# SELINA Solutions for Class 9 Maths Chapter 25 - Complementary Angles

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## Chapter 25 - Complementary Angles Exercise Ex. 25

Question 1 Solution 1 Question 2 Solution 2 Question 3 Solution 3 Question 4 Solution 4 Question 5 Solution 5 Question 6 Solution 6 Question 7 Solution 7 Question 8 Solution 8 Question 9 Solution 9 Question 10 Solution 10 Question 11 Solution 11 Question 12 Solution 12 Question 13 Solution 13 Question 14 Solution 14 Question 15

Show that:

tan 10° tan 15° tan 75° tan 80° = 1

Solution 15

L.H.S.

= tan 10° tan 15° tan 75° tan 80°

= tan (90° - 80°) tan (90° - 75°) tan 75° tan 80°

= cot 80° cot 75 ° tan 75° tan 80°

= (cot 80° tan 80°)(cot 75° tan 75°)

= (1)(1)

= 1

= R.H.S.

Question 16

Show that:

sin 42° sec 48° + cos 42° cosec 48° = 2

Solution 16 Question 17

Express each of the following in terms of angles between 0°and 45°:

(i) sin 59°+ tan 63°

(ii) cosec 68°+ cot 72°

(iii)cos 74°+ sec 67°

Solution 17 Question 18

For triangle ABC, show that:

(i) (ii) Solution 18

(i) We know that for a triangle ABC A + B + C = 180° (ii) We know that for a triangle ABC A + B + C = 180° B + C = 180° - A Question 19

Evaluate:

(i)  (iii)  (iv) (v) (vi) (vii) (viii) (ix) Solution 19

(i)  (iii)  (iv)  (v)  (vi)  (vii)  (viii)  (ix)  Question 20 Solution 20 Question 21

In each case, given below, find the value of angle A, where 0° A 90°.

sin (90° - 3A).cosec 42° = 1

Solution 21 Question 22

In each case, given below, find the value of angle A, where 0° A 90°.

cos (90° - A).sec 77° = 1

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