The shadow of a tower standing on a level pla | vedclass

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(C) Let the height of the tower be $h$ and the length of the shadow when the Sun's elevation is $60^{\circ}$ be $x \, m$.Let the tower be $SQ$,where $

Vedclass · Accepted Answer

$25 \sqrt{3}$

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(C) Let the height of the tower be $h$ and the length of the shadow when the Sun's elevation is $60^{\circ}$ be $x \, m$.Let the tower be $SQ$,where $S$ is the top and $Q$ is the base.When the elevation is $60^{\circ}$,the shadow length is $RQ = x$.In $	riangle S R Q$,$	an 60^{\circ} = \frac{SQ}{RQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}$ .......$(i)$When the elevation is $30^{\circ}$,the shadow length is $PQ = PR + RQ = 50 + x$.In $	riangle S P Q$,$	an 30^{\circ} = \frac{SQ}{PQ} = \frac{h}{50 + x}$.$\frac{1}{\sqrt{3}} = \frac{h}{50 + x} \Rightarrow 50 + x = h\sqrt{3}$.Substituting $x = \frac{h}{\sqrt{3}}$ from $(i)$:$50 + \frac{h}{\sqrt{3}} = h\sqrt{3} \Rightarrow 50 = h\sqrt{3} - \frac{h}{\sqrt{3}}$.$50 = h \left( \frac{3 - 1}{\sqrt{3}} ight) \Rightarrow 50 = h \left( \frac{2}{\sqrt{3}} ight)$.$h = \frac{50 \sqrt{3}}{2} = 25 \sqrt{3} \, m$.Thus,the height of the tower is $25 \sqrt{3} \, m$.

The shadow of a tower standing on a level plane is found to be $50 \, m$ longer when the Sun's elevation is $30^{\circ}$ than when it is $60^{\circ}$. Find the height of the tower (in $m$).

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