Question

Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height `10sqrt(3)` m

Answer 1

Height of the tower (AC) = `10sqrt(3)` m

Distance between the base of the tower and point of observation (AB) = 30 m

Let the angle of elevation ∠ABC be θ

In the right ∆ABC, tan θ = `"AC"/"AB"`

= `(10sqrt(3))/30`

= `sqrt(3)/3`

tan θ = `1/sqrt(3)`

= tan 30°

∴ Angle of inclination is 30°