Question

A pilot flying at an altitude of 3.7 km sights 2 control towers directly in

Answer 1

Part A; The plane will have to travel 70.61 km to fly over the first tower

Part B; The plane will have to travel 5.28 km to fly over the second tower

Step-by-step explanation:

Step 1; Assume the plane is x km away from the control tower. We know it is flying at a height of 3.7 km and an angle of depression of 3°. So a right-angled triangle can be formed using these measurements. The triangle's opposite side measures 3.7 kilometers while the opposite side measures x kilometers. The angle of the triangle is 3°

Step 2; Since we have the length of the opposite side and the angle of the triangle, we can determine the tan of an unknown angle.  

tan 3°= $\frac{3.7}{x}$, x =  $\frac{3.7}{tan 3}$, tan 3° = 0.0524,

x = 70.6106 kilometers.

So the plane must travel 70.61 km to fly over the first tower.

Step 3; Assume the plane is y km away from the control tower. We know it is flying at a height of 3.7 km and an angle of depression of 35°. So a right-angled triangle can be formed using these measurements. The triangle's opposite side measures 3.7 kilometers while the opposite side measures y kilometers. The angle of the triangle is 35°

Step 4; Since we have the length of the opposite side and the angle of the triangle, we can determine the tan of an unknown angle.  

tan 35°= $\frac{3.7}{y}$, y =  $\frac{3.7}{tan35}$, tan 35° = 0.7002,

y = 5.2842 kilometers.

So the plane must travel 5.28 km to fly over the second tower.