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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