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 51°. How far apart are the cars?

The cars are approximately 11,806 ft apart.

Step-by-step explanation:

Refer to the diagram below.

Let the Airplane be at C and the Cars be at Points A and B respectively.

We want to determine the distance AB between the two cars.

In Triangle ACD

Tan 34°=5150/AD

AD X Tan 34° = 5150

AD = 5150/Tan 34°= 7635.19ft

In Triangle CDB

Tan 51°=5150/DB

DB X Tan 51° = 5150

DB = 5150/Tan 51°= 4170.39ft

AB= AD+DB= 7635.19+4170.39 =11805.58ft

The cars are approximately 11,806 ft apart.

the answer is 1 becuz if it were 1.0 that would rip apart the entire so ya. : )

~potato.

copyright potato 2019.