Answer:
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
[tex]\tan(x_1)=\frac{H}{a}\\\\a=\frac{H}{\tan(x_1)}\\\\a=\frac{5150}{\tan(34)}=7635.2\ ft[/tex]
Similarly for triangle ABD,
[tex]b=\frac{H}{\tan(x_2)}\\\\b=\frac{5150}{\tan(58)}=3218.1\ ft[/tex]
Now, separation between the two cars is the sum of the distances 'a' and 'b'. So,
Separation = [tex]a+b=7635.2+3218.1=10853.3\ ft[/tex]
Therefore, the cars are 10853.3 feet apart.
