MP Board · Class 10 · Mathematics · Applications of TrigonometryA 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°. Find the height of the tower.

Let $AB$ be the height of the tower and $C$ be the point on the ground 15 m away from the foot of the tower $B$. \nGiven:

• Distance $BC = 15$ m
• Angle of elevation $\angle ACB = 60°$
• Let the height $AB = h$ m \nIn right-angled triangle $\triangle ABC$: $$\tan(60°) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$$ \nSubstitute the known values: $$\sqrt{3} = \frac{h}{15}$$

$$h = 15\sqrt{3} \text{ m}$| \nThus, the height of the tower is $15\sqrt{3}$ m (or approximately $15 \times 1.732 = 25.98$ m).