<p>In a triangle ABC if \(3 \sin A + 4 \cos B = 6\); \(4 \sin B + 3 \cos A = 1\) then possible value(s) of ∠C be:</p>
Step-by-Step Solution
Key Concept: Square and add trigonometric equations to find sin(A+B), then use A+B+C=π
<p>Square and add the two given equations: \((3 \sin A + 4 \cos B)^2 + (4 \sin B + 3 \cos A)^2 = 36 + 1\)</p><p>Expanding: \(9 + 16 + 24(\sin A \cos B + \sin B \cos A) = 37\)</p><p>This gives \(\sin(A + B) = 1\), so \(A + B = \frac{\pi}{2}\)</p><p>Therefore \(C = \frac{\pi}{2} - (A + B) = \frac{\pi}{6}\) (from the constraint equations)</p>
Correct Answer: B