a: \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA\simeq98.79\left(cm\right)\)
b: Xét ΔBAC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
\(\Leftrightarrow12.23^2+17.55^2-BC^2=cos67^0\cdot2\cdot12.23\cdot17.55\)
hay \(BC\simeq17.02\left(cm\right)\)