In triangle $MNC$, $MNC = 64^\circ$. $A$ is a point inside the triangle such that $AM = AC$, $AMN = 42^\circ$, $ACN = 26^\circ$, $AMC = x^\circ$ and $CAM = y^\circ$. We want to find the values of $x$ and $y$.

1. Problem Description

In triangle $MNC$, $MNC = 64^\circ$. $A$ is a point inside the triangle such that $AM = AC$, $AMN = 42^\circ$, $ACN = 26^\circ$, $AMC = x^\circ$ and $CAM = y^\circ$. We want to find the values of $x$ and $y$.

2. Solution Steps

First, find the angle $NMC$. We are given that $AMN = 42^\circ$. Since $NMC = 42^\circ$ and $MNC = 64^\circ$, we can find the angle $MCN$.

In triangle $MNC$, we have $NMC + MNC + MCN = 180^\circ$.

So, $42^\circ + 64^\circ + MCN = 180^\circ$.

$MCN = 180^\circ - 42^\circ - 64^\circ = 180^\circ - 106^\circ = 74^\circ$.

Since $ACN = 26^\circ$, we have $MCA = MCN - ACN = 74^\circ - 26^\circ = 48^\circ$.

In triangle $AMC$, $AM = AC$. This implies that triangle $AMC$ is an isosceles triangle.

Hence, $AMC = ACM$. So $x^\circ = 48^\circ$. Thus $x = 48$.

The sum of angles in triangle $AMC$ is $180^\circ$, so $AMC + ACM + CAM = 180^\circ$.

$48^\circ + 48^\circ + y^\circ = 180^\circ$.

$96^\circ + y^\circ = 180^\circ$.

$y = 180 - 96 = 84$.

Therefore, $x = 48$ and $y = 84$.

3. Final Answer

$x = 48$

$y = 84$