Answer:
2,810 feet
Explanation:
Take a look at the diagram
There are two triangles formed with a common side AB
AB represents the vertical height of the plane = 5150 feet
AC represents the distance of car 1 from O
In ΔABC, m∠A = 90°, m∠ABC = 38°
Since the three angles of triangle must add up to 180°
90 + 38 +m∠C = 180
128 + m∠C = 180°
m∠C = 180 - 128 = 52°
In a right angle triangle where one side is the angle ∅, the ratio of the side opposite this angle to the side adjacent to the angle is tan∅
Using the values provided we get
tan(52°) = 5150/AC
AC= 5150/tan(52) = 4,023.62 feet ≈ 4024 feet
This is the horizontal distance of the first car from the plane
Similarly for ΔABD
m∠D = 180 - (90 + 53) = 37°
tan(37) = 5150/AD
AD= 5150/tan(53) = 6,834.28 feet ≈ 6834 feet
This is the horizontal distance of the second car from the plane
The distance between the two cars = 6,834 - 4,024
= 2,810 feet