Find the exterior and interior angles x and y for the two triangles shown.

The sum of the interior angles in a triangle is $180^\circ$.

$$60^\circ + 80^\circ + y = 180^\circ$$ $$140^\circ + y = 180^\circ$$ $$y = 180^\circ - 140^\circ$$ $$y = 40^\circ$$

The angle $x$ is an exterior angle of the triangle. An exterior angle is equal to the sum of the two opposite interior angles.

$$x = 60^\circ + 80^\circ$$ $$x = 140^\circ$$

Alternatively, $x$ and $y$ form a linear pair, so $x + y = 180^\circ$.

$$x + 40^\circ = 180^\circ$$ $$x = 180^\circ - 40^\circ$$ $$x = 140^\circ$$

The values are:
$x = \boxed{\text{140}^\circ}$
$y = \boxed{\text{40}^\circ}$

:

The square symbol indicates a right angle, which is $90^\circ$. The sum of the interior angles in a triangle is $180^\circ$.

$$41^\circ + 90^\circ + y = 180^\circ$$ $$131^\circ + y = 180^\circ$$ $$y = 180^\circ - 131^\circ$$ $$y = 49^\circ$$

The angle $x$ is an exterior angle of the triangle. An exterior angle is equal to the sum of the two opposite interior angles.

$$x = 41^\circ + 90^\circ$$ $$x = 131^\circ$$

Alternatively, $x$ and $y$ form a linear pair, so $x + y = 180^\circ$.

$$x + 49^\circ = 180^\circ$$ $$x = 180^\circ - 49^\circ$$ $$x = 131^\circ$$

The values are:
$x = \boxed{\text{131}^\circ}$
$y = \boxed{\text{49}^\circ}$