AB is a vertical pole. The end A is on the level ground, C is the middle point of AB and P is a point on the level ground. The portion CB subtends an angle β at P. If AP=nAB; then show that tanβ=n/(2n2+1).
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Solution
Let AB=2x so that AC=CB=x By given condition AP=n.AB=n.2x, and let ∠CPA=α then tanα=x2nx=12n tan(α+β)=2x2nx=1n. ∴tanβtan{α+β)−α} =tan(α+β)−tanα1+tan(α+β)tanα =1n−12n1+1n.12n=n2n2+1