Just the sides given, can you find the angle?

I use the cosine rule I think solve like this is easier.

We can use Heron's formula expressed in another way, to find the area of $\triangle ABC$ :

$4[ABC]=\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$

Now, we substitute the expression we are given:

$4[ABC]=\sqrt{(a^2+b^2+c^2)^2-4c^2(a^2+b^2)}$

Try to simplify it:

$4[ABC]=\sqrt{a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2-4a^2c^2-4b^2c^2} \\ 4[ABC]=\sqrt{a^4+b^4+c^4+2a^2b^2-2a^2c^2-2b^2c^2} \\ 4[ABC]=\sqrt{(a^2+b^2-c^2)^2} \\ 4[ABC]=|a^2+b^2-c^2|$

Now, by law of sines, we now that $[ABC]=\dfrac{ab\sin \angle C}{2}$ , so $\sin \angle C=\dfrac{2[ABC]}{ab}$ , and substituting the area:

$\sin \angle C=\dfrac{|a^2+b^2-c^2|}{2ab}$

By law of cosines, we know that $\cos \angle C=\dfrac{a^2+b^2-c^2}{2ab}$ , and since $a$ and $b$ are positive, we can conclude that:

$\sin \angle C=|\cos \angle C|$

All we have to do is solve that equation:

$\sin^2 \angle C=\cos^2 \angle C \\ \tan^2 \angle C=1 \\ \tan \angle C=\pm 1 \\ \angle C=\dfrac{\pi}{4} \text{ or } \angle C=\dfrac{3\pi}{4}$

We see that both of these solutions work, so those are the possible values.