Question

The angle of elevation of a cliff from fixed point is $\theta$. After going up a distance of $k$ meters towards the top of the cliff at are angle of $\varphi$, it is found that the angle of elevations is $\alpha$. Show that height of a cliff is $\frac{k(\cos \varphi -\sin \varphi \cot \alpha )}{\cot \theta -\cot \alpha }$

Answer 1

The pictorial representation of the given data is represented by the figure

In $\Delta DEF,\frac{DE}{DF}=\sin \varphi ,\frac{EF}{DF}=\cos \varphi$

$\Rightarrow DE=k\sin \varphi ,EF=k\cos \varphi$ $[DF=kkm]$

Let $AB=x$

$\therefore AC=AB-BC=AB-DE=x-k\sin \varphi$

In $\Delta ACD,\frac{AC}{CD}=\tan \alpha$

$\Rightarrow \frac{x-k\sin \varphi }{CD}=\tan \alpha$

$\Rightarrow CD=(x-k\sin \varphi )\cot \alpha$

$BF=EF+BE=EF+CD=k\cos \varphi +(x-k\sin \varphi )\cot \alpha$

In $\Delta ABF,\frac{AB}{BF}=\tan \theta$

$\Rightarrow \frac{x}{k\cos \varphi +(x-k\sin \varphi )\cot \alpha }=\tan \theta$

$\Rightarrow x\cot \theta =k\cos \varphi +x\cot \alpha -k\sin \varphi \cot \alpha$

$\Rightarrow (\cot \theta -\cot \alpha )x=k(\cos \varphi -\sin \varphi \cot \alpha )$

$\Rightarrow x=\frac{k(\cos \varphi -\sin \varphi \cot \alpha )}{\cot \theta -\cot \alpha }$

Thus the height of the cliff is $\frac{k(\cos \varphi -\sin \varphi \cot \alpha )}{\cot \theta -\cot \alpha }$