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(D) Let the height of the tower be $h$ and the distance from the second point to the base of the tower be $x$.In the right-angled triangle formed by t

Vedclass · Accepted Answer

$25.95$

Vedclass · Answer

(D) Let the height of the tower be $h$ and the distance from the second point to the base of the tower be $x$.In the right-angled triangle formed by the tower and the second point,we have $	an(60^{\circ}) = \frac{h}{x}$.Since $	an(60^{\circ}) = \sqrt{3}$,we get $x = \frac{h}{\sqrt{3}}$.In the larger right-angled triangle formed by the tower and the first point,the distance from the base is $x + 30$.We have $	an(30^{\circ}) = \frac{h}{x + 30}$.Since $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$,we get $\frac{1}{\sqrt{3}} = \frac{h}{x + 30}$,which implies $x + 30 = h\sqrt{3}$.Substituting $x = \frac{h}{\sqrt{3}}$ into the equation: $\frac{h}{\sqrt{3}} + 30 = h\sqrt{3}$.Multiplying by $\sqrt{3}$: $h + 30\sqrt{3} = 3h$,so $2h = 30\sqrt{3}$.Thus,$h = 15\sqrt{3} \approx 15 	imes 1.732 = 25.98\, m$. Given the options,the closest value is $25.95\, m$.

From a point on the ground,the angle of elevation of the top of a tower is found to be $30^{\circ}$. Walking $30\, m$ towards the tower from that point,the angle of elevation of the top of the tower is found to be $60^{\circ}$. Find the height of the tower (in $m$).

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