JEE Advance - Mathematics (1980 - No. 28)
(ii) $$AB$$ is vertical pole. The end $$A$$ is on the level ground. $$C$$ is the middle point of $$AB$$. $$P$$ is a point on the level ground. The portion $$CB$$ subtends an angle $$\beta $$ at $$P$$. If $$AP = n\,AB,$$ then show that tan$$\beta $$ $$ = {n \over {2{n^2} + 1}}$$
The height of the tower is abc tan(θ) / 4Δ and tan(β) = n / (2n^2 + 1)
The height of the tower is abc tan(θ) / Δ and tan(β) = n / (n^2 + 1)
The height of the tower is 4abc tan(θ) / Δ and tan(β) = 2n / (2n^2 + 1)
The height of the tower is abc tan(θ) / 4Δ and tan(β) = 2n / (2n^2 + 1)
The height of the tower is abc tan(θ) / 2Δ and tan(β) = n / (2n^2 + 1)
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