A(1;0)
B(0;1)
C(0;0)
M(1/3; 1/3)
vector{АМ}=(-2/3;1/3); |vector{АМ}|=sqrt((-2/3)^2+(1/3)^2)=sqrt(5/9)
vector{BМ}=(1/3;-2/3); |vector{BМ}|=sqrt((1/3)^2+(-/3)^2)=sqrt(5/9)
∠ BMA= ∠ (vector{АМ},vector{BМ})
cos∠ (vector{АМ},vector{BМ})=(vector{АМ}*vector{BМ})/(|vector{АМ}|*|vector{BМ}|)=
=(-2/9)+(-2/9)/sqrt(5/9)*sqrt(5/9)=-4/5
∠ (vector{АМ},vector{BМ})= arccos(-4/5)