We are given a triangle with a line segment inside that is parallel to one of the sides. We are given four angle measures: $68^\circ$, $(3x-15)^\circ$, $2x^\circ$, and $(y^2)^\circ$. We want to find the values of $x$ and $y$.

1. Problem Description

We are given a triangle with a line segment inside that is parallel to one of the sides. We are given four angle measures: $68^\circ$, $(3x-15)^\circ$, $2x^\circ$, and $(y^2)^\circ$. We want to find the values of $x$ and $y$.

2. Solution Steps

Since the line segment is parallel to the base of the larger triangle, the angle with measure $2x^\circ$ is corresponding to the $68^\circ$ angle, which implies that $2x = 68$. Solving for $x$:

$2x = 68$

$x = \frac{68}{2}$

$x = 34$

Now, we know the angles in the large triangle are $68^\circ$, $(3x-15)^\circ$, and $(y^2)^\circ$. Since the angles in a triangle add up to $180^\circ$, we have:

$68 + (3x-15) + y^2 = 180$

Substitute $x = 34$:

$68 + (3(34)-15) + y^2 = 180$

$68 + (102 - 15) + y^2 = 180$

$68 + 87 + y^2 = 180$

$155 + y^2 = 180$

$y^2 = 180 - 155$

$y^2 = 25$

Taking the square root of both sides, we get:

$y = \pm 5$

However, since angles are positive values, we have to consider only the positive value for y, then $y = 5$.

3. Final Answer

$x = 34$

$y = 5$