How to find the magnitude of the angle <BAC (3 points)

How to find the magnitude of the angle A=(-6,1)
B=(6,-5)
C=(4,11)

Hi Catherine,

You can find out the vector AC, and its size:

AC = C - A = (4,11) - (-6,1) = (4--6,11-1) = (10,10).

|AC| = sqrt(10^2 + 10^2) = sqrt(200) = 10 sqrt(2).

You can do the same for the vector AB:

AB = B - A = (6,-5) - (-6,1) =(6--6,-5-1) = (12,-6).

|AB| = sqrt(12^2 + (-6)^2) = sqrt(144 + 36) = sqrt(180) = 6 sqrt(5).

You can also find the dot product of AB and AC:

AB * AC = 10 * 12 + 10 * -6 = 120 - 60 = 60.

The angle between two vectors is known to be:

arccos((AB * AC) / (|AB| |AC|)).

= arccos(60 / [10 sqrt(2) * 6 sqrt(5)]).

= arccos(60 / 60 sqrt(10)).

= arccos(1 / sqrt(10)).

= approximately 71.57 degrees or 1.25 radians.