The angle of elevation of the top of an unfin | vedclass

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Question

(D) Let $PQ$ be the height of the unfinished tower and $R$ be the point on the ground at a distance of $120\, m$ from the base $Q$. Given $QR = 120\,

Vedclass · Accepted Answer

$87.84$

Vedclass · Answer

(D) Let $PQ$ be the height of the unfinished tower and $R$ be the point on the ground at a distance of $120\, m$ from the base $Q$. Given $QR = 120\, m$ and $\angle PRQ = 45^{\circ}$. In $\Delta PQR$,$	an 45^{\circ} = \frac{PQ}{QR} \implies 1 = \frac{PQ}{120} \implies PQ = 120\, m$. Let the tower be raised by $SP$ height so that the new angle of elevation is $\angle SRQ = 60^{\circ}$. In $\Delta SQR$,$	an 60^{\circ} = \frac{SQ}{QR} = \frac{SP + PQ}{QR}$. $\sqrt{3} = \frac{SP + 120}{120}$. $SP + 120 = 120\sqrt{3}$. $SP = 120(\sqrt{3} - 1) = 120(1.732 - 1) = 120 	imes 0.732 = 87.84\, m$.

The angle of elevation of the top of an unfinished tower at a point $120\, m$ from its base is $45^{\circ}$. How much higher must the tower be raised so that its angle of elevation at the same point be $60^{\circ}$? (in $m$)

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