Solving for the unknown magnitudes of two vectors when given the equilibrant vector.

I have two vectors with a known direction and unknown magnitudes.

$V_1 = A(m) @ 20°$, $V_2 = B(m) @ 300°$ and $V_3 = 166m @ 140°$ such that $V_1 + V_2 + V_3 = 0$

I'll tell you what I do know. From my point of view $V_3$ is the equilibrant vector being that $V_1$ and $V_2$ add up to be $0$. What I don't know/remember is how to solve for two unknowns. I know there are up to two methods for solving for two unknowns (substitution, and elimination). Also, I can break the vectors down into their individual $x$ and $y$ components.

$V_1 = (A \cos(20°),A \sin(20°))$

$V_2 = (B \cos(60°), -B \sin(60°))$

$V_3 = (-166\cos (40°), 166\sin(40°))$

However, I am unsure of where to go from here. Thanks!

The most straightforward way to do the problem is to write $\vec{V}_1+\vec{V}_2=-\vec{V}_3$ in components and therefore obtain two equations in the two unknowns $A,B$. Some tedious algebra then obtains the correct answer.
But obvious answers are boring. Instead, suppose we take the cross product of $\vec{V}_1$ with $\vec{V}_1+\vec{V}_2+\vec{V}_3=0$. Since the cross product is anticommutative, we can rearrange this to $$\vec{V}_1\times \vec{V}_2 = \vec{V}_3\times \vec{V}_1\implies |V_1||V_2|\sin\theta_{21}=|V_3||V_1|\sin\theta_{13}\implies |V_2|=|V_3|\frac{\sin(\theta_1-\theta_3)}{\sin(\theta_2-\theta_1)}$$ where $\theta_1,\theta_2,\theta_3$ are the angles specified for the three vectors. A similar approach obtains $|V_1|.$
• For a geometric approach, note that the form of the result obtained for $|V_2|$ suggests that the Law of Sines will be useful... Sep 13, 2016 at 20:02