# Find the side length of triangle ABC;

Let ABC be an equilateral triangle, and let P be a point in the interior of the triangle. Given that PA = 3, PB = 4, and PC = 5, find the side length of ABC.

Relatively simple problem I think, but I can't quite get the right way to solve this. Please nothing fancy shmancy involving graphs or who-knows-what. I should be able to solve this with pretty basic trigonometry.

a few thoughts that may be wrong, but people on here ask me to provide my own thoughts first: At first glance I thought Law of Cosines might help, but I didn't go much places with it. I haven't officially learned trig, so I'm not much good at this.

Rotate the triangle 60 degrees around $A$ so that vertex $C$ gets rotated to vertex $B$, and let $P'$ be the image of $P$ under the rotation. Then $P'A=3$, and $\angle PAP' = 60^{\circ}$, so $\triangle PAP'$ is equilateral, and $PP'=3$. Furthermore, $P'B=5$, and $PB=4$, so $\triangle P'PB$ is a 3-4-5 triangle, and $\angle P'PB = 90^{\circ}$. This should tell you that $\angle APB=150^{\circ}$. Take it from there...
Hint: Let $P=(0,0)$ be the origin. We know that A, B, and C are points situated on circles centered in P, of radii $3$, $4$, and $5$, respectively. Let $A=(3,0)$. We must determine the value of angles $\beta$ and $\gamma$, so that $B=(4\cos\beta,~4\sin\beta)$ and $C=(5\cos\gamma,~5\sin\gamma)$ form an equilateral triangle with A.