Question

In each case, given below, find the value of
angle A, where 0° <A< 90°.
(i) sin (90° - 3A).cosec 42° = 1
(ii) cos (90° - A) sec 77° = 1​

Answer 1

2. cos 90-A = 1/sec77°

cos 90 -A = cos 77°

90-A = 77

A = 90-77 = 13°

Answer 2

Answer:

(i) 16

(ii) 13

Step-by-step explanation:

(i) Sin (90 - 3A). cosec 42 = 1

= sin (90 - 3A) = (1)/(cosec 42)

= cos 3A = (1)/(cosec (90 - 42)

= cos 3A = (1)/(sec 48)

= cos 3A = cos 48

= 3A = 48

= A = 16.

(ii) cos (90 - A). sec 77 = 1

= cos(90 - A) = (1)/(sec77)

= sin A = (1)/(sec(90 - 77)

= sin A = (1)/(cosec 13)

= sin A - sin 13  

= A = 13

Hope u understud.

Mark me as brainiest

Thank You