1. **Problem statement:** We have a tower with base point $P$ on horizontal ground. Points $A$ and $B$ lie on the same straight line with $P$, and the angles of elevation from $A$ and $B$ to the top of the tower are complementary (sum to $90^\circ$). Given $PA = a$ and $PB = b$, we need to show the height $h$ of the tower is $\sqrt{ab}$. 2. **Setup and notation:** Let the height of the tower be $h$. The angles of elevation from $A$ and $B$ are $\theta$ and $90^\circ - \theta$ respectively, since they are complementary. 3. **Using tangent for angles of elevation:** From point $A$, $\tan \theta = \frac{h}{a}$. From point $B$, $\tan(90^\circ - \theta) = \frac{h}{b}$. 4. **Recall complementary angle identity:** $\tan(90^\circ - \theta) = \cot \theta = \frac{1}{\tan \theta}$. 5. **Substitute and relate:** $$\frac{h}{b} = \frac{1}{\tan \theta} = \frac{1}{\frac{h}{a}} = \frac{a}{h}$$ 6. **Cross multiply:** $$h^2 = ab$$ 7. **Solve for height:** $$h = \sqrt{ab}$$ **Final answer:** The height of the tower is $\sqrt{ab}$.