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:


Step by Step Explanation:
  1. The situation given in the question is represented by the image given below.
    D B C A h (Tower) a x α β
    Let ^@ AB ^@ be a tower of height ^@h^@.
  2. 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} @^
  3. 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} @^
  4. 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} @^
  5. Thus, the height of the tower is ^@ \dfrac { a \tan \alpha \tan \beta } { \tan \beta - \tan \alpha } \space meters. ^@

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