### 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 $\dfrac { a \tan \alpha \tan \beta } { \tan \beta - \tan \alpha }.$

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 $\dfrac { a \tan \alpha \tan \beta } { \tan \beta - \tan \alpha } \space meters.$