/√3)2

= ¾ + 1/3

= (9 + 4)/12

= 13/12

6. If sin 3A = 1 and 0 ≤ A ≤ 90o, find:

(i) sin A 

(ii) cos 2A

(iii) tan2A – 1/cos2 A

Solution:

Given sin 3A = 1

Sin 3A = sin 90o

3A = 90o

A = 30o

(i) according to the question we have,

Sin A = sin 30o

Sin A = ½

(ii) cos 2A = cos 2(30o)

= cos 60o

= ½

(iii) According to the question we have,

Concise Selina Solutions for Class 9 Maths Chapter 23 - Image 25

= 1/3 – 4/3

= -3/3

= -1

Selina Solutions for Class 9 Maths Chapter 23- Trigonometrical Ratios of Standard Angles

The Chapter 23, Trigonometrical Ratios of Standard Angles is composed of 3 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.

23.1 Trigonometrical Ratios of Angles 30o and 60o.

23.2 Trigonometrical Ratios of Angle 45o

23.3 Solving a trigonometric equation

Selina Solutions for Class 9 Maths Chapter 23- Trigonometrical Ratios of Standard Angles

In Chapter 23 of Class 9, the students are taught about the Trigonometrical Ratios of Standard Angles. The chapter also includes the evaluation of an expression involving trigonometric ratios and problems related to it. Study Chapter 23 of Selina textbook to understand more about Trigonometrical Ratios of Standard Angles. Learn the Selina Solutions for Class 9 effectively to come out with flying colours in the examinations.

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