Find the scalar value and direction of the electric field

[HJKL]

Homework Statement

Two charged balls are placed in point A and B and the distance between them is 9,54cm. Each of the balls are charged with 8,0 x 10^-8 C. Find the scalar value and direction of the electric field in point C placed 5 cm from A and 6 cm from B.

Homework Equations

Cosine Rule: a^2 = b^2 + c^2 - 2ab x cosθ
E = K_e x (q/r^2)

The Attempt at a Solution

[/B]
I have calculated the electric field of A and B.
E_A = 287680 N/C
E_B = 199778 N/C

After that I used cosine rule to calculate E_C and got E_C = 255341 N/C. This is correct.

But when I try finding the direction I calculate the angle between E_C and E_A (β) like this:
β = cos^-1 ((E_A^2 + E_B^2-E_C^2)/(2E_AE_B)) = 60 degrees

The solution says that I should calculate the angle between E_C and E_A (β), but that the equation looks like this:
β = cos^-1 ((E_C^2 + E_A^2-E_B^2)/(2E_CE_A)) = 42,65 degrees

Why am I supposed to use that equation?

 

[Nguyen Son]

The wrong thing is your Cosine rule at the first line of your "relevant equation". Please check again

 

[HJKL]

Oh! No, its not that. I just wrote it down wrong here I've used c^2 = a^2 + b^2 - 2ab * cosθ, and I get the first part right. I just don't get why I'm supposted to use this equation β = cos^-1 ((E_C^2 + E_A^2-E_B^2)/(2E_C E_A)) to find the direction.

 

[Nguyen Son]

But...the answer in the solution is a correct answer, just compare to the Cosine rule
You wrote down the correct cosine rule, as above, but when you put $E_{A}, E_{B}, E_{C}$ into your calculation, I see that you put them in the wrong place

 

[HJKL]

I was a bit confused about which angles goes with which side, but I get it now. Thank you!

 

[Nguyen Son]

Hope this figure can help you