Question

A building stands under a bridge of length $l$, spanning over the velly. If $\alpha$ and $\beta$ are the measure of the angles of depression of the top of the building , as observed from the two ends of the bridge, prove that the bridge is at the height $\frac{l}{\cot \alpha +\cot \beta }$ above the top of the building.

Answer 1

$\begin{array}{c}In\Delta AOP\\ \Rightarrow \frac{\lambda }{AO}=\tan \alpha \\ AO=\frac{\lambda }{\tan \alpha }\to \left(i\right)\\ andIn\Delta POB\\ \Rightarrow \frac{\lambda }{OB}=\tan \beta \\ \Rightarrow OB=\frac{\lambda }{\tan \beta }\to \left(ii\right)\end{array}$

Adding equation (i) and (ii),

$\begin{array}{c}\Rightarrow OA+OB=\frac{\lambda }{\tan \beta }+\frac{\lambda }{\tan \alpha }\\ \Rightarrow \ell =\lambda \cot \beta =\lambda \cot \alpha \\ \Rightarrow \ell =\lambda \left(\cot \beta +\cot \alpha \right)\\ \therefore \lambda =\frac{\ell }{\left(\cot \beta +\cot \alpha \right)}.\end{array}$

Hence, this is the answer.