# 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?

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Dec 14, 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 $\Delta C D F , \text{ } \frac{C F}{D F} = \tan \alpha$

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

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

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

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

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

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

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