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

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(B) Let the height of the tower be $h$ and the length of the shadow when the angle of elevation is $45^{\circ}$ be $x$.In $	riangle PQS$,$	an 45^{\c

Vedclass · Accepted Answer

$30(\sqrt{3}+1)$

Vedclass · Answer

(B) Let the height of the tower be $h$ and the length of the shadow when the angle of elevation is $45^{\circ}$ be $x$.In $	riangle PQS$,$	an 45^{\circ} = \frac{h}{x} \implies 1 = \frac{h}{x} \implies x = h$.In $	riangle PQR$,$	an 30^{\circ} = \frac{h}{x + 60}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{h + 60}$.$h + 60 = h\sqrt{3}$.$60 = h(\sqrt{3} - 1)$.$h = \frac{60}{\sqrt{3} - 1}$.Rationalizing the denominator: $h = \frac{60(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{60(\sqrt{3} + 1)}{3 - 1} = \frac{60(\sqrt{3} + 1)}{2} = 30(\sqrt{3} + 1)\, m$.

The shadow of a tower standing on a level plane is found to be $60\, m$ longer when the angle of elevation of the sun is $30^{\circ}$ than when it is $45^{\circ}$. Find the height of the tower.

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