Question

1. To approach the runway, a small plane must begin a 6° descent starting from a height of 1000 feet above the ground. To the nearest foot, how many feet from the runway is the airplane at the start of this approach?

Answer 1

Answer:

The plane was 9514 ft. distant from the runway.

Step-by-step explanation:

See the attached diagram.

Now, B is the initial position of the plane which is 1000 ft. above A.

Now, Δ ABC is a right triangle having ∠ ACB = 6°

So, applying trigonometry, we can write

$\tan 6^{\circ} = \frac{AB}{AC} = \frac{1000}{x}$

⇒ x = 9514.36 ≈ 9514 ft.

Therefore, the plane was 9514 ft. distant from the runway. (Answer)