Answer:
16.69°
Step-by-step 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,
[tex]\tan\theta=\dfrac{60}{200}\\\\\theta=\tan^{-1}\dfrac{60}{200}\\\\=16.69^{\circ}[/tex]
So, the required angle is 16.69°.
