Two vertical poles are 150 m apart and the h | vedclass

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(B) Let the height of the shorter pole be $h$ and the height of the taller pole be $3h$. The distance between the poles is $150 \ m$. The observer is

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$25 \sqrt{3}$

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(B) Let the height of the shorter pole be $h$ and the height of the taller pole be $3h$. The distance between the poles is $150 \ m$. The observer is at the midpoint,so the distance from the observer to each pole is $75 \ m$.Let the angle of elevation of the shorter pole be $	heta$. Since the angles are complementary,the angle of elevation of the taller pole is $90^\circ - 	heta$.For the shorter pole: $	an 	heta = \frac{h}{75}$.For the taller pole: $	an(90^\circ - 	heta) = \frac{3h}{75} \Rightarrow \cot 	heta = \frac{3h}{75} = \frac{h}{25}$.Multiplying the two equations: $	an 	heta \cdot \cot 	heta = \left(\frac{h}{75}ight) \cdot \left(\frac{h}{25}ight)$.$1 = \frac{h^2}{1875} \Rightarrow h^2 = 1875$.$h = \sqrt{1875} = \sqrt{625 	imes 3} = 25 \sqrt{3} \ m$.

Two vertical poles are $150 \ m$ apart and the height of one is three times that of the other. If from the middle point of the line joining their feet,an observer finds the angles of elevation of their tops to be complementary,then the height of the shorter pole (in meters) is

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