Question

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

• A. ${30}^{\circ }$
• B. ${150}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$

$3\sin A+4\cos B=6\cdots \left(1\right)$
$4\sin B+3\cos A=1\cdots \left(2\right)$
On squaring and adding $\left(1\right)\&\left(2\right)$, we get
$9+16+24(\sin A\cos B+\cos A\sin B)=37\Rightarrow \sin (A+B)=\frac{1}{2}\Rightarrow \sin (\pi -C)=\frac{1}{2}\Rightarrow C={150}^{\circ }\text{or}{30}^{\circ }$

When $C={150}^{\circ }$
For $A={30}^{\circ },B=0^{\circ }$,
$3\sin A+4\cos B=\frac{3}{2}+4<6$
So, for $A<{30}^{\circ },B>0^{\circ }$ with $A+B={30}^{\circ }$,
$3\sin A+4\cos B<\frac{3}{2}+4<6$

$\therefore C={30}^{\circ }$