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(C) Let the height of the tower be $AC = 100 \ m$ and the point of observation be $B$. The angle of elevation is $\angle ABC = 30^{\circ}$.In the righ

Vedclass · Accepted Answer

$200$

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(C) Let the height of the tower be $AC = 100 \ m$ and the point of observation be $B$. The angle of elevation is $\angle ABC = 30^{\circ}$.In the right-angled triangle $ABC$,we need to find the distance between the top of the tower $(A)$ and the point of observation $(B)$,which is the hypotenuse $AB$.Using the trigonometric ratio:$\sin 30^{\circ} = \frac{	ext{Opposite}}{	ext{Hypotenuse}} = \frac{AC}{AB}$$\frac{1}{2} = \frac{100}{AB}$$AB = 100 	imes 2 = 200 \ m$.Therefore,the distance between the top of the tower and the point of observation is $200 \ m$.

To an observer on the ground,the angle of elevation of the top of a tower is $30^{\circ}$. If the height of the tower is $100 \ m$,find the distance between the top of the tower and the point of observation (in meters).

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