NCERT Solutions for Class 10 Math Chapter 8 - Introduction To Trigonometry

Share with your friends

Mathematics NCERT Grade 10, Chapter 8: Introduction to Trigonometry: In the beginning, a quote is given about trigonometry in an eye-catching way. Some real-life examples like finding the height of Qutub Minar, finding the width of the river, and finding the altitude of the ground from the hot air balloon are given in order to explain the need for trigonometry. In this chapter, students will study the trigonometric ratios of the angle i.e ratios of the sides of a right triangle with respect to its acute angles. The discussion of the trigonometric ratios will be restricted to acute angles only.
In section 8.2 various trigonometric ratios are explained. Trigonometric ratios are given in this section:

SINE
COSINE
TANGENT
COSECANT
SECANT
COTANGENT

Solved examples are given to explain how to find trigonometric ratios and how to prove certain given relations.

Special note is given stating that the values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle if the angle remains the same.

In exercise 8.1 students have to determine certain trigonometric ratios.
In the next section, trigonometric ratios of some specific angles are given. Value of ratios of angle 0 $°$ , 30 $°$ , 45

°

, 60 $°$ , 90 $°$ are given in table 8.1. Students must learn these values as they are the most important part of this chapter and are essential to solve the problems not only in this chapter but are useful in higher grades as well.
Exercise 8.2 contains 4 different types of questions based on trigonometric ratios.
In section 8.4 trigonometric ratios of complementary angles are given. Total 6 relations are given in this section. These relations must be on the fingertips of students in order to score better. Exercise 8.3 is based on the same.
The next part of the chapter is about trigonometric identities. Three important identities are highlighted in this section. These identities are true for all angles (A) such that A $\in$ [0 $°$ , 90 $°$ ]. Using the same students will be able to solve the problems of exercise 8.4 including the problems which need certain given statements to prove. In the end key points of chapter are discussed.

Page No 181:

Question 1:

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

(i) sin A, cos A

(ii) 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 =

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 1)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 1

Page No 181:

Question 2:

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 =

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 2)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 2

Page No 181:

Question 3:

If sin A =, 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 =

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 3)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 3

Page No 181:

Question 4:

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

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 4)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 4

Page No 181:

Question 5:

Given sec θ =, 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

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 5)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 5

Page No 181:

Question 6:

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)

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 6)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 6

Page No 181:

Question 7:

If cot θ =, evaluate

(i) (ii) cot² θ

Answer:

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 = =

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 7)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 7

Page No 181:

Question 8:

If 3 cot A = 4, Check whether

Answer:

It is given that 3cot A = 4

Or, cot A =

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 =

∴

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 8)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 8

Page No 181:

Question 9:

In ΔABC, right angled at B. If, find the value of

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

(ii) cos A cos C − sin 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

(ii) cos A cos C − sin A sin C

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 9)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 9

Page No 181:

Question 10:

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

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 10)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 10

Page No 181:

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) sec A =for some value of angle A.

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

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

(v) 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.

(iii) 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.

(iv) cot A is not the product of cot and A. It is the cotangent of ∠A.

Hence, the given statement is false.

(v) 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

Video Solution for introduction to trigonometry (Page: 181 , Q.No.: 11)

NCERT Solution for Class 10 math - introduction to trigonometry 181 , Question 11

Page No 187:

Question 1:

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)

Video Solution for introduction to trigonometry (Page: 187 , Q.No.: 1)

NCERT Solution for Class 10 math - introduction to trigonometry 187 , Question 1

Page No 187:

Question 2:

Choose the correct option and justify your choice.

(i)

(A). sin60°

(B). cos60°

(C). tan60°

(D). sin30°

(ii)

(A). tan90°

(B). 1

(C). sin45°

(D). 0

(iii) sin2A = 2sinA is true when A =

(A). 0°

(B). 30°

(C). 45°

(D). 60°

(iv)

(A). cos60°

(B). sin60°

(C). tan60°

(D). sin30°

Answer:

(i)

Out of the given alternatives, only

Hence, (A) is correct.

(ii)

Hence, (D) is correct.

(iii)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)

Out of the given alternatives, only tan 60°

Hence, (C) is correct.

Video Solution for introduction to trigonometry (Page: 187 , Q.No.: 2)

NCERT Solution for Class 10 math - introduction to trigonometry 187 , Question 2

Page No 187:

Question 3:

If 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°

Video Solution for introduction to trigonometry (Page: 187 , Q.No.: 3)

NCERT Solution for Class 10 math - introduction to trigonometry 187 , Question 3

Page No 187:

Question 4:

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

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

(ii) The value of sinθ increases as θ increases

(iii) The value of cos θ increases as θ increases

(iv) sinθ = cos θ for all values of θ

(v) 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) cos 0° = 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) sin θ = cos θ 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.

Video Solution for introduction to trigonometry (Page: 187 , Q.No.: 4)

NCERT Solution for Class 10 math - introduction to trigonometry 187 , Question 4

Page No 189:

Question 1:

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

Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 1)

NCERT Solution for Class 10 math - introduction to trigonometry 189 , Question 1

Page No 189:

Question 2:

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

Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 2)

NCERT Solution for Class 10 math - introduction to trigonometry 189 , Question 2

Page No 189:

Question 3:

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°

Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 3)

NCERT Solution for Class 10 math - introduction to trigonometry 189 , Question 3

Page No 189:

Question 4:

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°

Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 4)

NCERT Solution for Class 10 math - introduction to trigonometry 189 , Question 4

Page No 189:

Question 5:

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°

Video Solution for introduction to trigonometry (Page: 189 , Q.No.: 5)

NCERT Solution for Class 10 math - introduction to trigonometry 189 , Question 5

Page No 190:

Question 6:

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

Video Solution for introduction to trigonometry (Page: 190 , Q.No.: 6)

NCERT Solution for Class 10 math - introduction to trigonometry 190 , Question 6

Page No 190:

Question 7:

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°

Video Solution for introduction to trigonometry (Page: 190 , Q.No.: 7)

NCERT Solution for Class 10 math - introduction to trigonometry 190 , Question 7

Page No 193:

Question 1:

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,

Therefore,

Also,

Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 1)

NCERT Solution for Class 10 math - introduction to trigonometry 193 , Question 1

Page No 193:

Question 2:

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

Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 2)

NCERT Solution for Class 10 math - introduction to trigonometry 193 , Question 2

Page No 193:

Question 3:

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)

Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 3)

NCERT Solution for Class 10 math - introduction to trigonometry 193 , Question 3

Page No 193:

Question 4:

Choose the correct option. Justify your choice.

(i) 9 sec² A − 9 tan² A =

(A) 1

(B) 9

(D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

(A) 0

(B) 1

(D) −1

(iii) (secA + tanA) (1 − sinA) =

(A) secA

(B) sinA

(D) cosA

(iv)

(A) sec²A

(B) −1

(D) tan²A

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.

Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 4)

NCERT Solution for Class 10 math - introduction to trigonometry 193 , Question 4

Page No 193:

Question 5:

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

Answer:

(i)

(ii)

(iii)

$= \frac{\sin^{2} θ}{\cos θ (\sin θ - \cos θ)} - \frac{\cos^{2} θ}{\sin θ (\sin θ - \cos θ)}$

= secθ cosec θ +

= R.H.S.

(iv)

= R.H.S

(v)

Using the identity cosec² = 1 + cot²,

L.H.S =

= cosec A + cot A

= R.H.S

(vi)

(vii)

(viii)

(ix)

Hence, L.H.S = R.H.S

(x)

Video Solution for introduction to trigonometry (Page: 193 , Q.No.: 5)

NCERT Solution for Class 10 math - introduction to trigonometry 193 , Question 5

View NCERT Solutions for all chapters of Class 10