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In $\triangle A B C$, 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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The correct answer is:
$\frac{\pi}{6}$
$3 \sin A+4 \cos B=6$...(i)
$4 \sin B+3 \cos A=1$...(ii)
On adding Eqs. (i) and (ii) after taking square,
$\begin{aligned} & \left(9 \sin ^2 A+24 \sin A \cdot \cos B+16 \cos ^2 B\right)+ \\ & \left(16 \sin ^2 B+24 \sin B \cdot \cos A+9 \cos ^2 A\right)=37\end{aligned}$
$\begin{aligned} \Rightarrow \quad 9\left(\sin ^2 A+\cos ^2 A\right)+16\left(\cos ^2 B\right. & \left.+\sin ^2 B\right) \\ & +24 \sin (A+B)=37\end{aligned}$
$\begin{aligned} & \Rightarrow \quad 9+16+24 \sin (A+B)=37 \\ & \Rightarrow \quad 24 \sin (A+B)=12 \\ & \Rightarrow \sin (A+B)=\frac{1}{2} \\ & \Rightarrow A+B=\frac{\pi}{6}, \frac{5 \pi}{6}\end{aligned}$
$\begin{aligned} & \because \quad A+B>C \\ & \therefore \quad A+B=\frac{5 \pi}{6}\end{aligned}$
Hence, $C=\pi-\frac{5 \pi}{6}=\frac{\pi}{6} \quad[\because A+B+C=\pi]$
$4 \sin B+3 \cos A=1$...(ii)
On adding Eqs. (i) and (ii) after taking square,
$\begin{aligned} & \left(9 \sin ^2 A+24 \sin A \cdot \cos B+16 \cos ^2 B\right)+ \\ & \left(16 \sin ^2 B+24 \sin B \cdot \cos A+9 \cos ^2 A\right)=37\end{aligned}$
$\begin{aligned} \Rightarrow \quad 9\left(\sin ^2 A+\cos ^2 A\right)+16\left(\cos ^2 B\right. & \left.+\sin ^2 B\right) \\ & +24 \sin (A+B)=37\end{aligned}$
$\begin{aligned} & \Rightarrow \quad 9+16+24 \sin (A+B)=37 \\ & \Rightarrow \quad 24 \sin (A+B)=12 \\ & \Rightarrow \sin (A+B)=\frac{1}{2} \\ & \Rightarrow A+B=\frac{\pi}{6}, \frac{5 \pi}{6}\end{aligned}$
$\begin{aligned} & \because \quad A+B>C \\ & \therefore \quad A+B=\frac{5 \pi}{6}\end{aligned}$
Hence, $C=\pi-\frac{5 \pi}{6}=\frac{\pi}{6} \quad[\because A+B+C=\pi]$
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