# The angle of elevation of a cliff from a fixed point A is theta. After going up a distance of 'k' m towards the top of the cliff, at an angle of phi, it is found that the angle of elevation is alpha. Find height of cliff in terms of k?

Nov 20, 2016

Let CB be the cliff . From point A the angle of elevation of the peak C of the cliff is $\angle C A B = \theta$'. After going up a distance , $A D = k$ $m$ towards the top of the cliff at an angle,$\angle D A E = \phi$,it ls found the angle of elevation $\angle C D F = \alpha$.

DF and DE are perpendiculars drawn from D on CB and AB.

Now $D E = k \sin \phi \mathmr{and} A E = k \cos \phi$

Let
h="CB the height of the cliff" and b=BA

For $\Delta C A B , \text{ } \frac{C B}{B A} = \tan \theta$

$\implies \frac{b}{h} = \cot \theta \implies b = h \cot \theta$

Now $D F = B E = B A - A E = b - k \cos \phi$

$C F = C B - F B = C B - D E = h - k \sin \phi$

For $\Delta C D F , \text{ } \frac{C F}{D F} = \tan \alpha$

$\implies \frac{h - k \sin \phi}{b - k \cos \phi} = \tan \alpha$

$\implies \frac{h - k \sin \phi}{h \cot \theta - k \cos \phi} = \tan \alpha$

$\implies h - h \tan \alpha \cot \theta = k \sin \phi - k \cos \phi \tan \alpha$

$\implies h = \frac{k \left(\sin \phi - \cos \phi \tan \alpha\right)}{1 - \tan \alpha \cot \theta}$

$\implies h = \frac{k \sin \theta \left(\sin \phi \cos \alpha - \cos \phi \sin \alpha\right)}{\sin \theta \cos \alpha - \cos \theta \sin \alpha}$

$\implies h = \frac{k \sin \theta \sin \left(\phi - \alpha\right)}{\sin \left(\theta - \alpha\right)}$

$\implies h = \frac{k \sin \theta \sin \left(\alpha - \phi\right)}{\sin \left(\alpha - \theta\right)}$