Question

If $\cos \alpha =\frac{12}{13}$ and $\sin \beta =\frac{4}{5}$ then find $\sin (\alpha +\beta ).$

• A. $\frac{63}{65}$
• B. $\frac{58}{65}$
• C. $\frac{47}{65}$
• D. $\frac{39}{65}$

Answer 1

Answer: (A)

$\frac{63}{65}$

If $\cos \alpha =\frac{12}{13}$
Then, adjacent side of the right angled triangle containing angle $\alpha =12$ and hypotenuse $=15$
Using pythagoras theorem, we get the opposite side $=5$

Now we have, $\sin \alpha =\frac{Opposite}{Hypotenuse}=\frac{5}{13}$

If $\sin \beta =\frac{4}{5}$
Then, opposite side of the right angled triangle containing angle $\beta =4$ and hypotenuse $=5$
Using pythagoras theorem, we get the adjacent side $=3$

Now we have, $\cos \beta =\frac{Adjacent}{Hypotenuse}=\frac{3}{5}$

We know that $\sin (A+B)=\sin A\cos B+\cos A\sin B$
$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta =\frac{5}{13}\times \frac{3}{5}+\frac{12}{13}\times \frac{4}{5}=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}$