Using cross product to find angle between two vectors

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Homework Statement


Find the angle between
[tex]\begin{align*}<br /> \vec{A} = 10\hat{y} + 2\hat{z} \\<br /> and \\<br /> \vec{B} = -4\hat{y}+0.5\hat{z} <br /> \end{align*}[/tex]
using the cross product.

The answer is given to be 161.5 degrees.

Homework Equations


[tex] \left| \vec{A} \times \vec{B} \right| = \left| \vec{A} \right| \left| \vec{B} \right|sin(\theta)<br /> [/tex]

The Attempt at a Solution


[tex] \left| \vec{A} \times \vec{B} \right| =[/tex] [tex]\left|<br /> \begin{array}{ccc}<br /> \hat{x} & \hat{y} & \hat{z} \\<br /> 0 & 10 & 2 \\<br /> 0 & -4 & 0.5<br /> \end{array} \right| = \left| 13\hat{x} \right| = 13[/tex]

The magnitude of A cross B is 13.

Next we find the magnitude of vectors A and B:
[tex]\left| \vec{A} \right| = \sqrt{10^2+2^2} = \sqrt{104} = 10.198039[/tex]
and
[tex]\left| \vec{B} \right| = \sqrt{(-4)^2+(\frac{1}{2})^2} = \sqrt{16.25} = 4.0311289[/tex]

multiplying the previous two answers we get:
41.109609

So now we should have:
[tex]\frac{13}{41.109609} = sin(\theta)[/tex]

Solving for theta, we get:
18.434951 degrees.


This is frustrating: 180-18.434951 = the correct answer. I'm not quite sure where I'm going wrong here.

I must be making the same mistake repeatedly. Another problem was the same thing, but with the numbers changed, and I also got the 180-{the answer I was getting} = {the correct answer}, but when I tried the example using the SAME methodology, I got the correct answer.

Can someone please share some relevant wisdom in my direction?
 
on Phys.org
I can plot them, and I can see the angle, but I'm interested in calculating the angle.
When I use the dot product I get the correct result, but I cannot see where my mistake is while using the cross product.
 
Oh wow; I didn't even consider that the answer wasn't unique.
Thanks!