The angles of elevation of the top of a tree 220 meters high from two points lie on the same plane are 30° and 45°. What is the distance (in metres) between the two points?

Solution:

In ΔABC

tan45° = AB/BC

⇒ 1 = 220/BC

⇒ BC = 220 meter

In ΔABD

⇒ tan30° = AB/BD

⇒ 1/√3 = 220/BD

⇒ BD = 220√3 meter

The distance between the two points (CD) = BD – BC

⇒ The distance between the two points = 220√3 – 220

⇒ The distance between the two points = 220(1.732 – 1) = 161.05 meter