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(C) Let $AB$ be the wall and $AD$ be the initial position of the ladder. The foot of the ladder is at $D$,where $BD = 10 \, ft$.In $	riangle ABD$,$\a

Vedclass · Accepted Answer

$7.32$

Vedclass · Answer

(C) Let $AB$ be the wall and $AD$ be the initial position of the ladder. The foot of the ladder is at $D$,where $BD = 10 \, ft$.In $	riangle ABD$,$\angle ADB = 60^{\circ}$ and $\angle B = 90^{\circ}$.$\cos 60^{\circ} = \frac{BD}{AD} \implies \frac{1}{2} = \frac{10}{AD} \implies AD = 20 \, ft$.$	an 60^{\circ} = \frac{AB}{BD} \implies \sqrt{3} = \frac{AB}{10} \implies AB = 10\sqrt{3} \, ft$.When the ladder slips,the new position is $EC$,where $EC = AD = 20 \, ft$ (length of the ladder remains constant).In $	riangle EBC$,$\angle ECB = 30^{\circ}$ and $\angle B = 90^{\circ}$.$\sin 30^{\circ} = \frac{EB}{EC} \implies \frac{1}{2} = \frac{EB}{20} \implies EB = 10 \, ft$.The distance the ladder slipped down from the top of the wall is $AE = AB - EB = 10\sqrt{3} - 10 = 10(\sqrt{3} - 1) \, ft$.Using $\sqrt{3} \approx 1.732$,$AE = 10(1.732 - 1) = 10(0.732) = 7.32 \, ft$.

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