Question

Evaluate:

The value of $\frac{3\sin 3B+2\cos (2B+5^0)}{2\cos 3B-\sin (2B-{10}^0)}$, when $B={20}^0$ is $3\sqrt{3}+2\sqrt{2}$

If true then enter $1$ and if false then enter $0$

Answer 1

$\frac{3\sin 3B+2\cos (2B+5^0)}{2\cos 3B-\sin (2B-{10}^0)}$; when $B={20}^0$
= $\frac{3\sin {60}^0+2\cos \left({45}^0\right)}{2\cos {60}^0-\sin \left({30}^0\right)}$
= $\frac{3\times \frac{\sqrt{3}}{2}+2\times \frac{1}{\sqrt{2}}}{2\times \frac{1}{2}-\frac{1}{2}}$
= $\frac{\frac{3\sqrt{3}+2\sqrt{2}}{2}}{\frac{1}{2}}$
= $3\sqrt{3}+2\sqrt{2}$