In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.1. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.

In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A

(ii) sin C, cos C

Solution:

In right angle ∆ABC, we have

AB = 24 cm, BC = 7 cm

∴ Using Pythagoras theorem,

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

⇒ AC^{2} = 24^{2} + 7^{2} = 576 + 49 = 625 = 25^{2}

⇒ AC = 25 cm

Question 2.

In the figure, find tan P – cotR?

Solution:

In right angle ∆PQR

Using the Pythagoras theorem, we get

QR^{2} = PR^{2} – PQ^{2}

⇒ QR^{2} = 13^{2} – 12^{2} = (13 – 12)(13 + 12) = 1 × 25 = 25

∴ QR = \(\sqrt{25}\) = 5 cm

Now, tanP = \(\frac{Q R}{P Q}=\frac{5}{12}\) , cotR = \(\frac{Q R}{P Q}=\frac{5}{12}\)

∴ tanP – cotR = \(\frac{5}{12}-\frac{5}{12}\) = 0.

Question 3.

If sinA = \(\frac{3}{4}\), calculate cosA and tanA.

Solution:

Let us consider, the right angle ∆ABC, we have

Perpendicular = BC and Hypotenuse = AC

Question 4.

Given 15 cot A = 8, find sin A and sec A.

Solution:

In the right angle triangle ABC, we have 15 cot A = 8

Question 5.

Given sec θ = \(\frac { 13 }{ 12 } \), calculate all other trigonometric ratios.

Solution:

Question 6.

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Solution:

From equations (i) and (ii) we get:

\(\frac{\mathrm{CD}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{BF}}\)

⇒ ∆CDA ~ ∆EFB [By SSS similarity]

⇒ ∠A = ∠B Hence Proved

Question 7.

If cot θ = \(\frac { 7 }{ 8 }\), evaluate:

(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right)}\)

(ii) cot²θ

Solution:

Question 8.

If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.

Solution:

In right angled ∆ABC, ∠B = 90°

For ∠A, we have:

Base = AB and perpendicular = BC. Also, Hypotenuse = AC

Question 9.

In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\) find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Solution:

In right ∆ABC, ∠B = 90°

For ∠A, we have

Base = AB, Perpendicular = BC,

Hypotenuse = AC

Question 10.

In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.

Solution:

In right ∆PQR, Q = 90°

PR + QR = 25 cm and PQ = 5 cm

Let QR = x cm ⇒ PR = (25 – x) cm

∴ By Pythagoras theorem, we have

PR^{2} = QR^{2} + PQ^{2}

⇒ (25 – x)^{2} = x^{2} + 5^{2}

⇒ 625 – 50x + x^{2} = x^{2} + 25

⇒ – 50x = – 600

⇒ x = \(\frac{-600}{-50}\) = 12 i.e., QR = 12 cm

⇒ RP = 25 – 12 = 13 cm

Question 11.

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) secA = \(\frac{12}{5}\) for some value of angle A.

(iii) cosA is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sinθ = \(\frac{4}{3}\) for some angle θ.

Solution:

(i) False

∵ A tangent of an angle is the ratio of sides other than hypotenuse, which may be equal or unequal to each other.

(ii) True

∵ cos A is always less than 1.

∴ \(\frac{1}{\cos A}\) i.e., sec A will always be greater than 1.

(iii) False

∵ ‘cosine A’ is abbreviated as ‘cosA’

(iv) False

∵ ‘cot A’ is a single and meaningful term whereas ‘cot’ alone has no meaning.

(v) False

∵ \(\frac{4}{3}\) is greater than 1 and sinθ cannot be greater than 1

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.1. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.

In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A

(ii) sin C, cos C

Solution:

In right angle ∆ABC, we have

AB = 24 cm, BC = 7 cm

∴ Using Pythagoras theorem,

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

⇒ AC^{2} = 24^{2} + 7^{2} = 576 + 49 = 625 = 25^{2}

⇒ AC = 25 cm

Question 2.

In the figure, find tan P – cotR?

Solution:

In right angle ∆PQR

Using the Pythagoras theorem, we get

QR^{2} = PR^{2} – PQ^{2}

⇒ QR^{2} = 13^{2} – 12^{2} = (13 – 12)(13 + 12) = 1 × 25 = 25

∴ QR = \(\sqrt{25}\) = 5 cm

Now, tanP = \(\frac{Q R}{P Q}=\frac{5}{12}\) , cotR = \(\frac{Q R}{P Q}=\frac{5}{12}\)

∴ tanP – cotR = \(\frac{5}{12}-\frac{5}{12}\) = 0.

Question 3.

If sinA = \(\frac{3}{4}\), calculate cosA and tanA.

Solution:

Let us consider, the right angle ∆ABC, we have

Perpendicular = BC and Hypotenuse = AC

Question 4.

Given 15 cot A = 8, find sin A and sec A.

Solution:

In the right angle triangle ABC, we have 15 cot A = 8

Question 5.

Given sec θ = \(\frac { 13 }{ 12 } \), calculate all other trigonometric ratios.

Solution:

Question 6.

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Solution:

From equations (i) and (ii) we get:

\(\frac{\mathrm{CD}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{BF}}\)

⇒ ∆CDA ~ ∆EFB [By SSS similarity]

⇒ ∠A = ∠B Hence Proved

Question 7.

If cot θ = \(\frac { 7 }{ 8 }\), evaluate:

(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right)}\)

(ii) cot²θ

Solution:

Question 8.

If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.

Solution:

In right angled ∆ABC, ∠B = 90°

For ∠A, we have:

Base = AB and perpendicular = BC. Also, Hypotenuse = AC

Question 9.

In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\) find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Solution:

In right ∆ABC, ∠B = 90°

For ∠A, we have

Base = AB, Perpendicular = BC,

Hypotenuse = AC

Question 10.

In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.

Solution:

In right ∆PQR, Q = 90°

PR + QR = 25 cm and PQ = 5 cm

Let QR = x cm ⇒ PR = (25 – x) cm

∴ By Pythagoras theorem, we have

PR^{2} = QR^{2} + PQ^{2}

⇒ (25 – x)^{2} = x^{2} + 5^{2}

⇒ 625 – 50x + x^{2} = x^{2} + 25

⇒ – 50x = – 600

⇒ x = \(\frac{-600}{-50}\) = 12 i.e., QR = 12 cm

⇒ RP = 25 – 12 = 13 cm

Question 11.

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) secA = \(\frac{12}{5}\) for some value of angle A.

(iii) cosA is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sinθ = \(\frac{4}{3}\) for some angle θ.

Solution:

(i) False

∵ A tangent of an angle is the ratio of sides other than hypotenuse, which may be equal or unequal to each other.

(ii) True

∵ cos A is always less than 1.

∴ \(\frac{1}{\cos A}\) i.e., sec A will always be greater than 1.

(iii) False

∵ ‘cosine A’ is abbreviated as ‘cosA’

(iv) False

∵ ‘cot A’ is a single and meaningful term whereas ‘cot’ alone has no meaning.

(v) False

∵ \(\frac{4}{3}\) is greater than 1 and sinθ cannot be greater than 1