5 points

An airplane approaches an airport. At a certain time, the plane is 1020m

high. Its angle of elevation measured from the airport is 20.5 degrees.

How far is the plane from the airport to the nearest meter? *

A-2728

B-2913

C-1089

D-3

2728.11m

Step-by-step explanation:

The height of the plane is 1020m

The airport makes an angle of elevation of 20.50° with the plane

The distance of the plane from the airport = ?

See attached document for better illustration of the position of the plane and airport

Using SOHCAHTOA,

tanΘ = opposite/ adjacent

θ = 20.50°

opposite = 1020m

adjacent = ? = x

Tanθ = opposite/ adjacent

Tan20.50 = 1020 / x

X = 1020 / tan20.50

X = 1020 / 0.37388

X = 2728.11m

The distance of the plane from the airport is 2728.11m

the answer is 3/5

step-by-step explanation:

first divide 24 divided by 8 because that is the least common multiple

then divide 40 by 8

then you will get 3/5.