In triangle ABC,if 3 sin A + 4 cos B = 6 and | vedclass

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(D) Given equations are: $3 \sin A + 4 \cos B = 6$ ... $(i)$ $4 \sin B + 3 \cos A = 1$ ... (ii) Squaring and adding both equations: $(3 \sin A + 4 \co

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$\frac{\pi}{6}$

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(D) Given equations are: $3 \sin A + 4 \cos B = 6$ ... $(i)$ $4 \sin B + 3 \cos A = 1$ ... (ii) Squaring and adding both equations: $(3 \sin A + 4 \cos B)^2 + (4 \sin B + 3 \cos A)^2 = 6^2 + 1^2$ $(9 \sin^2 A + 16 \cos^2 B + 24 \sin A \cos B) + (16 \sin^2 B + 9 \cos^2 A + 24 \sin B \cos A) = 37$ $9(\sin^2 A + \cos^2 A) + 16(\sin^2 B + \cos^2 B) + 24(\sin A \cos B + \cos A \sin B) = 37$ $9(1) + 16(1) + 24 \sin(A + B) = 37$ $25 + 24 \sin(A + B) = 37$ $24 \sin(A + B) = 12$ $\sin(A + B) = \frac{1}{2}$ Since $A+B+C = \pi$,we have $\sin(A+B) = \sin(\pi - C) = \sin C$. Therefore,$\sin C = \frac{1}{2}$. Since $A, B, C$ are angles of a triangle,$C$ can be $\frac{\pi}{6}$ or $\frac{5\pi}{6}$. However,if $C = \frac{5\pi}{6}$,then $A+B = \frac{\pi}{6}$,which makes $\sin A$ and $\sin B$ very small,contradicting the initial equations. Thus,$C = \frac{\pi}{6}$.

In $\triangle ABC$,if $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$,then the $\angle C$ is

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