The problem asks us to find the size of angle $q$ and angle $r$ in the given triangle. We are given the angles $56^{\circ}$ and $48^{\circ}$ inside the triangle. Angle $q$ is an interior angle of the triangle, and angle $r$ is an exterior angle.

1. Problem Description

The problem asks us to find the size of angle $q$ and angle $r$ in the given triangle. We are given the angles $56^{\circ}$ and $48^{\circ}$ inside the triangle. Angle $q$ is an interior angle of the triangle, and angle $r$ is an exterior angle.

2. Solution Steps

a) To find the size of angle $q$, we use the fact that the sum of the interior angles in a triangle is $180^{\circ}$.

Therefore,

$56^{\circ} + 48^{\circ} + q = 180^{\circ}$

$104^{\circ} + q = 180^{\circ}$

$q = 180^{\circ} - 104^{\circ}$

$q = 76^{\circ}$

b) To find the size of angle $r$, we can use the fact that the angles $q$ and $r$ form a straight line, so their sum is $180^{\circ}$.

$q + r = 180^{\circ}$

Since we found $q = 76^{\circ}$,

$76^{\circ} + r = 180^{\circ}$

$r = 180^{\circ} - 76^{\circ}$

$r = 104^{\circ}$

Alternatively, we can use the exterior angle theorem, which states that the exterior angle of a triangle is equal to the sum of the two opposite interior angles. Thus,

$r = 56^{\circ} + 48^{\circ} = 104^{\circ}$

3. Final Answer

a) The size of angle $q$ is $76^{\circ}$.

b) The size of angle $r$ is $104^{\circ}$.