Question:

<p><strong>Step 1:</strong> Square both given equations:</p><p>$(3\sin P + 4\cos Q)^2 = 36$</p><p>$9\sin^2 P + 24\sin P \cos Q + 16\cos^2 Q = 36$ ... (i)</p><p>$(4\sin Q + 3\cos P)^2 = 1$</p><p>$16\sin^2 Q + 24\sin Q \cos P + 9\cos^2 P = 1$ ... (ii)</p><p><strong>Step 2:</strong> Add equations (i) and (ii):</p><p>$9\sin^2 P + 9\cos^2 P + 16\sin^2 Q + 16\cos^2 Q + 24(\sin P \cos Q + \sin Q \cos P) = 37$</p><p>$9(1) + 16(1) + 24\sin(P+Q) = 37$</p><p>$25 + 24\sin(P+Q) = 37$</p><p>$\sin(P+Q) = \frac{12}{24} = \frac{1}{2}$</p><p><strong>Step 3:</strong> Since P, Q, R are angles in a triangle: $P + Q + R = \pi$</p><p>Therefore: $P + Q = \pi - R$</p><p>$\sin(\pi - R) = \frac{1}{2}$</p><p>$\sin R = \frac{1}{2}$</p><p>Since $0 < R < \pi$ in a triangle: $R = \frac{\pi}{6}$ or $R = \frac{5\pi}{6}$</p><p>Checking feasibility with original constraints, $R = \frac{5\pi}{6}$ is the answer.</p><p>∴ Answer: B</p>