Question

1. The angle of elevation of the top of a pole from a point on the level ground and 15 m away
from the pole is 30°. Find the height of the pole.
Take √3
= 1.732)​

Answer 1

Given, BC is 15 and angle of elevation on ground is 30

0

Let the height of the pole be h

In right angled ΔABC.

tan30 = AB/BC

1/√3 = h/15

h = 15/√3

∴h = 5√3

Answer 2

GiveN:-

• The Angle of elevation of the top from a point P on the same line as that of base is 30°.
• Distance of the point form the base of the pole = 15 m.

To FinD:-

• Height of the pole?

Solution:-

Refer to the attachment...

We have the distance between the point and the base of the pole and the angle of elevation. We need to find the height of pole.

If the distance between their bottoms is base, then the height of the pole will be perpendicular. So, we can use the tangent here.

⇒ tan Θ = Height of the pole / Base length

⇒ tan 30° = AB / BP

⇒ 1/√3 = AB / 15

⇒ AB√3 = 15

⇒ AB = 15 / √3

⇒ AB = 5√3 m

Now taking √3 = 1.732

⇒ AB = 86.6 m

Hence:-

• The height of the pole is 86.6 m