Question

1. an airplane takes off 200 yards in front of a 60 foot building. at what angle of elevation must the plane take off in order to avoid crashing into the building? assume that the airplane flies in a straight line and the angle of elevation remains constant until the airplane flies over the building.

Explanation:

Step1: Convert yards to feet

Since \(1\) yard \( = 3\) feet, \(200\) yards \(=200\times3 = 600\) feet.

Step2: Use the tangent function

The tangent of an angle of elevation \(x\) in a right - triangle is given by \(\tan x=\frac{\text{opposite}}{\text{adjacent}}\). Here, the opposite side is the height of the building (\(h = 60\) feet) and the adjacent side is the horizontal distance (\(d=600\) feet). So, \(\tan x=\frac{60}{600}\).

Step3: Calculate the angle

\(\tan x = 0.1\). Then \(x=\arctan(0.1)\). Using a calculator, \(x\approx5.71^{\circ}\)

Answer:

The angle of elevation is approximately \(5.71^{\circ}\)