An airplane is flying at an elevation of 5150 ft, directly above a straight highway. Two motorists are driving cars on the highway, both on one side of the plane.

Question

An airplane is flying at an elevation of 5150 ft, directly above a straight highway. Two motorists are driving cars on the highway, both on one side of the plane. If the angle of depression to one car is 36° and that to the other is 51°, how far apart are the cars? (Round your answer to the nearest foot.)

Answer 1

For these type of questions, it's good to make a sketch of the situation.

Draw the airplane. Then draw the ground as a horizontal line below the airplane. Draw a line straight down from the airplane to the ground, and mark that line as 5150 feet. Now draw a line (up in the sky) through the airplane parallel to the ground.

First let's do the car at 51 degrees. Draw a line from the airplane down and to the right at angle of 51 degrees with the top line. (It doesn't have to be exact.) That's the angle of depression. Now you have triangle with the airplane and two points on the ground.

The top angle of the triangle is 90-51 = 39 degrees. From SOH CAH TOA, we know that tan(39) = Opposite Side / 5150. So the Opposite Side = 4170.4 feet. That's how far that car is ahead of the plane.

Do the same calculation for the car at 36 degrees. (You can draw it you want to. This line is shallower and goes further to the right.) 90-36 = 54 and tan(54) = Opposite Side / 5150. So, the Opposite Side = 7088.4. That's how far that second car is ahead of the plane.

So the cars are 7088.4 - 4170.4 = 2917 feet apart.