jmc

Let $t$ be the number of degrees in the measure of angle $BAC$ (which is what we want to compute).

Since $AX=XY$, we have $\angle AYX=\angle YAX=\angle BAC=t^{\circ }$. Then, since the sum of angles in $△AXY$ is $180^{\circ }$, we have $\angle AXY=(180-2t{)}^{\circ }$.

Angles $\angle AXY$ and $\angle BXY$ add to form a straight angle, so they are supplementary; $\angle BXY=(180-(180-2t){)}^{\circ }=(2t{)}^{\circ }$.

Since $XY=YB$, we have $\angle XBY=\angle BXY=(2t{)}^{\circ }$. Since the sum of angles in $△XYB$ is $180^{\circ }$, we have $\angle XYB=(180-4t{)}^{\circ }$.

Angles $\angle AYX$, $\angle XYB$, and $\angle BYC$ add to form a straight angle, so their sum is $180^{\circ }$. Therefore, $\angle BYC=(180-t-(180-4t){)}^{\circ }=(3t{)}^{\circ }$.

Since $YB=BC$, we have $\angle YCB=\angle BYC=(3t{)}^{\circ }$. Since the sum of angles in $△YBC$ is $180^{\circ }$, we have $\angle YBC=(180-6t{)}^{\circ }$.

Finally, $\angle ABC=\angle XBY+\angle YBC=(2t{)}^{\circ }+(180-6t{)}^{\circ }=(180-4t{)}^{\circ }$. We know this is equal to $120^{\circ }$, so we solve the equation $$180-4t=120$$ to obtain $t=\boxed{15}$.