Question

In a $\Delta ABC,$ $2\sin A+3\cos B=3\sqrt{2}$ and $3\sin B+2\cos A=1.$ Then the value of angle $C$ is

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${120}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

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

$2\sin A+3\cos B=3\sqrt{2}...\left(1\right)$
$3\sin B+2\cos A=1...\left(2\right)$
Squaring eqn(1) and (2), then add
$\Rightarrow 4+9+12\sin (A+B)=19$
$\Rightarrow \sin (A+B)=\frac{1}{2}$
$\Rightarrow A+B={30}^{\circ }\text{or}{150}^{\circ }$

If $A+B={30}^{\circ }\Rightarrow B={30}^{\circ }-A$
From eqn(1)
$2\sin A+3\cos ({30}^{\circ }-A)=3\sqrt{2}$
$\Rightarrow 7\sin A+3\sqrt{3}\cos A=6\sqrt{2}...\left(3\right)$

From eqn(2)
$3\sin ({30}^{\circ }-A)+2\cos A=1$
$\Rightarrow 7\cos A-3\sqrt{3}\sin A=2...\left(4\right)$

From eqn(3) and (4), we have
$\cos A=\frac{7+9\sqrt{6}}{38}\approx 0.8<\frac{\sqrt{3}}{2}$
$\Rightarrow \cos A<\cos {30}^{\circ }$
$\Rightarrow A>{30}^{\circ }$
$\Rightarrow A+B\ne {30}^{\circ }$
$\therefore A+B={150}^{\circ }$
$\Rightarrow C={30}^{\circ }$