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\(cos B =\frac {c^2+a^2-b^2}{2ca}\) \(cos β = \frac {19^2+11^2-23^2}{2(19)(11)}=\frac{361+121-529}{418}=\frac{-47}{418}=-0.112\) \( β ≅ 96.45^\circ\) Then, use the law of sines to find the size of the smallest angle \((α)\): \(\frac {sin A}{a}=\frac {sin B}{b}=\frac {sin C}{c}\) \(\frac {sin\ α }{11}=\frac {sin\ 96.45}{23}\) \( sin\ α =\frac{11× sin\ 96.45}{23}=\frac {11 × 0.99}{23}=\frac {10.89}{23}=0.47\) \( α ≅ 28.37^\circ\) […]
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