An airplane over the pacific sights an atoll at 8 degree angle of depression. If the plane is 520hm above water, how many kilometers is it from a point 520 m directly above the center of the atoll?

Help please :]

Assume the Earth is flat.

Draw a figure with flat horizontal Earth surface along the bottom. Put three points on the figure:
P (airplane)
A (atoll on the ground)
B (point 520 m above the atoll)

Draw the (right) triangle PAB. The angle at P is 8 degrees, aimed below horizontal to point A

It should be obvious that
BA = 520
PB = 520/tan 8 = 3700 m

To find the distance from the airplane to the point directly above the center of the atoll, we can use trigonometry. Specifically, we can use the tangent function.

The tangent of the angle of depression (θ) is equal to the opposite side divided by the adjacent side. In this case, the opposite side is the difference in height between the airplane and the point directly above the center of the atoll (520 m), and the adjacent side is the distance we want to find.

Let's solve for the distance, denoted as 'd':

tan(θ) = opposite/adjacent
tan(8°) = 520 m/d

First, let's convert 520 meters to kilometers since the answer is requested in kilometers. We divide by 1000:

520 m / 1000 = 0.52 km

Now we can rewrite the equation with the values:

tan(8°) = 0.52 km / d

To solve for 'd', we isolate it by taking the reciprocal of both sides:

1 / tan(8°) = d / 0.52 km

Now we can calculate the distance 'd':

d = 0.52 km / tan(8°)

Using a calculator:

d ≈ 3.388 km

Therefore, the airplane is approximately 3.388 kilometers away from the point directly above the center of the atoll.