The shadow of a tower standing on a level ground is found to be $40 m$ longer when the altitude is $30^{o}$ . When it is $60^{o}$ degree, find the height of the tower.

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Solution

Given $S_{2} = S_{1} + 40$
$S_{1} = \frac{H}{tan 60 °}$ and $S_{2} = \frac{H}{tan 30 °}$
$= \frac{H}{\sqrt{3}} = \sqrt{3} H$
$\sqrt{3} H = \frac{H}{\sqrt{3}} + 40$
$H (\frac{2}{\sqrt{3}}) = 40$
$H = 20 \sqrt{3} m$

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The shadow of a tower standing on a level ground is found to be 40 m longer when the altitude is 30o. When it is 60o degree, find the height of the tower.

Given S2=S1+40 S1=Htan60° and S2=Htan30°=H√3=√3H√3H=H√3+40H(2√3)=40H=20√3m

The shadow of a tower standing on a level ground is found to be $40 m$ longer when the altitude is $30^{o}$ . When it is $60^{o}$ degree, find the height of the tower.

Given $S_{2} = S_{1} + 40$
$S_{1} = \frac{H}{tan 60 °}$ and $S_{2} = \frac{H}{tan 30 °}$
$= \frac{H}{\sqrt{3}} = \sqrt{3} H$
$\sqrt{3} H = \frac{H}{\sqrt{3}} + 40$
$H (\frac{2}{\sqrt{3}}) = 40$
$H = 20 \sqrt{3} m$