Question

A window of a house is h metre above the ground . From the window , the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be $\alpha$ and $\beta$ respectively. Prove that the height of the house is $h\left(1+\tan \alpha \tan \beta \right)$ metres.

Answer 1

DISCLAIMER: The question given in the book is incorrect. We are proving the height of the house to be $h\left(1+\tan \alpha cot\beta \right)$

Let the window be at point A.
To prove: BD = $h\left(1+\tan \alpha cot\beta \right)$
In ∆ABC,
$$\begin{gathered} \tan \alpha =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{BC}{x} \\ \Rightarrow x=BC\tan \alpha .....\left(\mathrm{i}\right) \end{gathered}$$
In ∆ACD,
$$\begin{gathered} \tan \beta =\frac{CD}{AC}=\frac{h}{x} \\ \Rightarrow x=\frac{h}{\tan \beta }.....\left(\mathrm{ii}\right) \end{gathered}$$
From (i) and (ii)
$$\begin{gathered} BC\tan \alpha =\frac{h}{\tan \beta } \\ \Rightarrow BC=\frac{h\tan \alpha }{\tan \beta } \end{gathered}$$
Height of house = BC + CD
$$\begin{gathered} =\frac{h\tan \alpha }{\tan \beta }+h \\ =h\left[\frac{\tan \alpha }{\tan \beta }+1\right] \\ =h\left[\tan \alpha cot\beta +1\right] \end{gathered}$$