Question

The angle of elevation of the top of a tower from a point on the ground , which is 30m away from the foot of the tower , is 30 degree . Find the height of the tower.​

Answer 1

Answer:

we will take

$tan(30)$

which is equal to

$\frac{1}{ \sqrt{3} }$

let height be x

so

$\frac{1}{ \sqrt{3} } = \: \frac{x}{30} \\ \frac{1}{3} = \frac{ {x}^{2} }{900} \\ {x}^{2} = 300 \\ x = 10 \sqrt{3}$

that should be the answer

according to me

hope it helps

Answer 2

Given :

• The angle of elevation = 30°
• The angle elevate 30m away from the foot of the tower.

To Find :

• The height of the tower.

Reference of Figure :

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.5,5.8){ \bf{A} }\put(11.1,5.8){ \bf{B} }\put(11.05,9.1){ \bf{C} }\put(6,6){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(5,3){5}}\put(6.6,6.1){)}\put(6.8,6.15){ \bf{30^{\circ}} }\put(8.1,5.3){ \sf{30 m} }\end{picture}$

We have,

• BC = height of the tower.
• AB = 30 m.
• ∠CAB = 30°

In ∆ABC :

$\longrightarrow$ Tan θ = Perpendicular/Base

$\longrightarrow$ Tan 30° = h/30

$\longrightarrow$ 1/√3 = h/30

$\longrightarrow$ 30/√3 = h

$\longrightarrow$ (3 × 10)/√3 = h

$\longrightarrow$ (√3 × √3 × 10)/√3 = h

$\longrightarrow$ 10√3 = h

OR

$\longrightarrow$ h = 17.32 m.

Hence the height of the tower = 17.32 m.