Question

The angle of elevation of the top a cliff from a fixed point $A$ is $\theta$. After going up a distance of $K$ meters towards the top of the cliff at an angle of $\varphi$. It is found that the angle of elevation is $\alpha$. Show that the height of the cliff in meters is $K$
$\frac{K\cos \varphi -\sin \varphi .\cot \alpha }{\cot \theta -\cos \alpha }$

Answer 1

Let $CD$ is a hill. An angle of elevation at point $A$ and $B$ are $\varphi ,\theta$ and $\alpha$ respectively.

$AB=K$ m

$\angle BAM=\varphi$

$\angle DAC=\theta$ and $\angle DBN=\alpha$

Let $CD=h$ m

From right angled $\Delta AMB$

$\cos \varphi =\frac{AM}{AB}$

$AM=AB\cos \varphi =K\cos \varphi$ …(i)

and $\sin \varphi =\frac{BM}{AB}$

$BM=AB\sin \varphi =K\sin \varphi$ …(ii)

From right angled $\Delta ACD$,

$\cos \theta =\frac{AC}{DC}$

$AC=DC\cot \theta =h\cot \theta$ …(iii)

$\because MC=AC–AM$

From eq. (i) and (iii),

$MC=h\cot \theta –K\cos \varphi$

$\because MC=BN$

So, $BN=h\cot \theta –K\cos \varphi$ …..(iv)

From right angled $\Delta BND$,

$\tan \alpha =\frac{DN}{BN}$

$DN=BN\tan \alpha$

$MC\tan \alpha$

$=(h\cot \theta –K\cos \varphi )\tan \alpha$

$h=DN+NC$

$h=h\cot \theta \tan \alpha –K\cos \varphi \tan \alpha +K\sin \varphi$

$h[1–\cot \theta \tan \alpha ]=K[\sin \varphi –\cos \varphi \tan \alpha ]$

$h[1-\frac{\cot \theta }{\cot \alpha }]=K[\sin \varphi -\cos \varphi .\frac{1}{\cot \alpha }]$

$h\left[\frac{\cot \alpha -\cot \theta }{\cot \alpha }\right]=K\left[\frac{\sin \varphi \cot \alpha -\cos \varphi }{\cot \alpha }\right]$

$h=\frac{K(\sin \varphi \cot \alpha -\cos \varphi )}{\cot \alpha -\cot \theta }$

$h=\frac{K(\cos \varphi -\sin \varphi \cot \alpha )}{\cot \theta -\cot \alpha }$