Question

to approach a runway, a plane must begin a 7° descent starting from a height of 3 miles above the ground. to the nearest mile, how many miles from the runway is the airplane to start the approach?

Answer 1

to make a draw of the situation helps you greatly, we have a rectangle-triangle formed from the ground up 3 miles to the aircraft, from that high point the hypothenuse of the triangle goes straight to the ground, with a 7° angle of descending, we only need to find out the other side of the triangle, which is the amount of miles from the runway that the descending takes, so we have:

tan 7° = x/3

calling x the descending distance, and 3 the height, so we just solve for x and find it:

x = 3 tan 7°
x = 0.37 miles

that is the descending distance.

Answer 2

Explanation:

It is given that to approach a runway, a plane must begin a 7° descent starting from a height of 3 miles above the ground.

Thus, we have AB=3 miles and ∠C=7°.

Using the trigonometry, we have

`(AB)/(BC)=tan7^(\circ)`

⇒
`(3)/(BC)=tan7^(\circ)`

⇒
`BC=(3)/(tan7^(\circ))`

⇒
`BC=(3)/(0.122)`

⇒
`BC=24.6miles`

Thus, From 24.6 miles the runway is the airplane to start the approach.