Question

A building is 30 feet high. At a distance away from the building, an observer notices that the angle of elevation to the top of the building is 41º. How far is the observer from the base of the building?

Answer 1

As per the statement:

Height of the building(h)= 30 feet

Angle of elevation
`(\theta)`= 41°

We have to find the distance of observer from the base of the building.

Using tangent ratio:

`\tan \theta= \frac{\text{Opposite side}}{\text{Adjacent side}}`

You can see the diagram as shown below.

Opposite side = height of the building = 30 feet.

Adjacent side = distance of observer from the base of the building = x (let)

Then;

`\tan 41^(\circ)= (30)/(x)`

0.86928673781 =
`(30)/(x)`

By cross multiply we have;

`0.86928673781x = 30`

Divide by 0.86928673781 both sides we get;

x = 34.51 m

Therefore, the distance of observer from the base of the building is 34.51 m