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

isaiah walks 3 miles due north, turns and then 5 miles due east. how far is he from his starting point?

a^2 + b^2 = c^2

let c = distance from starting point.

(3)^2 + (5)^2 = c^2

9 + 25 = c^2

34 = c^2

sqrt{34} = sqrt{c^2}

sqrt{34} = c