Let $[u,v,w]$ be a positively oriented spherical triangle in $S^2$ If $a=\cos^{-1}(\frac13)$, $\beta = \frac{2\pi}{3}$, $\gamma = \frac{\pi}{3}$, find $b$.

We have that:

1. $a=\cos^{-1}(\frac13)\implies u\cdot v = \frac13$
2. $\beta = \frac{2\pi}{3}\implies \cos\left(\frac{2\pi}{3}\right)=-\frac12=\frac{(u\cdot w)-(u\cdot v)(u\cdot w)}{|u\times v||v\times w|}$
3. $\gamma = \frac{\pi}3\implies \cos\left(\frac{\pi}3\right)=\frac12=\frac{(v\cdot u)-(w\cdot v)(u\cdot w)}{|w\times v||u\times w|}$

Hence,

$$\frac{(v\cdot w)-\frac13(u\cdot w)}{|u\times v|}=\frac{\frac13 - (w\cdot v)(u\cdot w)}{|u\times w|}$$

But what do I do next? Have no idea at all. Can someone please help? Of course, it is clear that one needs to find v\cdot w and then $\cos^{-1}(v\cdot w)$, but I don't see how to do so.