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

Question

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

Aicharya · Accepted Answer

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

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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