**In the adjoining figure, AP is a man of height 1.8 m and BQ is a building 13.8 m high. If the man sees the top of the building by focusing his binoculars at an angle of 30**°** to the horizontal, find the distance of the man from the building.**

**Answer :**

Height of man AP = 1.8 m

Height of building BQ = 13.8 m

Angle of elevation from the top of the building to the man = 30°

AB = x then PC = x

QC = 13.8 – 1.8 = 12 m

In right △ PQC

tan θ = QC/PC

tan 30° = 12/x

1/√3 = 12/x

x = 12 √3 m

Hence,

the distance of the man from the building is 12 √3 m.

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