Question
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
-
5π/6
-
π/6
-
π/4
-
3π/4
Solution
B.
π/6
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
