Nepal
The angle of elevation of the top of a tower as observed from a point on the ground is ^@ \alpha ^@ and on moving ^@ a \space meters ^@ towards the tower, the angle of elevation is ^@ \beta ^@. Prove that the height of the tower is ^@ \dfrac { a \tan \alpha \tan \beta } { \tan \beta - \tan \alpha }. ^@
Answer:
- The situation given in the question is represented by the image given below.
Let ^@ AB ^@ be a tower of height ^@h^@. - In the right-angled triangle ^@ABC^@, we have @^ \begin{aligned} & \cot \beta = \dfrac { BC } { BA } \\ \implies & \cot \beta = \dfrac { x } { h } \\ \implies & x = h \cot \beta = \dfrac { h } { \tan \beta } && \ldots \text{(i)} \end{aligned} @^
- In the right-angled triangle ^@ABD^@, we have @^ \begin{aligned} & \cot \alpha = \dfrac { BD } { BA } \\ \implies & \cot \alpha = \dfrac { x + a } { h } \\ \implies & x + a = h \cot \alpha = \dfrac { h } { \tan \alpha } \\ \implies & h = (x + a) \tan \alpha && \ldots \text{(ii)} \end{aligned} @^
- Now, let us substitute the value of ^@ x ^@ in ^@ eq \space \text{(ii)} ^@. @^ \begin{aligned} & h = \bigg(\dfrac { h } { \tan \beta } + a\bigg) \tan \alpha \\ \implies & h = \dfrac { h \tan \alpha } { \tan \beta } + a \tan \alpha \\ \implies & h \tan \beta = h \tan \alpha + a \tan \alpha \tan \beta \\ \implies & h (\tan \beta - \tan \alpha) = a \tan \alpha \tan \beta \\ \implies & h = \dfrac { a \tan \alpha \tan \beta } { \tan \beta - \tan \alpha } \end{aligned} @^
- Thus, the height of the tower is ^@ \dfrac { a \tan \alpha \tan \beta } { \tan \beta - \tan \alpha } \space meters. ^@
