A vertical pole consists of two parts,the low | vedclass

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(B) Let the total height of the pole be $H$. The lower part is $\frac{H}{3}$ and the upper part is $\frac{2H}{3}$.Let the point be $P$ at a distance $

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$20 \ m$ and $60 \ m$

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(B) Let the total height of the pole be $H$. The lower part is $\frac{H}{3}$ and the upper part is $\frac{2H}{3}$.Let the point be $P$ at a distance $d = 20 \ m$ from the base.Let $\alpha$ be the angle subtended by the lower part at $P$,and $\beta$ be the angle subtended by the whole pole at $P$.The angle subtended by the upper part is $	heta = \beta - \alpha$,where $	an 	heta = \frac{1}{2}$.From the geometry,$	an \alpha = \frac{H/3}{d} = \frac{H}{3d}$ and $	an \beta = \frac{H}{d}$.Using the formula $	an(\beta - \alpha) = \frac{	an \beta - 	an \alpha}{1 + 	an \beta 	an \alpha} = \frac{1}{2}$.Substituting the values: $\frac{\frac{H}{d} - \frac{H}{3d}}{1 + (\frac{H}{d})(\frac{H}{3d})} = \frac{1}{2}$.$\frac{2H/3d}{1 + H^2/3d^2} = \frac{1}{2} \Rightarrow \frac{2H}{3d} 	imes \frac{3d^2}{3d^2 + H^2} = \frac{1}{2}$.$\frac{2Hd}{3d^2 + H^2} = \frac{1}{2} \Rightarrow 4Hd = 3d^2 + H^2$.$H^2 - 4dH + 3d^2 = 0$.Given $d = 20$,$H^2 - 4(20)H + 3(20^2) = 0 \Rightarrow H^2 - 80H + 1200 = 0$.$(H - 20)(H - 60) = 0$.Thus,$H = 20 \ m$ or $H = 60 \ m$.

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