Question

. A tower stands vertically on the ground. From a point on the ground, which is 25m away

from the foot of the tower, the angle of elevation of the top of the tower is found to be

30°.Find the height of the tower.​

Answer 1

(refer to the attachment)

Given :

A tower stands vertically on the ground. From a point on the ground, which is 25m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 30°.

To Find :

The height of the tower.

Solution :

It is said that a tower stands on a point on the ground which is 25 m away from the foot of the tower.

A is the top of the tower and from A the angle of elevation is 30°.

AB is the height of the tower.

So,

According to the diagram,

• BC = 25 m
• ∠ABC = 90°
• AB = h m
• ∠ ACB = 30°

Using tanθ,

⇒ tanθ = Height/Base

where,

• θ = 30°
• tan 30° = 1/√3
• Height = AB = h m
• Base = BC = 25 m

Substituting the required values,

⇒ 1/√3 = h/25

⇒ 25 × 1 = √3 × h

⇒ 25 = √3h

⇒ 25/√3 = h

Taking √3 = 1.732,

⇒ 25/1.732 = h

⇒ 14.43 = h

∴ h = 14.43.

The height of the tower is 14.43 m.

Explore More :

• Trigonometric Table :

$\Large{\begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$