Two cables pull on a ring with forces represented by the vectors a and b…

Step 1. The dot product. $\lvert\vec a + \vec b\rvert^2 = s^2 + s^2 + 2\,\vec a\cdot\vec b = s^2$, so $\vec a\cdot\vec b = -\dfrac{s^2}{2}$.

Step 2. The first angle. $\cos\theta = \dfrac{-s^2/2}{s^2} = -\dfrac{1}{2}$, so $\theta = 120^\circ$.

Step 3. The difference. $\lvert\vec a - \vec b\rvert^2 = s^2 + s^2 + s^2 = 3s^2$, so $\lvert\vec a - \vec b\rvert = \sqrt{3}\,s \approx 1.73\,s$.

Step 4. The second angle. $\vec a\cdot(\vec a - \vec b) = s^2 + \dfrac{s^2}{2} = \dfrac{3s^2}{2}$. Then $\cos\varphi = \dfrac{3s^2/2}{s \times \sqrt{3}\,s} = \dfrac{\sqrt{3}}{2}$, so $\varphi = 30^\circ$.