A person of height 6 ft wants to pluck a fru | vedclass

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(D) Let the height of the person be $CD = 6 \ ft$.Let the height of the tree be $AB = \frac{26}{3} \ ft$.The distance between the person and the tree

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(D) Let the height of the person be $CD = 6 \ ft$.Let the height of the tree be $AB = \frac{26}{3} \ ft$.The distance between the person and the tree is $BD = \frac{8}{\sqrt{3}} \ ft$.Let $E$ be a point on $AB$ such that $CE$ is parallel to $BD$. Thus,$BE = CD = 6 \ ft$.The height of the fruit above the person's eye level is $AE = AB - BE = \frac{26}{3} - 6 = \frac{26 - 18}{3} = \frac{8}{3} \ ft$.The horizontal distance is $CE = BD = \frac{8}{\sqrt{3}} \ ft$.In the right-angled triangle $\Delta AEC$,$	an 	heta = \frac{AE}{CE}$.$	an 	heta = \frac{8/3}{8/\sqrt{3}} = \frac{8}{3} 	imes \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.Since $	an 	heta = \frac{1}{\sqrt{3}}$,we have $	heta = 30^{\circ}$.

$A$ person of height $6 \ ft$ wants to pluck a fruit which is on a $\frac{26}{3} \ ft$ high tree. If the person is standing $\frac{8}{\sqrt{3}} \ ft$ away from the base of the tree,then at what angle should he throw a stone so that it hits the fruit (in $^{\circ}$)?

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