Let point A be at (1,1). Let point B be at (2,3) and point C at (4,0).

Let theta = angle BAC. Then we can write cos theta = frac{x}{sqrt{2}} for some value of x. What is x?

Guest May 9, 2019

#1**+1 **

Distance from A to C = sqrt [ (1- 4)^2 + (1 - 0)^2 ] = sqrt [ (-3)^2 + (1)^2 ] = sqrt (10)

Distance from A to B = sqrt [ (1 - 2)^2 + (1 - 3)^2 ] = sqrt [ (-1)^2 + (-2)^2 ] = sqrt (5)

Distance fromB to C = sqrt [ (4 - 2)^2 + (3 - 0)^2 ] = sqrt [ 2^2 + 3^2 ] = sqrt [13]

So........using the Law of Cosines we have that

[sqrt (13)]^2 - [sqrt(10)]^2 - [sqrt (5)]^2

________________________________ = cos theta

- 2 sqrt (10)*sqrt(5)

[ 13 - 10 - 5 ]

__________ = cos theta

-2 sqrt (50)

[-2]

________ = cos theta

-2 sqrt (50)

1

_______ = cos theta

sqrt (50)

1

_______ = cos theta

5 sqrt (2)

So x = 1/5

CPhill May 10, 2019