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 ^@.
Answer:
- 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} ^@.
