How to find the angles between three vectors in a 3D space? 
My book asked me to find the resultant of the 3 vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$. I can find the resultant of $\vec{A}$ and $\vec{B}$ or $\vec{A}$ and $\vec{C}$ using $$R=\sqrt{A^2+B^2+2AB\cos\alpha}$$ (since I know $\alpha$), but I don't know how to find the resultant of $\vec{R}$ and $\vec{C}$ (since I don't know the $\alpha$).
How should I approach this math?
 A: As @NinadMunshi said, we don't know the angle between $\vec A$ and $\vec C$, so we can't figure out the resultant of $\vec R$ and $\vec C$. But, under the assumption, that the angle between the vector $\vec B$ and the projection of $\vec C$ on the $z$ axis is 90°, we can add with components.
Let $x$ be on the right, $y$ be up and $z$ be behind us. Then, the vectors are
$$\vec A = (0,6,0)\rm\, N$$
$$\vec B = (5,0,0)\rm\, N$$
$$\vec C = (0,-\sqrt{3},1)\rm\, N$$
$\vec C$ is like that because $\cos{30°}\cdot 2\rm\, N=\sqrt{3}\rm\, N$ and $\sin{30°}\cdot 2\rm\, N=1\rm\, N$.
We can find resultant of $\vec A$ and $\vec B$:
$$\vec R=\vec A + \vec B = (0,6,0)\rm\, N + (5,0,0)\rm\, N = (5,6,0)\rm\, N$$
Let $\vec P$ be the resultant of $\vec R$ and $\vec C$. Then,
$$\vec P=\vec R + \vec C = (5,6,0)\rm\, N + (0,-\sqrt{3},1)\rm\, N = \underline{(5,6-\sqrt{3},1)\rm\, N}$$
But yeah, that's under the assumption that the angle between the vector $\vec B$ and the projection of $\vec C$ on the $z$ axis is 90°. Otherwise, we can't answer it (not enough information).
