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Hey there stay home and stay safe.

According to HomeworkLib guidelines I have to solve only the first question when multiple questions are given.So I am solving the first question.

\triangle ABC $ is a triangle given where\\$ \angle CAB=43.8\degree\\ BC=538 ft\\ AC=338 ft\\ $We'll use sine rule to find the value of AB$\\ \frac{AB}{sin\angle C}=\frac{AC}{sin\angle B}=\frac{BC}{sin\angle A}\\ \Rightarrow \frac{AB}{sin\angle C}=\frac{338}{sin\angle B}=\frac{538}{sin(43.8\degree)}\\ sin\angle B=\frac{338sin(43.8\degree)}{538}=0.419255184\\ \Rightarrow \angle B=24.78757307\\ $We know the sum of all angles of a triangle is 180$\degree\\ So,\angle C=180\degree-(43.8+24.78757307)\degree=111.41242693\degree\\ Now,AB=\frac{BCsin\angle C}{sin\angle A}=\frac{538sin(111.41242693\degree)}{sin(43.8\degree)}=723.6442683\approx723.64\ ft

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