Question

If the value of $\frac{3\sin 3B+2\cos (2B+5^{\circ })}{2\cos 3B-\sin (2B-{10}^{\circ })}$; when $B=$ ${20}^{\circ }$ is $3\sqrt{m}+2\sqrt{2}$, then find the value of $m$.

Answer 1

$\frac{3\sin 3B+2\cos (2B+5^{\circ })}{2\cos 3B-\sin (2B-{10}^{\circ })}$; when B = ${20}^{\circ }$
when, $B={20}^{\circ }$
$\frac{3\sin {60}^{\circ }+2\cos \left({45}^{\circ }\right)}{2\cos {60}^{\circ }-\sin \left({30}^{\circ }\right)}$

$=\frac{3\left(\frac{\sqrt{3}}{2}\right)+2\left(\frac{1}{\sqrt{2}}\right)}{2\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)}$

$=\frac{\left(\frac{3\sqrt{3}}{2}\right)+\left(2\frac{\sqrt{2}}{2}\right)}{\left(\frac{2}{2}\right)-\left(\frac{1}{2}\right)}$

$=3\sqrt{3}+2\sqrt{2}$