Given a triangle ABC where $\measuredangle BAC = 120^\circ$, $BC = \sqrt{37}$, and area $3\sqrt{3}$, find the lengths AB and AC.

Here is my attempt: enter image description here

At first I thought of doing it by getting BA by $BA = BP - AP$ then use the two lengths and angle formula for triangle area having $\sin \alpha$ as my variable, but the problem is that I cannot define BP in terms of $sin\alpha$, and get an equation with $\cos \alpha$ instead where they have different factors that prevent me from applying any trigonometric identity.
I've also tried doing $3\sqrt{3} = \frac{PC \cdot BA}{2}$, but I run into the same problem.

Any other solution I've tried (like the Pythagorean theorem) leaves me with a trigonometric identity equation where I cannot extract anything useful.



Let $AB=y$,

By cosine rule $$x^2+y^2-2xy \cos 120^\circ = 37$$

From the area, we know that

$$\frac12 xy \sin 120^\circ = 3\sqrt3$$


and $$x^2+y^2+12=37$$


Can you solve for $x$ and $y$ from $(1)$ and $(2)$? (Further hint: I purpose write $25=5^2$, if you are familiar with certain identities, the answer could be very obvious).


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