Question

$\huge{\underline{\underline{\bf Question}}}$

the angle of elevation of a cliff from a fixed point is $\sf \theta$ after going up a distance of ke metres towards the top of the Cliff at an angle of it is found that the angle of elevation is alpha show that the height of the Cliff :-
$\sf k \dfrac{( \cos(ϕ) - \sin(ϕ)) }{ \cot( \theta) - \cot( \alpha ) }$
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Answer 1

Appropriate Question :-

The angle of elevation of a cliff from a fixed point is θ, after going up a distance of k metres towards the top of the Cliff at an angle phi, it is found that the angle of elevation is alpha. Show that the height of the Cliff is

$\sf \: \dfrac{k(cos \phi \: - \: sin \phi \: cot \alpha)}{cot \theta \: - \: cot \alpha} \: metre$

$\large\underline{\sf{Solution-}}$

Let assume that AB represents the height of the cliff such that AB = h metres.

Let assume that O be the fixed point.

So,

$\rm :\longmapsto\:\angle AOB = \theta \:$

$\rm :\longmapsto\:\angle AOC = \phi \:$

$\rm :\longmapsto\:OC = k \: metres$

Construction :- Let from C, drop CD and CE perpendicular on AB and OA respectively.

Further, given that

$\rm :\longmapsto\:\angle DCB = \alpha$

Now, In right triangle DCB

$\rm :\longmapsto\: \cot\alpha = \dfrac{DC}{DB}$

$\rm :\longmapsto\: \cot\alpha = \dfrac{AE}{AB - DA}$

$\rm :\longmapsto\: \boxed{\tt{ \cot\alpha = \dfrac{AO - OE}{h - CE}}} - - - (1)$

Now, In right triangle OCE

$\rm :\longmapsto\:sin \phi \: = \dfrac{CE}{OC}$

$\rm :\longmapsto\:sin \phi \: = \dfrac{CE}{k}$

$\rm \implies\:CE = k \: sin \phi \: - - - (2)$

Also,

$\rm :\longmapsto \: cos \phi \: = \: \dfrac{OE}{OC}$

$\rm :\longmapsto \: cos \phi \: = \: \dfrac{OE}{k}$

$\rm \implies\:OE \: = \: k \: cos \phi - - - - (3)$

Now, In right triangle OAB

$\rm :\longmapsto\:cot \theta \: = \: \dfrac{AO}{AB}$

$\rm :\longmapsto\:cot \theta \: = \: \dfrac{AO}{h}$

$\rm :\longmapsto\: \: AO = \: \: h \: cot \theta \: - - - - (4)$

On substituting the values from equation (2), (3), and (4) in equation (1), we get

$\rm :\longmapsto\: \cot\alpha = \dfrac{ h \: cot \theta- k \: cos \phi}{h - k \: sin \phi}$

$\rm :\longmapsto\:h \: cot \alpha - k \: sin \phi \: cot \alpha =h \: cot \theta - k \: cos \phi$

$\rm :\longmapsto\: k \: cos \phi - k \: sin \phi \: cot \alpha =h \: cot \theta - \: h \: cot \alpha$

$\rm :\longmapsto\: k \:( cos \phi \: - \: sin \phi \: cot \alpha ) \: =h \: (cot \theta - \: cot \alpha )$

$\bf\implies \:h = \dfrac{k(cos \phi \: - \: sin \phi \: cot \alpha)}{cot \theta \: - \: cot \alpha} \: metre$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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