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

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(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 $Q$.In $\Delta PQR$,$

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 $Q$.In $\Delta PQR$,$	an 45^{\circ} = \frac{PQ}{RQ}$.Since $	an 45^{\circ} = 1$,we have $PQ = RQ = 120 \ m$.Let the tower be raised to a height $P'$ such that the angle of elevation from $R$ becomes $60^{\circ}$.In $\Delta P'QR$,$	an 60^{\circ} = \frac{P'Q}{RQ}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $P'Q = RQ \cdot \sqrt{3} = 120 \sqrt{3} \ m$.The additional height required is $P'P = P'Q - PQ = 120 \sqrt{3} - 120 = 120(\sqrt{3} - 1) \ m$.Using $\sqrt{3} \approx 1.732$,we get $P'P = 120(1.732 - 1) = 120 	imes 0.732 = 87.84 \ m$.

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 (in $m$) must the tower be raised so that its angle of elevation is $60^{\circ}$ at the same point?

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