# B Deriving law of sines from cross product

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1. Aug 28, 2016

### Mr Davis 97

I am trying to derive the law of signs from the cross product.

First, we have three vectors $\vec{A} ~\vec{B} ~\vec{C}$ such that $\vec{A} + \vec{B} + \vec{C} = 0$. This creates a triangle. Then, we label the angles opposite the respective sides as a, b, and c. I am not sure where to go from here... We could take the cross product of each combination of $\vec{A}$ and $\vec{B}$, but these cross products aren't necessarily equal, so can't set them equal to derive the law of sines... Any help would be appreciated.

2. Aug 28, 2016

If you take $C \times (A+B+C)=0$ and $C \times C=0$ Then $|A||C|sin(\theta_1)=|B||C|sin(\theta_2)$ (neglecting a minus sign which is of little significance).Result is $sin(\theta_1)/|B|=sin(\theta_2)/|A|$

3. Aug 28, 2016

### Mr Davis 97

How do I get $sin(\theta_3)/|C|$ in there? That was my main problem.

4. Aug 28, 2016

That will show up if you take $B \times (A+B+C)=0$ (or $A \times (A+B+C)=0$). You can only do two angles at a time by this method. Comparing two of the equations will tie the 3rd one in there. e.g. $B \times$ gives you $sin(\theta_2)/|A|=sin(\theta_3)/|C|$.