Question

An engineer is surveying a building with a device that is 6 feet above the ground. He is 60 feet from the building and is looking with an angle of elevation of 45. How tall is the building ?

Answer 1

\(\sf tan\theta=\dfrac{opposite}{adjacent}=tan(45^o)=\dfrac{?}{60~ft}\)

\( \sf ?=tan(45^o)*(60 ft)\)

Add 6 ft because the device is 6 ft off the ground.