cứu em với ạ

\(\overrightarrow{AB}=\left(4;0\right)\)

\(\overrightarrow{AC}=\left(3;3\right)\)

\(\cos\widehat{A}=\dfrac{4\cdot3+3\cdot0}{\sqrt{4^2}+\sqrt{3^2+3^2}}=\dfrac{12}{4+3\sqrt{2}}=-24+18\sqrt{2}\)

=>Đề sai rồi bạn

\(AB=\sqrt{\left(3-1\right)^2+\left(4-2\right)^2}=2\sqrt{2}\)

\(AC=\sqrt{\left(6-1\right)^2+\left(-5-2\right)^2}=\sqrt{74}\)

\(BC=\sqrt{\left(6-3\right)^2+\left(-5-4\right)^2}=3\sqrt{10}\)

\(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{-\sqrt{37}}{37}\)

=>góc A=99 độ

AB/sinC=AC/sinB=BC/sinA

=>\(\dfrac{3\sqrt{10}}{sin99}=\dfrac{2\sqrt{2}}{sinC}=\dfrac{\sqrt{74}}{sinB}\)

=>góc C=17 độ; góc B=64 độ

tam giác ABC là tam giác vuông cân ==> góc A = 90 độ, góc B = góc C = 45 độ

a: \(\left\{{}\begin{matrix}x_G=\dfrac{2+4+2}{3}=\dfrac{8}{3}\\y_G=\dfrac{1+0+3}{3}=\dfrac{4}{3}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_I=\dfrac{2+4}{2}=3\\y_I=\dfrac{1+0}{2}=\dfrac{1}{2}\end{matrix}\right.\)

Ta có B A → = 3 ; − 1  và B C → = − 4 ; − 2 . Suy ra:

cos B A → , B C → = B A → . B C → B A → . B C → = 3. − 4 + − 1 . − 2 9 + 1 . 16 + 4 = − 2 2 ⇒ B ^ = B A → , B C → = 135 O .  

Chọn D.