Question

If $0<A<{90}^0:0<B<{90}^0;\sin A=\frac{12}{13}$ and $\cos B=\frac{4}{5}$ Find the value of $\sin (A+B)$

• A. $\frac{63}{65}$
• B. $\frac{33}{65}$
• C. $\frac{12}{65}$
• D. None of these

Answer 1

The correct option is A $\frac{63}{65}$

$\sin A=\frac{12}{13}so\cos A=\sqrt{1-{\sin }^{2}A}$
$=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$
$\cos B=\frac{4}{5}so\sin B=\sqrt{1-{\cos }^{2}A}$
$=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$
We get $\sin (A+B)=\sin A\cos B+\cos A\sin B$
$=\frac{12}{13}\times \frac{4}{5}+\frac{5}{13}\times \frac{3}{5}$
$=\frac{63}{65}$