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(D) Let $AB$ and $PQ$ be the two towers of equal height $h$ metres. Let $T$ be a point on the road $BQ$ such that $BT = x$ and $TQ = 100 - x$.In $\Del

Vedclass · Accepted Answer

$25$

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(D) Let $AB$ and $PQ$ be the two towers of equal height $h$ metres. Let $T$ be a point on the road $BQ$ such that $BT = x$ and $TQ = 100 - x$.In $\Delta ABT$, $	an 60^{\circ} = \frac{AB}{BT} = \frac{h}{x}$.Therefore, $x = \frac{h}{	an 60^{\circ}} = \frac{h}{\sqrt{3}}$.In $\Delta PQT$, $	an 30^{\circ} = \frac{PQ}{TQ} = \frac{h}{100 - x}$.Therefore, $100 - x = \frac{h}{	an 30^{\circ}} = h\sqrt{3}$.Substituting $x = \frac{h}{\sqrt{3}}$ into the equation $100 - x = h\sqrt{3}$:$100 = x + h\sqrt{3} = \frac{h}{\sqrt{3}} + h\sqrt{3}$.$100 = h \left( \frac{1 + 3}{\sqrt{3}} ight) = h \left( \frac{4}{\sqrt{3}} ight)$.$h = \frac{100 \sqrt{3}}{4} = 25 \sqrt{3} \, m$.

Two towers of equal height stand on either side of a wide road which is $100 \, m$ wide. At a point on the road between the towers, the angles of elevation of the tops of the towers are $60^{\circ}$ and $30^{\circ}$. Find their heights in metres. (in $\sqrt{3}$)

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