NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

## Introduction to Trigonometry Solutions – Class 10 Maths

**Exercise 8.1 :** Solutions of Questions on Page Number : **181**

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

**(i) sin A, cos A**

**(ii) sin C, cos C**

**Answer :**

Applying Pythagoras theorem for ΔABC, we obtain

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

= (24 cm)^{2} + (7 cm)^{2}

= (576 + 49) cm^{2}

= 625 cm^{2}

∴ 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

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

(13 cm)^{2} = (12 cm)^{2} + QR^{2}

169 cm^{2} = 144 cm^{2} + QR^{2}

25 cm^{2} = QR^{2}

QR = 5 cm

tan P – cot R =

**Q3 : ****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

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

(4k)^{2} = AB^{2} + (3k)^{2}

16k^{2} – 9k^{2} = AB^{2}

7k^{2} = AB^{2}

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

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

= (8k)^{2} + (15k)^{2}

= 64k^{2} + 225k^{2}

= 289k^{2}

AC = 17k

**Q5 : ****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}

169k^{2} = 144k^{2} + BC^{2}

25k^{2} = BC^{2}

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

CD^{2} = AC^{2} – AD^{2} … (3)

And, CD^{2} = BC^{2} – BD^{2} … (4)

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

AC^{2} – AD^{2} = BC^{2} – BD^{2}

⇒ (k BC)^{2} – (k BD)^{2} = BC^{2} – BD^{2}

⇒ k^{2} (BC^{2} – BD^{2}) = BC^{2} – BD^{2}

⇒ k^{2} = 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 =, 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

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

= (8k)^{2} + (7k)^{2}

= 64k^{2} + 49k^{2}

= 113k^{2}

AC =

(i)

(ii) cot^{2} θ = (cot θ)^{2} = =

**Q8 : ****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}

= 16k^{2} + 9k^{2}

= 25k^{2}

AC = 5k

cos^{2} A – sin^{2} A =

∴

**Q9 : ****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,

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

=

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

∴ AC = 2k

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

(ii) cos A cos C – sin 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

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

x^{2} = (5)^{2} + (25 – x)^{2}

x^{2} = 25 + 625 + x^{2} – 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.**

**(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

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

(12k)^{2} = (5k)^{2} + BC^{2}

144k^{2} = 25k^{2} + BC^{2}

BC^{2} = 119k^{2}

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

**Exercise 8.2 :** Solutions of Questions on Page Number : **187**

**Q1 : ****Evaluate the following**

**(i) sin60° cos30° + sin30° cos60°**

**(ii) 2tan ^{2}45° + cos^{2}30° – sin^{2}60°**

**(iii)**

**(iv)**

**(v)**

**Answer :**

(i) sin60° cos30° + sin30° cos60°

(ii) 2tan

^{2}45° + cos

^{2}30° – sin

^{2}60°

(iii)

(iv)

(v)

**Q2 : ****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.

**Q3 : ****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°

**Q4 : ****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°

As

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.

**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°

**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,

Therefore,

Also,

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

**Answer :**

We know that,

Also, sin^{2} A + cos^{2} A = 1

sin^{2} A = 1 – cos^{2} A

tan^{2}A + 1 = sec^{2}A

tan^{2}A = sec^{2}A – 1

**Q3 : ****Evaluate**

**(i)**

**(ii) sin25° cos65° + cos25° sin65°**

**Answer :**

(i)

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

= 1

(ii) sin25° cos65° + cos25° sin65°

= sin^{2}25° + cos^{2}25°

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

**Q4 : Choose the correct option. Justify your choice.**

**(i) 9 sec ^{2} A – 9 tan^{2} A =**

**(A) 1**

**(B) 9**

**(C) 8**

**(D) 0**

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

**(A) 0**

**(B) 1**

**(C) 2**

**(D) – 1**

**(iii) (secA + tanA) (1 – sinA) =**

**(A) secA**

**(B) sinA**

**(C) cosecA**

**(D) cosA**

**(iv)**

**(A) sec**

^{2}A**(B) – 1**

**(C) cot**

^{2}A**(D) tan**

^{2}A**Answer :**

(i) 9 sec

^{2}A – 9 tan

^{2}A

= 9 (sec

^{2}A – tan

^{2}A)

= 9 (1) [As sec

^{2}A – tan

^{2}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)

### NCERT Solutions for Class 10 Maths All Chapters

- Chapter 1 – Real Numbers
- Chapter 2 – Polynomials
- Chapter 3 – Pair of Linear Equations 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