Question

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?

Answer 1

16.69°

Explanation:

Given that,

Height of the building, h = 60 foot

Distance from the base of the building, b = 60 foot

We need to find the angle of elevation must the plane takes off in order to avoid crashing into the building. So,

`\tan\theta=(60)/(200)\\\\\theta=\tan^(-1)(60)/(200)\\\\=16.69^(\circ)`

So, the required angle is 16.69°.