Question

A plane is flying at an altitude of 12,000 m. From the pilot, the angle of depression to the airport is 32o. How far is the tower from a point directly beneath the plane?

Answer 1

This is an appllication of right angle trigonometry. 12000 m is the side opposite the 32 degree angle, and the desired horiz. distance is the unknown; let's call it x.

Then tan 32 degrees = opp / adj = 12000 m / x

0.625 = 12000 m / x

Then x = 12000 m / 0.625 = 19204 m (answer)

The tower is 19204 m from a point directly beneath the plane.

Answer 2

Using the slope concept, it is found that the tower is 32,970 feet from a point directly beneath the plane.

What is a slope?

The slope is given by the vertical change divided by the horizontal change, and it's also the tangent of the angle of depression.

In this problem:

• The vertical change is of 12,000 ft.

• The horizontal change is the distance d.

• The angle is of 20º.

Hence:

$\tan{20^\circ} = \frac{12000}{d}$

$d = \frac{12000}{\tan{20^\circ}}$

$d = 32970$

The tower is 32,970 feet from a point directly beneath the plane.

More can be learned about the slope concept at https://brainly.com/question/18090623