PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

 PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

PSEB 10th Class 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:
(i) We are to find sin A .cos A AB = 24 cm; BC = 7 cm
By using Pythagoras Theorem,

     

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
Let ABC be any right angled triangle where A is an acute angle with right angle at B.

   

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, show that LA = LB.
Solution:
Let ABC be any triangle, where ∠A and ∠B are acute angles. To find cos A and cos B.

          

Question 10.
In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
Given: ∆PQR, right angled at Q

PR + QR = 25 cm
PQ = 5 cm
In right angled triangle PQR
By using Pythagoras Theorem,
(PR)2 = (PQ)2 + (RQ)2
or (PR)2 = (5)2 + (RQ)2
[∴ PR + QR = 25, QR = 25 – PR]
or (PR)2 = 25 + [25 – PR]2
or (PR)2 = 25 + (25)2 + (PR)2 – 2 × 25 × PR
or (PR)2 = 25 + 625 + (PR)2 – 50
or (PR)2 – (PR)2 + 50 PR = 650
or 50 PR = 650
or PR = 650/50
or PR = 13 cm
QR = 25 – PR
QR = (25 – 13) cm
or QR = 12 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) sec A = 12/5 for some value of angle A.
(iii) cos A is abbreviation used for cosecant of angle A.
(iv) cot A is product of cot and A.
(v) sin θ = 4/3 for some angle θ.
Solution:
(i) False
∵ tan 60° = √3 = 1.732 > 1.

(ii) True; sec A = 12/5 = 240 > 1
∵ Sec A is always greater than 1.

(iii) False.
Because cos A is used for cosine A.

(iv) False.
Because cot A is cotangent of the angle A not the product of cot and A.

(v) False; sin θ = 4/3 = 1.666 > 1
Because sin θ is always less than 1.

Ex 8.2

Question 1.
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan2 45° + cos2 30° – sin2 60°

  

Question 2.
Choose the correct option and justify your choice.

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

   

Ex 8.3

Question 1.
Evaluate:

(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
[∵ cos (90° – 0) = sin O]
= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59°
=cosec 31° – sec (90° – 31°)
= cosec 31° – cosec 31°
[∵ sec (90° – θ) = cosec θ].

Question 2.
Show that:
(i) tan 4 tan 230 tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:

(ii) L.H.S.= cos 38° cos 52° – sin 38° sin 52°
= cos 38° × cos (90 – 38°) – sin 38° × sin (90° – 38°)
= cos 38° × sin 38° – sin 38° × cos 38
= 0.
∴ L.H.S. = RH.S.

Question 3.
If tan 2A = cot (A – 18°) where 2A is an acute angle, find the value of A.
Solution:
Given: tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°)
[cot (90° – θ) = tan θ]
⇒ 90°- 2A = A – 18°
⇒ 3A = 108°
⇒A = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
Given that: tan A = cot B
⇒ tan A = tan(90° – B)
[∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B.
⇒ A + B = 90°..

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
Given that: sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A — 20°)
[∵ cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20°
⇒ 5A = 110°
⇒ A = 22°.

Question 7.
Express sin 67° + cos 75° in terms of Trigonometric ratios of angles between 0° and 45°.
Solution:
Given that: sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
[∵ sin(90° – θ) = cos θ and cos (90° – θ) = sin θ].

Ex 8.4

Question 1.
Express the trigonometric ratios of sin A, sec A and tan A in terms of cot A.
Solution:

  

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° × cos (90° – 25°)
+ cos 25° × sin (90° – 25°)
[∵ cos (90° – θ) = sin θ
sin(90° – θ) = cos θ].
= sin 25° × sin 25° + cos 25° × cos 25°
= sin2 25° + cos2 25° = 1.

Question 4.
Choose the correct option. Justify your choice:
(i) 9 sec2 A – 9 tan2 A =
(A) 1
(B) 9
(C) 8
(D) 0.

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(A) θ
(B) 1
(C) 2
(D) – 1.

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A.