Solution

In the given figure, side $$BC$$ of $$\triangle ABC$$ is produced to $$D$$
Consider the $$\triangle ABC$$
We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles
$$\therefore$$ $$\angle ABC+\angle BAC=\angle ACD$$
$${68}^{o}+x={130}^{o}$$
$$x={130}^{o}-{68}^{o}$$
$$x={62}^{o}$$
Also, we know that the sum of all the angles of a triangles is $${180}^{o}$$
$$\therefore$$ $$x+y+{68}^{o}={180}^{o}$$
$${62}^{o}+y+{68}^{o}={180}^{o}$$
$$y+{130}^{o}={180}^{o}$$
$$y={180}^{o}-{130}^{o}$$
$$y={50}^{o}$$
Hence, the value of $$x$$ is $${62}^{o}$$ and value of $$y$$ is $${50}^{o}$$