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Question:

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

5π6

3π4

π6

π4

Solution:

Consider,
3sinP + 4cosQ = 6 (i)
4sinQ + 3cosP = 1 (ii)
Squaring and adding (i) and (ii),
(3sinP + 4cosQ)² + (4sinQ + 3cosP)² = 6² + 1²
9sin²P + 16cos²Q + 24sinPcosQ + 16sin²Q + 9cos²P + 24sinQcosP = 37
9(sin²P + cos²P) + 16(sin²Q + cos²Q) + 24(sinPcosQ + cosPsinQ) = 37
9(1) + 16(1) + 24sin(P+Q) = 37
25 + 24sin(P+Q) = 37
24sin(P+Q) = 12
sin(P+Q) = 1/2
P + Q = π/6 or 5π/6
Since P, Q, R are angles of a triangle,
P + Q + R = π
If P + Q = π/6, then R = π - π/6 = 5π/6
If P + Q = 5π/6, then R = π - 5π/6 = π/6
Therefore, R = π/6 or 5π/6