In a ∆PQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1
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.