### The angle of elevation of the top of a tower from the two points $P$ and $Q$ at distances of $a$ and $b$ respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is $\sqrt{ab}$ where $a > b$.

Step by Step Explanation:
1. Let $AB$ be the tower of height $h$ and $\angle APB$ be $\theta.$
As $\angle APB$ and $\angle AQB$ are complementary angles, $\angle AQB = 90^\circ - \theta.$

The image below represents the given situation.
2. Now, from right-angled triangle $APB$, we have \begin{aligned} & \tan \theta = \dfrac { AB } { PB } \\ \implies & \tan \theta = \dfrac { h } { a } \\ \implies & h = a \tan \theta && \ldots \text{(i)} \end{aligned}
3. Now, from right-angled triangle $AQB$, we have \begin{aligned} & \tan(90^\circ - \theta) = \dfrac { AB } { BQ } \\ \implies & \cot \theta = \dfrac { h } { b } && [\tan(90^\circ - \theta) = \cot \theta] \\ \implies & h = b \space \cot \theta = \dfrac{ b } { \tan \theta } && \ldots \text{(ii)} \end{aligned}
4. On multiplying $eq \space \text{(i)}$ and $eq \space \text{(ii)}$, we get \begin{aligned} & h^2 = ab \\ & h = \sqrt{ab} \end{aligned}
5. Therefore, the height of the tower is $\sqrt{ab}$.