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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?

 Feb 2, 2019
 #1
avatar+96080 
+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

 

 

cool cool cool

 Feb 2, 2019
edited by CPhill  Feb 2, 2019

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