Question

In triangle $ABC$, we are given that $3\sin A+4\cos B=6$ and $4\sin B+3\cos A=1$. Then the measure of the angle $C$ is.

• A. ${30}^{\circ }$
• B. ${150}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

Answer: (B)

${150}^{\circ }$

$3\sin A+4\cos B=6....\left(i\right)4\sin B+3\cos A=1....\left(ii\right)$

squaring and adding $\left(i\right)$ and $\left(ii\right)$

$(9{\sin }^{2}A+16{\cos }^{2}B+24\sin A\cos B)+(16{\sin }^{2}B+9{\cos }^{2}A+24\sin B\cos A)=379({\sin }^{2}A+{\cos }^{2}A)+16({\sin }^{2}B+{\cos }^{2}B)+24(\sin A\cos B+\sin B\cos A)=3725+24\sin (A+B)=3724\sin (A+B)=12\sin (A+B)=\frac{12}{24}=\frac{1}{2}$

As $A+B+C=\pi$

$A+B=\pi -C\sin (\pi -C)=\frac{1}{2}\pi -C=\frac{\pi }{6}\Rightarrow C=\frac{5\pi }{6}={150}^o$

Hence, optiom $B$ is correct.