A man is standing on the top of a building 60 m high. the observe that the angle of depression of the top and the bottom of a tower has measures 30 and 60 Rs find the height of the tower.

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Solution

Man is at $60$ m high on building angle of depression of top $= 30$
bottom $= 60$
In $Δ$ ABC
$tan 60 = \sqrt{3} = \frac{B A}{A C}$
$A C = \frac{B A}{\sqrt{3}} = \frac{60}{\sqrt{3}} \frac{\sqrt{3}}{\sqrt{3}}$
$= 20 \sqrt{3}$
In $Δ$ BDE
$tan 30 = \frac{1}{\sqrt{3}} = \frac{D E}{B E}$
$\Rightarrow D E = \frac{B E}{\sqrt{3}} = \frac{20 \sqrt{3}}{\sqrt{3}} = 20$ m
Height of tower $=$ DE $+$ EC $= 80$ m.

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A man is standing on the top of a building 60 m high. the observe that the angle of depression of the top and the bottom of a tower has measures 30 and 60 Rs find the height of the tower.

Man is at 60m high on building angle of depression of top =30bottom =60In ΔABCtan60=√3=BAACAC=BA√3=60√3√3√3=20√3In ΔBDEtan30=1√3=DEBE⇒DE=BE√3=20√3√3=20mHeight of tower=DE+EC=80m.