The angle of elevation of the top of a tower from the two points | Math Question and Answer

Answer:

Step by Step Explanation:

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

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