Gujarat Board Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

Gujarat Board Textbook Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

Ex 8.1

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
Solution:
We have
AB = 24 cm
BC = 7 cm

(i) In right ΔABC
⇒ AC2 = AB2 + BC2
⇒ AC2 = 242 + 72
⇒ AC2 = 576 + 49

Question 10.
In APQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the value of sin P, cos P and tan P.
Solution:
In right triangle PQR,
PR2 = PQ2 + QR2 ……..(1)
(By Pythagoras theorem)
PR + QR = 25
⇒ PR = 25 – QR ……….(2)

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 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 = 4/3 for some angle.
Solution:
(i) False.

the value of the perpendicular may be longer than the base.

(ii) True

Because Hypotenuse is always greater than the base.

(iii) False
Since cos A is the abbreviation used for the cosine of angle A.

(iv) False
Since cot is meaningless without angle A.

(v) False
Since value of sin O is always less than 1.

Ex 8.2

Question 1.
(i) sin 60° cos 30° + sin 30° cos 60°

Ex 8.3

Question 1.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. (CBSE 2012)
Solution:
We have
tan 2A cot (A – 18°)
tan 2A = tan [90°- (A – 18°)]
tan 2A = tan [90° – A + 18°]
2A = 90° – A + 18°
3A = 108°
A = 108°/3
A = 36°

Question 4.
If tan A = cot B, prove that A+ B = 90°.
Solution:
We have
tan A = cot B
tanA = tan(90° – B) [cot θ = tan (90° – 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:
We have
sec 4A = cosec (A – 20°)
cosec (90° – 4A) = cosec (A – 20°)
90° – 4A = A – 20°
90° + 20° = 5A
5A = 110°
A = 110°/5 = 22°

Question 7.
Express sin 67° ÷ cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
We have
sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

Ex 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. (CBSE 2012)
Solution:

(ii) sin 25° cog 65° + cos 25° sin 65°
= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= 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 θ – 9 tan2 θ =
(a) 1
(b) 9
(c) 8
(d) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)=
(a) –
(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