# Unknown angle between two vectors

## Homework Statement

Scalar product of vectors AB = -6; vector product of AB = 9. Find the angle between vectors A and B.

## Homework Equations

Scalar product: cos(θ)*AB
Vector product: sin(θ)*AB
(sin(θ))/cos(θ) = tanθ

## The Attempt at a Solution

AB = -6/cos(θ)
AB = 9/sin(θ)
9/sin(θ) = -6/cos(θ)
9/-6 = (sin(θ))/cos(θ)
3/-2 = tan(θ)
tan^-1(3/-2) = -56° which I take to mean 56° from each other.
But their answer is 124° which is -56° + 180°. How can this be when the whole time we are dealing with one angle between two vectors?

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3/-2 = tan(θ)
tan θ = negative value means θ can only be 90<θ<180 or 270<θ<360.

I have a question also , why the vector product which is a vector can equal to 9? is that a magnitude of AxB?
Thank you

tan θ = negative value means θ can only be 90<θ<180 or 270<θ<360.
Ohhhh, I see, thanks.

I have a question also , why the vector product which is a vector can equal to 9? is that a magnitude of AxB?
Thank you
Yes, I should have mentioned that.