Consider points \(A, B, C\) in the following diagram:
Let \(\theta = \angle BAC\). Then we can \(\cos \theta = \dfrac{x}{\sqrt{2}}\) write
for some value of x. What is x?
A = (1, 1)
B = (2, 3)
C = (4, 0)
AB = sqrt [ (1 - 2)^2+ (3 - 1)^2 ] = sqrt [ 5]
CA = sqrt [(1 - 4)^2 + (1 - 0)^2 ] = sqrt [10]
BC = sqrt [ (2 - 4)^2 + (3 - 0)^2 ] = sqrt [ 13]
Using the Law of Cosines
cos (theta) = ( [13 - 10 - 5]/ [ - 2sqrt (50) ])
arcos = [ -2] / [-2 sqrt (50) ] = theta
arcos = 1 /sqrt (50) = theta
arcos =1 / [5sqrt(2) ] = (1/5)(1/sqrt(2) ) = (1/5)/sqrt (2) = theta
So
cos (theta) = (1/5) / sqrt(2)
So.... x = 1/5