A tower stands vertically on the ground. From a point on the ground, which is 2

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A tower stands vertically on the ground. From a point on the ground, which is 24 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^{\circ}$ . Find the height of the tower.

Options:

A .

26 \sqrt{3} m

B .

24 \sqrt{3} m

C .

24 m

D .

26 m

Answer: Option B
:
B

The above $Δ A B C$ represents the given situation.
AB = 24 m
In Triangle ABC,
$t a n 60^{\circ} = \sqrt{3} = \frac{C B}{A B} \Rightarrow A B \times \sqrt{3} = C B \Rightarrow C B = 24 \sqrt{3} m .$
Hence, the height of the tower = $24 \sqrt{3} m .$

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