Question

Use the law of cosine to find the value of 2⋅2⋅2 cosθ.
A. 7
B. -7
C. 3.5
D. -3.5

Answer 1

\[\large\rm c^2=a^2+b^2-2ab \cos \theta\]Where c is the side opposite our angle theta.
Plugging in these values gives us,\[\large\rm 1^2=2^2+2^2-2\cdot2\cdot2\cdot \cos \theta\]

Answer 2

So they're actually asking you to solve for this orange part here,\[\large\rm 1^2=2^2+2^2-\color{orangered}{2\cdot2\cdot2\cdot \cos \theta}\]

Answer 3

So I did 1=4+4-2*2*2*cosθ
1=8-6*cosθ
1=2*cosθ
1/2=cosθ
then I did the inverse of cosine times 1/2 and got pi/3.

Not really sure if I did that right.
@zepdrix

Answer 4

I don't know why you're taking it that far.
Leave the orange thing completely alone.
They didn't ask you to solve for \(\large\rm \cos\theta\),
they asked you to solve for \(\large\rm 2\cdot2\cdot2\cos\theta\)

Answer 5

\[\large\rm 1^2=2^2+2^2-\color{orangered}{2\cdot2\cdot2\cdot \cos \theta}\]\[\large\rm 1=4+4-\color{orangered}{2\cdot2\cdot2\cdot \cos \theta}\]\[\large\rm 1=8-\color{orangered}{2\cdot2\cdot2\cdot \cos \theta}\]Subtract 8 from both sides,
multiply each side by -1,
and then I guess you're done, ya?

Answer 6

Oh!! I can't believe I didn't see that. Thank you so much haha!

Answer 7

I was just confused on what it wanted me to solve.

Answer 8

:D