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 34°, and that to the other is 58°. How far apart are the cars?

The cars are 10853.3 feet apart.

Explanation:

Given:

Height of airplane (H) = 5150 ft

Angle of depression of car 1 (x₁) = 34°

Angle of depression of car 2 (x₂) = 58°

Consider two right angled triangles ABC and ABD as shown below.

AB = H = 5150 ft, ∠ACB (x₁) = 34°, ∠ADB (x₂) = 58°, BC = 'a', BD = 'b'

Now, from triangle ABC, using tangent of the angle, we have

Similarly for triangle ABD,

Now, separation between the two cars is the sum of the distances 'a' and 'b'. So,

Separation =

Therefore, the cars are 10853.3 feet apart.

probably not if net force is zero