\(a,\left\{{}\begin{matrix}\widehat{C}=90^0-40^0=50^0\\BC=\dfrac{AB}{\cos B}\approx7,8\left(cm\right)\\AC=BC\cdot\sin B\approx5\left(cm\right)\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\widehat{B}=90^0-35^0=55^0\\BC=\dfrac{AB}{\cos B}\approx17,4\left(cm\right)\\AC=BC\cdot\sin B\approx14,3\left(cm\right)\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}\widehat{C}=90^0-58^0=32^0\\AB=\sin C\cdot BC\approx10,6\left(cm\right)\\AC=\sin B\cdot BC\approx17\left(cm\right)\end{matrix}\right.\\ d,\left\{{}\begin{matrix}\widehat{B}=90^0-42^0=48^0\\AB=\sin C\cdot BC\approx54,9\left(cm\right)\\AC=\sin B\cdot BC\approx60,9\left(cm\right)\end{matrix}\right.\)
\(e,\left\{{}\begin{matrix}AB=\sqrt{BC^2-AC^2}=4\sqrt{39}\left(cm\right)\\\cos C=\dfrac{AC}{BC}=\dfrac{5}{8}\approx53^0\Rightarrow\widehat{C}\approx53^0\\\widehat{B}\approx90^0-53^0=37^0\end{matrix}\right.\)
\(f,\left\{{}\begin{matrix}BC=\sqrt{AB^2+AC^2}=3\sqrt{85}\left(cm\right)\\\tan B=\dfrac{AC}{AB}=\dfrac{5}{8}\approx\tan32^0\Rightarrow\widehat{B}\approx32^0\\\widehat{C}\approx90^0-32^0=58^0\end{matrix}\right.\)
