Question

If an acute angled $△ABC,\sin A=\frac{4}{5}$ and $\sin B=\frac{12}{13}$, then $\sin C=$

• A. $\frac{38}{65}$
• B. $\frac{56}{65}$
• C. $\frac{33}{65}$
• D. $\frac{16}{65}$

Answer 1

The correct option is B $\frac{56}{65}$

$\sin A=\frac{4}{5}and\sin B=\frac{12}{13}\sin C=(\pi -(A+B\left)\right)=\sin (A+B)\sin A\cos B+\sin B\cos A=\frac{4}{5}\times \frac{5}{13}\times \frac{12}{13}\times \frac{3}{5}[\cos B=\frac{\sqrt{{13}^2{-12}^2}}{13}=\frac{\sqrt{25}}{13}and\cos A=\frac{\sqrt{5^2-4^2}}{5}=\frac{\sqrt{9}}{5}]=\frac{20+36}{65}=\frac{56}{65}$

Answer B