Angle between two vectors given magnitudes and difference in magnitude

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Given: |a| = 31 , |b| = 23 and |a – b| = 42 , what is the angle between a and b and the magnitude of |a + b|

How would you solve this, and what kind of diagrams can I use to solve this question?

Edit: I got 78.92 as the angle between |a| and |b|. I got 34.87 as the magnitude of |a + b|

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Your answer for $|\textbf{a} + \textbf{b}|$ is correct. For angle between $\textbf{a}$ and $\textbf{b}$, it should be $\gt 90^0$ as $\cos\theta$ is negative. I think you missed the negative sign in your working.

$|\textbf{a} - \textbf{b}|^2 = |\textbf{a}|^2 + |\textbf{b}|^2 - 2 \ \textbf{a} \cdot \textbf{b} \implies \textbf{a} \cdot \textbf{b} = \frac{1}{2} (31^2 + 23^2 - 42^2) = -137$

So $|\textbf{a} + \textbf{b}|^2 = |\textbf{a}|^2 + |\textbf{b}|^2 + 2 \ \textbf{a} \cdot \textbf{b} = 31^2 + 23^2 - 274 = 1216$

$\implies |\textbf{a} + \textbf{b}| = 8 \sqrt{19}$

For angle between $\textbf{a}$ and $\textbf{b}$,

$\textbf{a} \cdot \textbf{b} = |a| \ |b| \cos\theta \implies \cos\theta = \displaystyle \small - \frac{137}{713}$

$\implies \theta \approx 101.1^0 $