In a ∆PQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to

- π/4
- 3π/4
- 5π/6
- π/6

**Answer**

3 sin P + 4 cos Q = 6 ...... (1)

4 sin Q + 3 cos P = 1 ...... (2)

From (1) and (2) ∠P is obtuse.

(3 sin P + 4 cos Q)^{2} + (4 sin Q + 3 cos P)^{2} = 37

9 + 16 + 24 (sin P cos Q + cos P sin Q) = 37

24 sin (P + Q) = 12

sin (P + Q) = 1/2

P + Q = 5π/6

**The correct option is D.**