A man is observing,from the top of a tower,a | vedclass

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(C) Let the height of the tower be $h$ and the speed of the boat be $v 	ext{ m/s}$.Let $C$ be the base of the tower.In $	riangle ADC$ (where $D$ is

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$10(\sqrt{3}+1)$

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(C) Let the height of the tower be $h$ and the speed of the boat be $v 	ext{ m/s}$.Let $C$ be the base of the tower.In $	riangle ADC$ (where $D$ is the top of the tower),$	an 30^{\circ} = \frac{h}{AC} \implies AC = h \cot 30^{\circ} = h\sqrt{3}$.In $	riangle BDC$,$	an 45^{\circ} = \frac{h}{BC} \implies BC = h \cot 45^{\circ} = h$.The distance $AB = AC - BC = h\sqrt{3} - h = h(\sqrt{3}-1)$.The boat covers distance $AB$ in $20 	ext{ seconds}$,so the speed $v = \frac{AB}{20} = \frac{h(\sqrt{3}-1)}{20}$.The time taken to travel from $B$ to $C$ is $t = \frac{BC}{v} = \frac{h}{\frac{h(\sqrt{3}-1)}{20}} = \frac{20}{\sqrt{3}-1}$.Rationalizing the denominator: $t = \frac{20(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{20(\sqrt{3}+1)}{3-1} = \frac{20(\sqrt{3}+1)}{2} = 10(\sqrt{3}+1) 	ext{ seconds}$.

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