An observer 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from her eyes is 45∘. What is the height of the tower? [2 MARKS]
Let AB be the tower of height h and CD be the observer of height 1.5 m at a distance of 28.5 m from the tower AB. [12 MARK]
In ΔAED, we have
tan45∘=h28.5 [1 MARK]
⇒1=h28.5
⇒h=28.5 m
∴h=AE+BE=AE+DC
=(28.5+1.5)m=30 m
Height of tower = h + 1.5
= 28.5 + 1.5
= 30 m [12 MARK]
Hence, the height of the tower is 30 m.