# The angle of elevation of a cliff from a point A on the ground and from point B ,50m vertically above point are alpha and beta respectively .The height of the cliff in m is?

let the height of the cliff be PQ = h m
the angle of elevation from point A is $\angle PAQ=\alpha$
the angle of elevation from point B is
let the point A is at distance of x m from the foot of the cliff .
in the right angled triangle PAQ;

now in the right angled triangle PBC;
$\mathrm{tan}\beta =\frac{PC}{BC}=\frac{PQ-CQ}{AQ}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\beta =\frac{h-50}{x}\phantom{\rule{0ex}{0ex}}h-50=x\mathrm{tan}\beta \phantom{\rule{0ex}{0ex}}h=50+x\mathrm{tan}\beta \phantom{\rule{0ex}{0ex}}h=50+h.cot\alpha .\mathrm{tan}\beta \phantom{\rule{0ex}{0ex}}h\left(1-cot\alpha .\mathrm{tan}\beta \right)=50\phantom{\rule{0ex}{0ex}}h.\left[1-\frac{1}{\mathrm{tan}\alpha }.\mathrm{tan}\beta \right]=50\phantom{\rule{0ex}{0ex}}h.\left[\frac{\mathrm{tan}\alpha -\mathrm{tan}\beta }{\mathrm{tan}\alpha }\right]=50\phantom{\rule{0ex}{0ex}}h=\frac{50.\mathrm{tan}\alpha }{\mathrm{tan}\alpha -\mathrm{tan}\beta }$

hope this helps you

• 2
50tan(beta)cot(alpha)+30
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