Answer:
[tex]\boxed{\text{48 ft}}[/tex]
Step-by-step explanation:
The angle of depression from the camera at D and the angle of elevation from the target cars are congruent.
1. Distance of car A from tower
Consider ∆ACD
tan25 = 50/AC
AC = 50/tan25 = 50/0.4663 = 107.2 ft
2. Distance of car B from tower
Consider ∆BCD
tan40 = 50/BC
AC = 50/tan40 = 50/0.8391 = 59.6 ft
3. Distance from cars A and B
AB = AC – BC = 107.2 – 59.6 = 48 ft
The distance between the two cars is [tex]\boxed{\textbf{48 ft}}[/tex].
