Question

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

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

Answer 1

The correct option is D ${30}^o$

Given
$3\sin A+4\cos B=\mathrm{6....}\left(i\right)$
$4\sin B+3\cos A=\mathrm{1....}\left(ii\right)$
By squaring Eqs. (i) and (ii) we get
$9{\sin }^{2}A+16{\cos }^{2}B+24\sin A.\cos B=\mathrm{36....}\left(iii\right)$
$16{\sin }^{2}B+9{\cos }^{2}A+24\sin B.\cos A=1....\left(iv\right)$
By adding Eqs (iii) and (iv) we get
$9({\sin }^{2}A+{\cos }^{2}A)+16({\sin }^{2}B+{\cos }^{2}A)+24(\sin A.\cos B+\cos A.\sin B)=37$
$\Rightarrow 9+16+24\sin (A+B)=37[\because {\sin }^2\theta +{\cos }^2\theta =1]$
$\Rightarrow 24\sin (A+B)=12$
$\Rightarrow \sin (A+B)=\frac{1}{2}$
$\therefore A+B={30}^{o}or{150}^o$
But, $A+B={30}^o$ is not possible
Because sum of two angles of a triangle is greater than third angle
$\therefore A+B={150}^o$
$\because$ Sum of angle of a triangle $={180}^o$
$\therefore A+B+C={180}^o$
$\Rightarrow C={180}^o-(A+B)={180}^o-{150}^o={30}^o$