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 units

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

Step2: Use the tangent function

In a right - triangle (where the height of the building is the opposite side and the horizontal distance is the adjacent side), the tangent of the angle of elevation \(x\) is given by \(\tan x=\frac{\text{opposite}}{\text{adjacent}}\). Here, the opposite side \(y = 60\) feet and the adjacent side \(z=600\) feet. So, \(\tan x=\frac{60}{600}=0.1\).

Step3: Find the angle

We know that if \(\tan x = 0.1\), then \(x=\arctan(0.1)\). Using a calculator, \(x\approx5.71^{\circ}\)

Answer:

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