The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object, the angle of elevation is found to be $\beta$. Then the height (in metres) of the object is

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Aicharya · Accepted Answer

Let the height of the object is $h$.

In $\triangle A B C$, $\begin{aligned} & \tan \alpha=\frac{h}{x+d} \\ & \Rightarrow \quad x+d=h \cot \alpha \\ & \end{aligned}$ And in $\triangle A B D$, $\begin{aligned} \tan \beta & =\frac{h}{x} \\ \Rightarrow \quad x & =h \cot \beta\end{aligned}$ On putting this value in Eq. (i), we get $\begin{array}{llll} h \cot \beta+d =h \cot \alpha \\ \Rightarrow h(\cot \alpha-\cot \beta) =d \\ \Rightarrow h =\frac{d}{\cot \alpha-\cot \beta}\end{array}$

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on…

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