Question

To approach the runway, a pilot of a small plane must begin a 10 descent starting from a height of 1983 feet above the ground. To the nearest tenth of a mile, how many miles from the runway is the airplane at the start of this approach?

Answer 1

sin10 = 1790/d

d = 1983/(sin 10)

d = 11665 feet

Since there is 5,280 feet in one mile we have

11665/5280 = 2.2 miles

hope this helps you: )

Answer 2

The distance between the runway and the airplane where it start this approach is 2.16 miles.

Explanation:

A pilot of a small plane must begin a 10 descent starting from a height of 1983 feet above the ground that is AB is the height of plane above the ground, AB= 1983 feet. and A is the point from where the pilot starts descent.

thus, ∠ACB = ∠DAC = 10°

We have to find the distance between the runway and the airplane where it start this approach that is we have to find length AC( in miles) . Let AC = x

Applying trigonometric ratio,

`\sin C=\frac{\text{Perpendicular}}{\text{Hypotenuse}}`

Substitute the values,

`\sin 10^(\circ)=(1983)/(x)`

`x=(1983)/(\sin 10^(\circ))`

`x=11422.811`(approx)

Thus, Distance between the runway and the airplane is 11422.81 feet.

Since , we have to find distance in miles

We know 1 mile = 5280 feet

1 feet =
`(1)/(5280)` miles.

11422.81 feet =
`(1)/(5280)* 11422.81` miles.

= 2.16 miles (approx)

Thus, the distance between the runway and the airplane where it start this approach is 2.16 miles.