[tex]\textit{Law of sines} \\\\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\cfrac{sin(B)}{\overline{AC}}=\cfrac{sin(C)}{\overline{AB}}\implies \cfrac{sin(38^o)}{9}=\cfrac{sin(C)}{11}\implies \cfrac{11\cdot sin(38^o)}{9}=sin(C) \\\\\\ sin^{-1}\left[ \cfrac{11\cdot sin(38^o)}{9} \right]=sin^{-1}\left[ sin(C) \right]\implies sin^{-1}\left[ \cfrac{11\cdot sin(38^o)}{9} \right]=\measuredangle C \\\\\\ 48.81^o\approx \measuredangle C~\hspace{10em} therefore\qquad 180-38 - 48.81 ~~\approx ~~\stackrel{\measuredangle A}{93.19^o}[/tex]
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