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 at a distance $d =

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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 at a distance $d = 20 \, m$ from the base. Let $\alpha$ be the angle subtended by the lower part and $\beta$ be the angle subtended by the whole pole at the point. The angle subtended by the upper part is $	heta = \beta - \alpha$,where $	an 	heta = \frac{1}{2}$.We have $	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}$,we get:$\frac{1}{2} = \frac{\frac{H}{d} - \frac{H}{3d}}{1 + (\frac{H}{d})(\frac{H}{3d})} = \frac{\frac{2H}{3d}}{1 + \frac{H^2}{3d^2}} = \frac{2Hd}{3d^2 + H^2}$.Thus,$3d^2 + H^2 = 4Hd$,which simplifies to $H^2 - 4dH + 3d^2 = 0$.Substituting $d = 20$,we get $H^2 - 4(20)H + 3(20^2) = 0$,so $H^2 - 80H + 1200 = 0$.Factoring the quadratic equation: $(H - 20)(H - 60) = 0$.Therefore,the possible heights of the pole are $H = 20 \, m$ and $H = 60 \, m$.

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