NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry. Furthermore, here we’ve provided you with the latest solution for Class 10 CBSE NCERT Maths Chapter 8 – Introduction to Trigonometry. As a result here you’ll find solutions to all the exercises. This NCERT Class 10 solution will help you to score good marks in your exam.

Students can refer to our solution for NCERT Class 10 Maths Chapter 8 – Introduction to Trigonometry. The Chapter 8 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise wise latest solution.

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise Wise Solution

Exercise 8.1 – Page 181 of NCERT
Exercise 8.2 – Page 187 of NCERT
Exercise 8.3 – Page 189 of NCERT
Exercise 8.4 – Page 193 of NCERT

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise 8.1

Here you’ll find NCERT Chapter 8 – Introduction to Trigonometry Exercise 8.1 Solution.
Exercise 8.: Solutions of Questions on Page Number: 181.

Q1 : In ΔABC right angled at B, AB = 24 cm, BC = 7 m. Determine

1. sin A, cos A
2. sin C, cos C

Answer :

Applying Pythagoras theorem for ΔABC, we obtain AC² = AB² + BC²

= (24 cm)² + (7 cm)²

= (576 + 49) cm²

= 625 cm²

∴ AC = cm = 25 cm

(i) sin A =

cos A =

(ii)

sin C =

cos C =

Q2 :

In the given figure find tan P – cot R

Answer :

Applying Pythagoras theorem for ΔPQR, we obtain PR² = PQ² + QR²

(13 cm)²  = (12 cm)² + QR² 169 cm² = 144 cm² + QR²

25 cm² = QR²

QR = 5 cm

tan P – cot R =

Q3 :

If sin A = 3/4, calculate cos A and tan A.

Answer :

Let ΔABC be a right-angled triangle, right-angled at point B.

Given that,

Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer. Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC² (4k)² = AB² + (3k)² 16k ² – 9k ² = AB²

7k ² = AB²

AB =

Q4 : Given 15 cot A = 8. Find sin A and sec A.

Answer Consider a right-angled triangle, right-angled at B

.

It is given that,

cot A =

Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer. Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (15k)²

= 64k² + 225k²

= 289k²

AC = 17k

Q5: Given sec θ = 13/12, calculate all other trigonometric ratios.

Answer :

Consider a right-angle triangle ΔABC, right-angled at point B

.

If AC is 13k, AB will be 12k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain (AC)² = (AB)² + (BC)²

(13k)² = (12k)² + (BC)² 169k² = 144k² + BC²

25k² = BC²

BC = 5k

Q6: If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Answer :
Let us consider a triangle ABC in which CD ⊥ AB

.

It is given that cos A = cos B

… (1)
We have to prove ∠A = ∠B. To prove this, let us extend AC to P such that BC = CP

.

From equation (1), we obtain

By using the converse of B.P.T,

CD||BP

⇒∠ACD = ∠CPB (Corresponding angles) … (3) And,
∠BCD = ∠CBP (Alternate interior angles) … (4)
By construction, we have BC = CP.

∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) …
(5) From equations (3), (4), and (5), we obtain

∠ACD = ∠BCD … (6)
In ΔCAD and ΔCBD,

∠ACD = ∠BCD [Using equation (6)]

∠CDA = ∠CDB [Both 90°]

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

Alternatively,

Let us consider a triangle ABC in which CD ⊥ AB.

It is given that, cos A = cos B

Let

⇒ AD = k BD … (1) And,
AC = k BC … (2)

Using Pythagoras theorem for triangles CAD and CBD, we obtain

CD² = AC² – AD² … (3)

And, CD² = BC² – BD² … (4)

From equations (3) and (4), we obtain

AC² – AD² = BC² – BD²

⇒ (k BC)² – (k BD)² = BC² – BD²

⇒ k² (BC² – BD²) = BC² – BD²

⇒ k² = 1

⇒ k = 1

Putting this value in equation (2), we obtain

AC = BC

⇒ ∠A = ∠B(Angles opposite to equal sides of a triangle)

Q7 :

If cot θ =7/8 , evaluate
(i)
(ii) cot² θ

Answer:

(i) Let us consider a right triangle ABC, right-angled at point B.

If BC is 7k, then AB will be 8k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k²

AC =

(i)

(ii) cot² θ = (cot θ)² = =

Q8 :

If 3 cot A = 4, Check whether

Answer :

It is given that 3cot A = 4

Or, cot A = 4/3

Consider a right triangle ABC, right-angled at point B

.

If AB is 4k, then BC will be 3k, where k is a positive integer. In ΔABC,

(AC)² = (AB)² + (BC)²

= (4k)² + (3k)²

= 16k² + 9k²

= 25k²

AC = 5k

cos² A – sin² A =

∴

Q9:

In ΔABC, right angled at B. If

find the value of

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

Answer :

If BC is k, then AB will be

where k is a positive integer.

In ΔABC,

AC² = AB² + BC²

=

= 3k² + k² = 4k²

∴ AC = 2k

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

Q10: In ΔPQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer :

Given that, PR + QR = 25

PQ = 5

Let PR be x.

Therefore, QR = 25 – x

Applying Pythagoras theorem in ΔPQR, we obtain

PR² = PQ² + QR²

x² = (5)² + (25 – x)²

x² = 25 + 625 + x² – 50x

50x = 650

x = 13

Therefore, PR = 13 cm QR = (25 – 13) cm = 12 cm

Q11: State whether the following are true or false. Justify your answer.

1. The value of tan A is always less than 1.
2. Sec A =for some value of angle A.
3. cos A is the abbreviation used for the cosecant of angle A.
4. cot A is the product of cot and A
5. sin θ = for some angle θ

Answer :

(i) Consider a ΔABC, right-angled at B

.

But > 1

∴tan A > 1

So, tan A < 1 is not always true.

Hence, the given statement is false.

(ii)

Let AC be 12k, AB will be 5k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

(12k)² = (5k)² + BC²

144k² = 25k²  + BC²

BC² = 119k²

BC = 10.9k

It can be observed that for given two sides AC = 12k and AB = 5k,

BC should be such that,

AC – AB < BC < AC + AB

12k – 5k < BC < 12k + 5k

7k < BC < 17 k

However, BC = 10.9k. Clearly, such a triangle is possible and hence, such value of sec A is possible. Hence, the given statement is true.

3. Abbreviation used for cosecant of angle A is cosec A. And cos A is the abbreviation used for cosine of angle A.
   Hence, the given statement is false.
4. cot A is not the product of cot and A. It is the cotangent of ∠A.
   Hence, the given statement is false.
5. sin θ =

We know that in a right-angled triangle,

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.
Hence, the given statement is false.

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise 8.2

Here you’ll find NCERT Chapter 8 – Introduction to Trigonometry Exercise 8.2 Solution.
Exercise 8.2: Solutions of Questions on Page Number: 187.

Q1: Evaluate the following

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan²45° + cos²30° – sin²60°

(iii)

(iv)

(v)

Answer :

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan²45° + cos²30° – sin²60°

(iii)

(iv)

(v)

Q2: Choose the correct option and justify your choice.

(i)

(A). sin60°
(B). cos60°
(C). tan60°
(D). sin30°

Answer :

(i)

Out of the given alternatives, only

Hence, (A) is correct.

(ii)

(A). cos60°
(B). sin60°
(C). tan60°
(D). sin30°

Hence, (C) is correct.

(iii) sin2A = 2sinA is true when A =
(A). 0°
(B). 30°
(C). 45°
(D). 60°

Answer: Out of the given alternatives, only A = 0° is correct.
As sin 2A = sin 0° = 0
2 sinA = 2sin 0° = 2(0) = 0
Hence, (A) is correct.

(iv)

(A). cos60°
(B). sin60°
(C). tan60°
(D). sin30°

Out of the given alternatives, only tan 60°

Hence, (C) is the correct option.

Q3: If tan(A + B) = √3 and If tan(A – B) = 1/√3;
and ; 0° < A + B ≤90°, A > B find A and B.
Answer :

⇒

⇒ A + B = 60 … (1)

⇒ tan (A – B) = tan30

⇒ A – B = 30 … (2)

On adding both equations, we obtain 2A = 90

⇒ A = 45

From equation (1), we obtain 45 + B = 60

B = 15

Therefore, ∠A = 45° and ∠B = 15°

Q4: State whether the following are true or false. Justify your answer.

1. sin(A + B) = sin A + sin B
2. The value of sin θ¸ increases as θ¸increases
3. The value of cos θ¸increases as θ¸increases
4. sin θ¸= cos θ¸ for all values of θ¸
5. cot A is not defined for A = 0°

Answer :

(i) sin(A + B) = sin A + sin B Let A = 30° and B = 60°

sin (A + B) = sin (30° + 60°)

= sin 90°

= 1

sin A + sin B = sin 30° + sin 60°

Clearly, sin (A + B) ≠sin A + sin B
Hence, the given statement is false.

(ii) The value of sin θ increases as θ increases in the interval of 0° < θ < 90° as sin 0° = 0

sin 90° = 1

Hence, the given statement is true.

(iii) cos0° = 1

cos90° = 0

It can be observed that the value of cos θ does not increase in the interval of 0° < θ < 90°. Hence, the given statement is false.

(iv) sinθ = cosθ for all values of θ.
This is true when θ = 45°

It is not true for all other values of θ.

As and

Hence, the given statement is false.

(v) cot A is not defined for A = 0°

As

= undefined
Hence, the given statement is true.

 

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise 8.3

Here you’ll find NCERT Chapter 8 – Introduction to Trigonometry Exercise 8.3 Solution.
Exercise 8.3: Solutions of Questions on Page Number: 189.

Q1: Evaluate

(i)

(ii)

(iii) cos 48° – sin 42°

(iv) cosec 31° – sec 59°

Answer:

(i)

(ii)

(iii)cos 48° – sin 42° = cos (90° – 42°) – sin 42°

= sin 42° – sin 42°

= 0

(iv) cosec 31° – sec 59° = cosec (90° – 59°) – sec 59°

= sec 59° – sec 59°

= 0

Q2: Show that

(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii)cos 38° cos 52° – sin 38° sin 52° = 0

Answer :

(i) tan 48° tan 23° tan 42° tan 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°)

= (1) (1)

= 1

(ii) cos 38° cos 52° – sin 38° sin 52°

= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°

= sin 52° sin 38° – sin 38° sin 52°

= 0

Q3: If tan 2A = cot (A- 18°), where 2A is an acute angle, find the value of A.

Answer: Given that,

tan 2A = cot (A- 18°)
cot (90° – 2A) = cot (A -18°)
90° – 2A = A- 18°
108° = 3A
A = 36°

Q4: If tan A = cot B, prove that A + B = 90°

Answer : Given that, tan A = cot B

tan A = tan (90° – B) A = 90° – B

A + B = 90°

Q5: If sec 4A = cosec (A- 20°), where 4A is an acute angle, find the value of A.

Answer: Given that,

sec 4A = cosec (A – 20°)

cosec (90° – 4A) = cosec (A – 20°) 90° – 4A= A- 20°

110° = 5A

A = 22°

Q6: If A, Band C are interior angles of a triangle ABC then show that

Answer: We know that for a triangle ABC,

∠ A + ∠B + ∠C = 180°

∠B + ∠C= 180° – ∠A

Q7: Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Answer: sin 67° + cos 75°

= sin (90° – 23°) + cos (90° – 15°)

= cos 23° + sin 15°

NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise 8.4

Here you’ll find NCERT Chapter 8 – Introduction to Trigonometry Exercise 8.4 Solution.
Exercise 8.4: Solutions of Questions on Page Number: 193.

Q1: Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Answer : We know that,

.will always be positive as we are adding two positive quantities

Therefore

We know that,

However,

Q2: Write all the other trigonometric ratios of ∠ A in terms of sec A.

Answer: We know that,

Also, sin² A + cos²  A = 1 sin² A = 1 – cos² A

tan²A + 1 = sec²A tan²A = sec²A – 1

Q3: Evaluate

(i)

(ii) sin25° cos65° + cos25° sin65°

Answer:

(i)

(As sin²A + cos²A = 1)

= 1

(ii) sin25° cos65° + cos25° sin65°

= sin²25° + cos²25°
= 1 (As sin²A + cos²A = 1)

Q4: Choose the correct option. Justify your choice.

1. 9 sec² A – 9 tan² A =
   1. 1
   2. 9
   3. 8
   4. 0
2. (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
   1. 0
   2. 1
   3. 2
   4. – 1
3. (secA + tanA) (1 – sinA) =
   1. secA
   2. sinA
   3. cosecA
   4. cosA
4. 
   1. sec² A
   2. -1
   3. cot² A
   4. tan² A

Answer :

(i) 9 sec²A – 9 tan²A

= 9 (sec²A – tan²A)

= 9 (1) [As sec² A – tan² A = 1]

= 9

Hence, alternative (B) is correct.

(ii)

(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Hence, alternative (C) is correct.

(iii) (secA + tanA) (1 – sinA)

= cosA

Hence, alternative (D) is correct.

(iv)

Hence, alternative (D) is correct.

Q5: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Answer :

(i)

(ii)

(iii)

We hope the NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry helps you. If you have any queries regarding chapter 8. Drop a comment below and we will get back to you at the earliest.

NCERT Class 10 Maths All Chapters Solution

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equation in Two Variables

Chapter 4: Quadratic Equations

Chapter 5: Arithmetic Progressions

Chapter 6: Triangles

Chapter 7: Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9: Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Areas Related to Circles

Chapter 13: Surface Areas and Volumes

Chapter 14: Statistics

Chapter 15: Probability