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

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(C) Let the height of the tower be $h \ m$. Let $AB$ be the shadow when the sun's altitude is $60^{\circ}$ and $AC$ be the shadow when the sun's altit

Vedclass · Accepted Answer

$25 \sqrt{3}$

Vedclass · Answer

(C) Let the height of the tower be $h \ m$. Let $AB$ be the shadow when the sun's altitude is $60^{\circ}$ and $AC$ be the shadow when the sun's altitude is $30^{\circ}$.Given,$BC = 50 \ m$.In $	riangle TAB$,$	an 60^{\circ} = \frac{h}{AB} \Rightarrow \sqrt{3} = \frac{h}{AB} \Rightarrow AB = \frac{h}{\sqrt{3}}$.In $	riangle TAC$,$	an 30^{\circ} = \frac{h}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + 50} \Rightarrow AB + 50 = h\sqrt{3}$.Substituting $AB = \frac{h}{\sqrt{3}}$ in the second equation:$\frac{h}{\sqrt{3}} + 50 = h\sqrt{3}$$50 = h\sqrt{3} - \frac{h}{\sqrt{3}}$$50 = h \left( \frac{3 - 1}{\sqrt{3}} ight) = h \left( \frac{2}{\sqrt{3}} ight)$$h = \frac{50 \sqrt{3}}{2} = 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 altitude is $30^{\circ}$ than when it is $60^{\circ} .$ Find the height of the tower. (In $m$)

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