A pilot flying at an altitude of 3.7 km sights a control tower directly in front of him. The angle of depression to the base of the tower is 55 degrees

Question

a) How far is the pilot from the tower? Round to the nearest hundredth b) if the pilot is traveling at a rate of 100 m/s, then how long does it take the pilot to fly over the tower?

Answer 1

Draw the situation...represented by a right triangle.  from the horizontal ground, draw the altitude and label it 3.7 km.

back on the ground, select a point to the right of where the altitude is drawn, to represent the base of the tower.

you have a right triangle with vertical leg 3.7km and horizontal leg distance d...this d is the distance that the airplane will fly,parallel to the ground, to get directly above the tower.  Since the angle of depression is 55 degrees, the adjacent acute angle in the triangle measures 35 degrees, and tan 35 = d/3.7.

 

this gives d =2.59 km.

 

now you can use D = RT to find how long the plane would take to get directly above the tower.  using R = 100 m/sec =

100/(1/3600) m/hr = 360000 m/hr = 360 km/hr, you will find that T = 0.0072 hr = 0.432 min = 25.9 sec

 

I BELIEVE THAT YOU SHOULD CHECK THIS WORK since it does not agree with the concise answer given by Skye in the Comment.