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(D) Let the height of the shorter pole be $h$ and the height of the taller pole be $2h$.Let the distance between the two poles be $x$.The midpoint of

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$\frac{x}{2 \sqrt{2}}$

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(D) Let the height of the shorter pole be $h$ and the height of the taller pole be $2h$.Let the distance between the two poles be $x$.The midpoint of the line joining their feet divides the distance $x$ into two equal parts,each of length $\frac{x}{2}$.Let the angles of elevation from the midpoint be $	heta$ and $90^\circ - 	heta$.For the shorter pole: $	an(90^\circ - 	heta) = \frac{h}{x/2} = \frac{2h}{x}$.Since $	an(90^\circ - 	heta) = \cot 	heta$,we have $\cot 	heta = \frac{2h}{x}$.For the taller pole: $	an 	heta = \frac{2h}{x/2} = \frac{4h}{x}$.Multiplying the two equations: $	an 	heta \cdot \cot 	heta = \left(\frac{4h}{x}ight) \cdot \left(\frac{2h}{x}ight)$.$1 = \frac{8h^2}{x^2}$.$h^2 = \frac{x^2}{8}$.$h = \sqrt{\frac{x^2}{8}} = \frac{x}{2\sqrt{2}}$.

Two poles are $x$ metres apart and the height of one is double than that of the other. If from the midpoint of the line joining their feet,an observer finds the angle of elevation of their tops to be complementary,then the height of the shorter pole is $\ldots \ldots \ldots \ldots$

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