The shadow of a tower standing on a level ground is found to be $40 \, m$ longer when the Sun's altitude is $30^{\circ}$ than when it is $60^{\circ} .$ Find the height of the tower.
(N/A) Let $AB$ be the tower of height $h \, m$ and $BC$ be the length of the shadow when the Sun's altitude is $60^{\circ}$. Let $BC = x \, m$.
When the Sun's altitude is $30^{\circ}$,the length of the shadow is $DB = (x + 40) \, m$.
In $\triangle ABC$,$\tan 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \dots (1)$.
In $\triangle ABD$,$\tan 30^{\circ} = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \implies h = \frac{x + 40}{\sqrt{3}} \dots (2)$.
Equating $(1)$ and $(2)$,we get $x\sqrt{3} = \frac{x + 40}{\sqrt{3}}$.
$3x = x + 40 \implies 2x = 40 \implies x = 20 \, m$.
Substituting $x = 20$ in $(1)$,$h = 20\sqrt{3} \, m$.
Thus,the height of the tower is $20\sqrt{3} \, m$ (or approximately $34.64 \, m$).
When the Sun's altitude is $30^{\circ}$,the length of the shadow is $DB = (x + 40) \, m$.
In $\triangle ABC$,$\tan 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \dots (1)$.
In $\triangle ABD$,$\tan 30^{\circ} = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \implies h = \frac{x + 40}{\sqrt{3}} \dots (2)$.
Equating $(1)$ and $(2)$,we get $x\sqrt{3} = \frac{x + 40}{\sqrt{3}}$.
$3x = x + 40 \implies 2x = 40 \implies x = 20 \, m$.
Substituting $x = 20$ in $(1)$,$h = 20\sqrt{3} \, m$.
Thus,the height of the tower is $20\sqrt{3} \, m$ (or approximately $34.64 \, m$).
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