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?

Guest Feb 2, 2019

#1**+1 **

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

CPhill Feb 2, 2019