Question
A tower stands vertically on the ground. From a point on the ground, which is $15 \ m$ 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.
✓
Answer
First, let us draw a simple diagram to represent the problem. Here $A B$ represents the tower, $C B$ is the distance of the point from foot of the tower and $\angle A C B$ is the angle of elevation. We need to determine the height of the tower, i.e., $AB$ . Also, $ACB$ is a triangle, right-angled at $B$ . To solve the problem, we choose the trigonometric ratio $\tan 60^{\circ}$ (or $\cot 60^{\circ}$ ), as the ratio involves $A B$ and $B C$.
Now, $\tan 60^{\circ}=\frac{ AB }{ BC }$
i.e., $\sqrt{3}=\frac{\mathrm{AB}}{15}$
i.e., AB = $15 \sqrt{3}$
Hence, the height of the tower is $15 \sqrt{3}$ m
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