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(A) Let $OP$ be the tower of height $h \, m$ and $PQ$ be the flagstaff of height $6 \, m$. Let the sun make an angle $	heta$ with the ground. Let $OA

Vedclass · Accepted Answer

$60^{\circ}$

Vedclass · Answer

(A) Let $OP$ be the tower of height $h \, m$ and $PQ$ be the flagstaff of height $6 \, m$. Let the sun make an angle $	heta$ with the ground. Let $OA = x$ be the shadow of the tower and $AB = 2 \sqrt{3} \, m$ be the shadow of the flagstaff.In $	riangle OAP$,$	an 	heta = \frac{OP}{OA} = \frac{h}{x}$.In $	riangle OBQ$,$	an 	heta = \frac{OQ}{OB} = \frac{h + 6}{x + 2 \sqrt{3}}$.Equating the two expressions for $	an 	heta$:$\frac{h}{x} = \frac{h + 6}{x + 2 \sqrt{3}}$$h(x + 2 \sqrt{3}) = x(h + 6)$$hx + 2 \sqrt{3}h = hx + 6x$$2 \sqrt{3}h = 6x$$\frac{h}{x} = \frac{6}{2 \sqrt{3}} = \sqrt{3}$.Since $	an 	heta = \frac{h}{x}$,we have $	an 	heta = \sqrt{3}$.Therefore,$	heta = 60^{\circ}$.

If a flagstaff of $6 \, m$ height placed on the top of a tower throws a shadow of $2 \sqrt{3} \, m$ along the ground,then the angle that the sun makes with the ground is

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