If sin (A + 2B) = √3/2 and cos (A + 4B) = 0, find A and B. (a) 30°

1.	If sin (A + 2B) = √3/2 and cos (A + 4B) = 0, find A and B. (a) 30°, 15° (b) 45°, 60° (c) 0, 90° (d) 45°, 45°
Answer» (a) Here, sin(A + 2B) = √3/2 sin(A + 2B) = sin 60° A + 2B = 60° ...(i) and cos(A + 4B) = 0 cos (A + 4B) = cos 90° A + 4B = 90° A = – 4B + 90° ...(ii) Put the value of A in eq. (ii), A + 2B = 60° or – 4B + 2B + 90° = 60° – 2B = 60° – 90° – 2B = – 30° B = 15° and A = – 4B + 90° = – 4(15°) + 90° A = 30°

If sin (A + 2B) = √3/2 and cos (A + 4B) = 0, find A and B. (a) 30°, 15° (b) 45°, 60° (c) 0, 90° (d) 45°, 45°

Answer»

(a) Here, sin(A + 2B) = √3/2

sin(A + 2B) = sin 60°

A + 2B = 60° ...(i)

and cos(A + 4B) = 0

cos (A + 4B) = cos 90°

A + 4B = 90°

A = – 4B + 90° ...(ii)

Put the value of A in eq. (ii),

A + 2B = 60°

or – 4B + 2B + 90° = 60°

– 2B = 60° – 90°

– 2B = – 30°

B = 15°

and A = – 4B + 90°

= – 4(15°) + 90°

A = 30°