Spherical triangle: given two angles and one length find another length

Question

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.

Answer 1

First we determine the unknown angle using the law of cosines. Then we can find the two unknown sides using Napier's analogies.

More details: Wikipedia

Answer 2

We have ASA (angle,side,angle) $ C=2 \pi/3, (\cos(a)=1/3,\sin(a)=\sqrt{8}/3),B=\pi/3$. It is easiest to calculate the angle opposite $a$ first, using the supplemental cosine rule, we have \begin{eqnarray*} \cos(A)=-\cos(B) \cos(C) +\sin(B) \sin(C) \cos(a)=- \frac{1}{2} \frac{1}{2} +\frac{\sqrt{3}}{2} \frac{\sqrt{3}}{2}\frac{1}{3} =\frac{1}{2} \end{eqnarray*} So $\cos(A)=\frac{1}{2}$ and $\sin(A) =\frac{\sqrt{3}}{2}$ . Now using the sine rule \begin{eqnarray*} \frac{\sin(b)}{\sin(B)}=\frac{\sin(a)}{\sin(A)} =\frac{\frac{\sqrt{8}}{3}}{\frac{\sqrt{3}}{2}}=\sqrt{\frac{32}{27}} \end{eqnarray*} So $\color{red}{\sin(b)=\sqrt{8}/3}$ and $\color{red}{\cos(b)=-1/3}$. Note that this spherical triangle is isosceles ($a \leftrightarrow c$).