Question

If ${270}^{\circ }<A<{360}^{\circ },{90}^{\circ }<B<{180}^{\circ },\cos A=\frac{5}{13},\tan B=-\frac{15}{8}$, then $\sin (A+B)=$

• A. $\frac{140}{221}$
• B. $\frac{171}{221}$
• C. $\frac{140}{171}$
• D. $\frac{221}{171}$

Answer 1

Answer: (B)

$\frac{171}{221}$

We have,

$\cos A=\frac{5}{13}$

Then,

$\sin A=\frac{12}{13}$

Since,

$\tan B=-\frac{15}{8}$

Then,

$\cos B=\frac{8}{17},\sin B=\frac{15}{17}$

Therefore,

$\sin (A+B)=\sin A.\cos B+\sin B.\cos A$

$\sin (A+B)=\frac{12}{13}.\frac{8}{17}+\frac{15}{17}.\frac{5}{13}$

$\sin (A+B)=\frac{96}{221}+\frac{75}{221}$

$\sin (A+B)=\frac{171}{221}$

Hence, this is the answer.