A vertical pole stands at a point A on the boundary of a circular park of radius a and subtends an angle α at another point B on the boundary. If the chord AB subtends angle α at the centre of the park, then the height of the pole is
A
2acosα2tanα
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B
2asinα2tanα
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C
2acosα2cotα
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D
2asinα2cotα
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Solution
The correct option is B2asinα2tanα Let O be centre of the park and h be the height of pole, i.e., AC=h
Since OA=OB=a, ∴∠OAB=∠OBA=β (say) β+β+α=180∘ ⇒β=90∘−α2
Using sine rule in △OAB, ABsinα=OAsinβ ⇒AB=a×sinαsin(90∘−α2) ⇒AB=2asinα2
Now, in △ABC, tanα=ACAB ∴h=2asinα2tanα