Correct option is (b) a sin 2α
Let EF be the TV tower such that
∠EAF = α, ∠EBF = 2α,
∠ECF = 3α and AB = a
In \(\Delta\)EBA,
ext. ∠EBF = ∠BEA + ∠BAE
⇒ 2α = ∠BEA + α
⇒ ∠BEA = α
⇒ \(\Delta\)BEA is isosceles
⇒ BE = a.
In rt. \(\Delta\)EBF, sin 2α = \(\frac{EF}{EB}\)
⇒ EF = EB sin 2α = a sin 2α