Question

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, and the angle of depression to one car is 35 ̊ and to the other is 52 ̊. How far apart are the cars?

Answer 1

Answer: 11378.583 feet (approx)

Step-by-step explanation:

Let from the foot of the line that shows the distance of the plane from the road the distance of first car is x feet and the distance of second car is y feet,

Thus, by the below figure,

We can write,

$tan 52^{\circ}=\frac{5150}{x}$

⇒ $x=\frac{5150}{tan 52^{\circ}}$

⇒ $x = 4023.62097651$

Similarly, by the below diagram,

$tan 35^{\circ}=\frac{5150}{y}$

⇒ $y=\frac{5150}{tan 35^{\circ}}$

⇒ $y = 7354.96223472$

Thus, the total distance between the car = x + y = 4023.62097651 + 7354.96223472= 11378.5832112≈11378.583 feet

Answer 2

Answer: 11378.58 ft

Step-by-step explanation:

1. Draw  a figure like the one attached, where A is the position of the airplane and B and C are the positions of the cars.

2. The distance between the cars is the length BC, then, you need to calculate the measure of BD and DC to add them and obtain BC.

3. You know the opposite side and the angles of depression, then, you can calculate the distance between the cars as following:

$BC=BD+DC\\BC=\frac{5150}{tan(35)}+\frac{5150}{tan(52)}\\BC=11378.58ft$