A tower subtends angles α,2αand 3α respectively at points A, B and C, all lying on a horizontal line through the foot of the tower. Then AB/BC=
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a
sin3αsin2α
b
1+2 cos2α
c
2+ cos2α
d
sin2αsinα
answer is B.
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Detailed Solution
using exterior angle property ∠AEB=α and ∠BEC= αusing sine rule From ΔABE, ABsinα=BEsinα,FromΔBCE,BCsinα=BEsin3α⇒ABBC=sin3αsinα=3sinα−4sin3αsinα =3−4sin2α=1+21−2sin2α=1+2cos2α