Question

In $\Delta$ ABC, $3\sin A+4\cos B=6and3\cos A+4\sin B=1$ then,the measure of an angle c in degree is

• A. $30$
• B. $60$
• C. $150$
• D. $30$ or $150$

Answer 1

The correct option is A $30$

In $△ABC,$

$3\sin A+4\cos B=6⟹(3\sin A+4\cos B{)}^2=6^2⟹9{\sin }^{2}A+16{\cos }^{2}B+24\sin A\cos B=36\cdots (1)$

$3\cos A+4\sin B=1⟹(3\cos A+4\sin B{)}^2=1⟹9{\cos }^{2}A+16{\sin }^{2}B+24\cos A\sin B=1\cdots (2)$

$\left(1\right)+\left(2\right)⟹9({\sin }^{2}A+{\cos }^{2}A)+16({\cos }^{2}B+{\sin }^{2}B)+24(\sin A\cos B+\cos A\sin B)=37$

$⟹9+16+24\sin (A+B)=37$

$⟹24\sin ({180}^{\circ }-C)=12$

$⟹\sin C=\frac{1}{2}$

$⟹C={30}^{\circ }$