Question

In a ∆ PQR , if 3sinP + 4cosQ = 6 and 4sinQ + 3cosP = 1, then the angle R is equal to 

(A) 6π/6 

(B) π/6  

(C) π/4 

(D) 3π/4

Answer 1

Correct option (B) π/6

Explanation:

3sinP + 4cosQ = 6 … (i)

4sinQ + 3cosP = 1 … (ii)

From (1) and (2) ∠ P is obtuse,

(3sinP + 4cosQ)² + (4sinQ + 3cosP)² = 37

⇒ 9 + 16 + 24(sinP cosQ + cosPsinQ) = 37

⇒  24 sin(P + Q) = 12

⇒ sin(P + Q) = 1/2 

⇒ P + Q = 5/6 

⇒ R = π/6