ta có AB=\(\sqrt{\left(-5+3\right)^2+\left(1-5\right)^2}=2\sqrt{5}\)
AC=\(\sqrt{\left(0+3\right)^2+\left(-4-5\right)^2}=3\sqrt{10}\)
BC=\(\sqrt{5^2+\left(-4-1\right)^2}=5\sqrt{2}\)
theo dinh li cos trong tam giac ta co
cosA=\(\frac{AB^2+AC^2-BC^2}{2AC\cdot AB}\)
=\(\frac{\left(2\sqrt{5}\right)^2+\left(3\sqrt{10}\right)^2-\left(5\sqrt{2}\right)^2}{2\cdot3\sqrt{10}\cdot2\sqrt{5}}=\frac{\sqrt{2}}{2}\)
=> BAC=45 do
\(cos\) Â cos (AB,AC)=\(\dfrac{\left(-8\right)\left(-3\right)+\left(-4\right)\left(-9\right)}{\sqrt{\left(-8\right)^2+\left(-4\right)^2}.\sqrt{\left(-3\right)^2+\left(-9\right)^2}}\) =\(\dfrac{\sqrt{2}}{2}\) vậy Â=(AB,AC)=\(45^o\) cos B(BC,BA)=\(\dfrac{5.8+\left(-5\right).4}{\sqrt{5^2+\left(-5\right)^2}.\sqrt{8^2+4^2}}\) =\(\dfrac{\sqrt{10}}{10}\) cos C (CA,CB)=\(\dfrac{3.\left(-5\right)+9.5}{\sqrt{3^2+9^2}.\sqrt{\left(-5\right)^2+5^2}}=\dfrac{\sqrt{5}}{5}\)
