Trigonometry-14

From Homeworkwiki

An observer 1.6 m tall is 30√3 m away from a tower. The angle of elevation from his eye

to the top of the tower is 30⁰. Determine the height of the tower.

Solution: AB is the observer, CD is the tower.

BC = 30 √3 m; BA = 1.6 m

∠EAD = 30⁰

AE = 30 √3 m

ED / AE = tan 30⁰

ED = AE X tan 30⁰

= 30 √3 X 1/√3 m

= 30 m

Therefore, height of the tower = CE + ED = 1.6 m + 30 m = 31.6 m