For example, the lateral edges of the pyramid $P-ABC$ form a $75^{\circ}$ angle with the base, the internal angles of $\triangle ABC$ are in the ratio $\angle A: \angle B: \angle C=1: 2: 9$, and $AC=3$. Find the height of the pyramid.
Let $P O \perp$ base $A B C$, with $O$ as the foot of the perpendicular, then $O$ is the circumcenter of $\triangle A B C$. Given that
$\angle A: \angle B: \angle C=1: 2: 9, \angle B=30^{\circ}$, $\angle C=135^{\circ}$, it is evident that $O$ is outside $\triangle A B C$, on $O C$, then $O C=\frac{3}{2 \sin 30^{\circ}}=3$. Also, $\angle P C O=75^{\circ}$, hence $P O$ $=O C \operatorname{tg} 75=3(2+\sqrt{3})$ (length units). (Author's unit: Tianjin Experimental High School)