Question

If $\beta$ is acute and $\sin \beta =\frac{4}{5}$, the value of
$\frac{\sin \alpha (3\cos \beta +4\sin \beta )+\cos \alpha (3\sin \beta -4\cos \beta )}{\sqrt{3}\sin \alpha }$ is equal to

• A. $\frac{5}{\sqrt{3}}$
• B. $\frac{\sqrt{3}(7+24\cot \alpha )}{15}$
• C. $\frac{5(\sqrt{3}+8\tan \alpha )}{17}$
• D. $\frac{125}{\sqrt{2}}$

Answer 1

The correct option is A $\frac{5}{\sqrt{3}}$

Given $\beta$ is acute angle

$\sin \beta =\frac{4}{5}⟹\cos \beta =\frac{3}{5}$

$\frac{\sin \alpha (3\cos \beta +4\sin \beta )+\cos \alpha (3\sin \beta -4\cos \beta )}{\sqrt{3}\sin \alpha }=\frac{\sin \alpha (3\times \frac{3}{5}+4\times \frac{4}{5})+\cos \alpha (3\times \frac{4}{5}-4\times \frac{3}{5})}{\sqrt{3}\sin \alpha }$

$=\frac{\sin \alpha \left(\frac{9+16}{5}\right)+\cos \alpha \left(\frac{12-12}{5}\right)}{\sqrt{3}\sin \alpha }$

$=\frac{\sin \alpha \left(5\right)+\cos \alpha \left(0\right)}{\sqrt{3}\sin \alpha }$

$=\frac{5}{\sqrt{3}}$