Question

A building is 18sqrt(3) meters high. Find the angle of elevation of its top from a point 54 meters away from its foot.​

Answer 1

we know tanø = perpendicular / base

here perpendicular = height of building which is 18√3

and base = 54m

and let ø be the angle of elevation

tanø = 18√3/54 = 1/√3 = tan30°

so, angle of elevation is 30°

Answer 2

Given:

The height of the building is 18$\sqrt{3}$ meters.

The distance from the foot of the building to the point is 54 meters.

To find:

The angle of elevation from 54 meters away from the foot of the building.

Solution:

Let us take the angle of elevation as ∅ and the point as B.

We know that,

Tan ∅ = perpendicular/ base

The perpendicular is the height of the building.

Perpendicular = Height of the building = OA= 18$\sqrt{3}$ meters

The value of the base is the distance of point B from the foot of the building.

Base = Distance from foot of the building to the Point B = OB = 54 meters.

Tan ∅  = OA/OB

= (18$\sqrt{3}$)/ 54

= 1/ $\sqrt{3$

Tan∅ = 1/$\sqrt{3}$

The value of tan 30°  = 1/$\sqrt{3}$

∅ = tan⁻¹( tan 30° )

∅ = 30°

Angle of elevation, ∅ is 30°