find the angles of a specific triangle

Question

$P$ 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 $y$?

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

Answer 1

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...