\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(4;2\right)\\\overrightarrow{AC}=\left(7;1\right)\\\overrightarrow{BC}=\left(3;-1\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}AB=2\sqrt{5}\\AC=5\sqrt{2}\\BC=\sqrt{10}\end{matrix}\right.\)
a. Ta có \(\frac{4}{7}\ne\frac{2}{1}\Rightarrow A;B;C\) ko thẳng hàng.
b.
\(cosB=\frac{AB^2+BC^2-AC^2}{2AB.BC}=-\frac{\sqrt{2}}{2}\Rightarrow B=135^0\)
\(cosC=\frac{AC^2+BC^2-AB^2}{2AC.BC}=\frac{2\sqrt{5}}{5}\Rightarrow C\approx26^033'\)