find the angles of a specific triangle 
$P$ is a point inside the equilateral triangle $ABC$. What are the angles of the new triangle created by $AP,BP$ and $CP$, according to $x$ and $y$?

I tried cosine law but it didn't work. $AB$ is equal to $AC$ and $BC$, others are unknown.

Say $AB=BC=CA=a$, $BP=b$, $\angle BAP=z$
It's easy to see that:
$$\angle BCP=180^\circ-(y+\angle CBP)$$
$$\angle CBP=60^\circ-\angle PBA=60^\circ - (180^\circ-x-z)=x+z-120^\circ$$
Therefore:
$$\angle BCP=180^\circ-(y+x+z-120^\circ)=300^\circ-(x+y+z)$$
From sine law applied to triangle ABP:
$$\frac{a}{\sin x}=\frac{b}{\sin z}$$
From sine law applied to triangle BCP:
$$\frac{a}{\sin y}=\frac{b}{\sin (300^\circ-(x+y+z))}$$
From the last two equations you get:
$$\frac{\sin x}{\sin y \sin z}=\frac{1}{\sin (300^\circ-x-y-z))}$$
$$\sin x\sin (300^\circ-x-y-z))=\sin y \sin z$$
You should be able to continue from here...
