If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are 30∘, 45∘, and 60∘ respectively, then the ratio AB:BC is
A
1:√3
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B
2:3
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C
√3:1
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D
√3:√2
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Solution
The correct option is C√3:1
∵ PB bisects ∠APC, therefore AB : BC = PA : PC
Also ΔAPQ,sin30∘=hPA⇒PA=2h
And in ΔCPQ,sin60∘=hPC⇒PA=2h√3 ∴AB:BC=2h:2h√3=√3:1