An airplane is flying at an elevation of 5150 ft, directly above a straight highway. two motorists are driving cars on the highway on opposite sides of the plane. the angle of depression to one car is 32°, and that to the other is 51°. how far apart are the cars? (round your answer to the nearest foot.) ft

Using trigonometric ratios we can get the distance;
For the first car; The distance from the point on the highway below the plane 
tan = opp/adj
tan(36°) = 5150/x
0.727 = 5150/x
0.727x = 5150
x = 7088.37
For the second car we also use tangent; the distance from the point on the highway below the plane will be; 
tan(56°) = 5150/y
1.483 = 5150/y
1.483y = 5150
y= 3473.72
The we can add the two distances to get how far apart the cars are;
7088.37 + 3473.72 = 10562.09 feet. 
= 10562.09 ft

Ihope you get it right bro