Чтобы найти сумму углов ∠ABC и ∠CAB, нужно:
cos(∠ABC) = (BA · BC) / (|BA| · |BC|)
BA · BC = 1*3 + (-2)*0 = 3
|BA| = sqrt(1^2 + (-2)^2) = sqrt(1 + 4) = sqrt(5)
|BC| = sqrt(3^2 + 0^2) = sqrt(9) = 3
cos(∠ABC) = 3 / (sqrt(5) * 3) = 1 / sqrt(5)
∠ABC = arccos(1 / sqrt(5)) ≈ 63.4°
AC · AB = 2*(-1) + 2*2 = -2 + 4 = 2
|AC| = sqrt(2^2 + 2^2) = sqrt(4 + 4) = sqrt(8) = 2*sqrt(2)
|AB| = sqrt((-1)^2 + 2^2) = sqrt(1 + 4) = sqrt(5)
cos(∠CAB) = 2 / (2*sqrt(2) * sqrt(5)) = 1 / sqrt(10)
∠CAB = arccos(1 / sqrt(10)) ≈ 71.6°
∠ABC + ∠CAB ≈ 63.4° + 71.6° = 135°
Ответ: 135