Two vertical poles are at a distance of 6.28 | vedclass

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(A) The limit of resolution of the human eye is approximately $	heta = 1' = \left(\frac{1}{60}ight)^{\circ}$.To see two objects separately,the angl

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$21.6$

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(A) The limit of resolution of the human eye is approximately $	heta = 1' = \left(\frac{1}{60}ight)^{\circ}$.To see two objects separately,the angle subtended by them at the eye must be at least $	heta$.Using the formula $	heta = \frac{d}{r}$,where $d$ is the distance between the poles and $r$ is the distance of the observer from the poles.First,convert $	heta$ into radians: $	heta = \frac{1}{60} 	imes \frac{\pi}{180} \, 	ext{rad}$.Given $d = 6.28 \, m = 6.28 	imes 10^{-3} \, km$.Rearranging the formula for $r$: $r = \frac{d}{	heta}$.Substituting the values: $r = \frac{6.28 	imes 10^{-3}}{\frac{1}{60} 	imes \frac{\pi}{180}}$.Using $\pi \approx 3.14$,we get $r = \frac{6.28 	imes 10^{-3} 	imes 60 	imes 180}{3.14} = 2 	imes 10^{-3} 	imes 60 	imes 180 = 21600 \, m = 21.6 \, km$.

Two vertical poles are at a distance of $6.28 \, m$ from each other. What should be the maximum distance of an observer from the poles to see them separately? (in $km$)

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