Evaluate :

$\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}}$; when B = 20°.

$\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}}\\[1em] = \dfrac{\text{3 sin (3 x 20°) + 2 cos((2 x 20°) + 5°)}}{\text{2 cos (3 x 20°) - sin((2 x 20°) - 10°)}}\\[1em] = \dfrac{\text{3 sin 60° + 2 cos(40° + 5°)}}{\text{2 cos 60° - sin(40° - 10°)}}\\[1em] = \dfrac{\text{3 sin 60° + 2 cos 45°}}{\text{2 cos 60° - sin 30°}}\\[1em] = \dfrac{3 \times \dfrac{\sqrt3}{2} + 2 \times \dfrac{1}{\sqrt2}}{2 \times \dfrac{1}{2} - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2}{\sqrt2}}{\dfrac{2}{2} - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2 \times \sqrt2}{\sqrt2 \times \sqrt2}}{\dfrac{2 - 1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2\sqrt2}{2}}{\dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3 + 2\sqrt2}{2}}{\dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3 + 2\sqrt2}{\cancel2}}{\dfrac{1}{\cancel2}}\\[1em] = 3\sqrt3 + 2\sqrt2$

Hence, $\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}} = 3\sqrt3 + 2\sqrt2.$