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 smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

explanation:

step-by-step explanation: