A vertical pole (more than 100 , ft high) con | vedclass

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(B) Let the total height of the pole be $h \, ft$. The lower portion is $\frac{h}{3}$ and the upper portion is $\frac{2h}{3}$.Let the point be $P$ at

Vedclass · Accepted Answer

$120$

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(B) Let the total height of the pole be $h \, ft$. The lower portion is $\frac{h}{3}$ and the upper portion is $\frac{2h}{3}$.Let the point be $P$ at a distance of $40 \, ft$ from the foot of the pole.Let $\alpha$ be the angle subtended by the lower portion at $P$,and $\beta$ be the angle subtended by the whole pole at $P$.Then $	an \alpha = \frac{h/3}{40} = \frac{h}{120}$ and $	an \beta = \frac{h}{40} = \frac{3h}{120}$.The upper portion subtends an angle $	heta = 	an^{-1}(\frac{1}{2})$ at $P$,so $	an 	heta = \frac{1}{2}$.Since $	heta = \beta - \alpha$,we have $	an 	heta = 	an(\beta - \alpha) = \frac{	an \beta - 	an \alpha}{1 + 	an \beta 	an \alpha}$.Substituting the values: $\frac{1}{2} = \frac{\frac{3h}{120} - \frac{h}{120}}{1 + (\frac{3h}{120})(\frac{h}{120})} = \frac{\frac{2h}{120}}{1 + \frac{3h^2}{14400}} = \frac{h/60}{1 + \frac{h^2}{4800}}$.$\frac{1}{2} = \frac{h}{60} 	imes \frac{4800}{4800 + h^2} = \frac{80h}{4800 + h^2}$.$4800 + h^2 = 160h \Rightarrow h^2 - 160h + 4800 = 0$.$(h - 120)(h - 40) = 0$.So $h = 120$ or $h = 40$.Since the pole is more than $100 \, ft$ high,$h = 120 \, ft$.

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