A vertical pole (more than $50 f t$ . high) consists of two portions, the lower being $\frac{1}{3} r d$ of the whole. If the upper portion subtends an angle ${tan}^{- 1} \frac{1}{2}$ at a point in a horizontal plane through the foot of the pole and distance $40 f t$ . from it, find the height of the pole.

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Solution

We are given that $tan θ = \frac{1}{2}$
Where $θ$ is the angle subtended by the upper portion.
Also $tan α = \frac{3 h}{40}, tan β = \frac{h}{40}$
and $θ = α - β, ∴ tan θ = \frac{tan α - tan β}{1 + tan α tan β}$
or $\frac{1}{2} = \frac{(3 h / 40) - (h / 40)}{1 + (3 h^{2} / 40^{2})} = \frac{80 h}{1600 + 3 h^{2}}$
or $3 h^{2} + 1600 = 160 h$
or $3 h^{2} - 160 h + 1600 = 0$ .
or $3 h^{2} - 120 h - 40 h + 1600 = 0$ .
or $(h + 40) (3 h - 40) = 0$ .
$∴ h = 40$ and hence $3 h = 120$ is the required length of the pole. $3 h \neq 40$ as the pole of given to be the more than $100 f t .$

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