Height problems

Height is adjacent to right angle. So line $AB = \frac {h}{tan 30^\circ}$ and line $AD = \frac {h}{tan28^\circ}$ . Using Pythagorean theorem on the base triangle,

$\Bigg(\frac{h}{tan28^\circ}\Bigg)^2 = \Bigg(\frac {h}{tan30^\circ}\Bigg)^2 + 150^2$ $\Bigg(\frac{h^2}{tan^{2}28^\circ}\Bigg) = \Bigg(\frac {h^2}{tan^{2}30^\circ}\Bigg) + 150^2$ $h^2tan^230^\circ = h^2tan^228^\circ + 150tan^230^\circ tan^228^\circ$ $(tan^230^\circ - tan^228^\circ)(h^2) = 150tan^230^\circ tan^228^\circ$ $h^2 = \frac{150tan^230^\circ tan^228^\circ}{tan^230^\circ - tan^228^\circ}$ $h = \sqrt {\frac{150tan^230^\circ tan^228^\circ}{tan^230^\circ - tan^228^\circ}}$ $\boxed{h = 204.67}$