Question

If $\sin (\alpha -\beta )=\frac{1}{2}$and $\cos (\alpha +\beta )=\frac{1}{2}$, where $\alpha$ and $\beta$ are positive acute angles, then

• A. $\alpha =45°,\beta =15°$
• B. $\alpha =15°,\beta =45°$
• C. $\alpha =60°,\beta =15°$
• D. None of these

Answer 1

The correct option is A

$\alpha =45°,\beta =15°$

Explanation for the correct option.

Given that, $\sin (\alpha -\beta )=\frac{1}{2}$and $\cos (\alpha +\beta )=\frac{1}{2}$.

Also, $\alpha ,\beta$ are positive acute angles.

$\sin \left(\alpha -\beta \right)=\sin {30}^{o}and\cos \left(\alpha +\beta \right)=\cos {60}^o$

Comparing arguments we get:

$$\begin{gathered} \alpha -\beta ={30}^o...\left(i\right) \\ \alpha +\beta ={60}^o...\left(ii\right) \end{gathered}$$
Add equation (i) and (ii),
$$\begin{gathered} 2\alpha =90° \\ \Rightarrow \alpha =45° \end{gathered}$$
Substituting $\alpha =45°$ in (ii),
$\begin{array}{rcl}45°+\beta & =& 60°\\ & \Rightarrow & \beta =15°\end{array}$
Hence, option A is correct.