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Now

Let

AO=H

CD=OB=$60$m

A’B=AB=$60 + h$ ………………..$\left( 1 \right)$

Now in triangle AOD

\[

\tan {30^0} = \dfrac{{AO}}{{OD}} = \dfrac{H}{{OD}} \\

H = \dfrac{{OD}}{{\sqrt 3 }} \\

\]

Because the value of $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$

Thus

$OD = \sqrt 3 H$

Now in triangle A’OD

$

\tan {60^0} = \dfrac{{OA'}}{{OD}} = \dfrac{{OB + BA'}}{{OD}} \\

\Rightarrow \sqrt 3 = \dfrac{{60 + 60 + H}}{{\sqrt 3 H}} = \dfrac{{120 + H}}{{\sqrt 3 H}} \\

$

Because the value of $\tan {60^0} = \sqrt 3 $

And according to equation 1

We get

A’B=AB=$60 + h$

Now

$

120 + H = 3H \\

\Rightarrow 120 = 2H \\

\Rightarrow H = 60m \\

$