Question

an airplane is flying at an elevation of 5150 feet, 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 degrees and to the other car is 52 degrees, how far apart are the cars?

Answer 1

7086.9 feet.

Step-by-step explanation:

We can see that the two triangles formed by the plane and the cars are similar, because they share a common angle (90 degrees) and have corresponding angles that are equal (the angles of depression). Therefore, we can use the proportionality of corresponding sides to find the distance between the cars. Let x be the distance from the plane to the car with 35 degrees angle of depression, and y be the distance from the plane to the car with 52 degrees angle of depression. Then we have:

• x / sin(35) = y / sin(52) = 5150 / sin(90)
• Cross-multiplying and solving for x and y, we get:
• x = 5150 x sin(35) / sin(90) x = 2957.8 feet
• y = 5150 x sin(52) / sin(90) y = 4129.1 feet
• The distance between the cars is the sum of x and y:
• d = x + y d = 2957.8 + 4129.1 d = 7086.9 feet

The answer is 7086.9 feet.