Given that triangle ABC has angles \( \angle A = 2^\circ \), \( \angle B = 57^\circ \), and \( \angle C = 49^\circ \), we can find the exterior angle at point A created by line \( p \) being parallel to side BC.
In this scenario, \( \angle 1 \) (the exterior angle) is equal to the sum of the two remote interior angles, which are \( \angle B \) and \( \angle C \). Therefore, we calculate:
\[ m\angle 1 = m\angle B + m\angle C = 57^\circ + 49^\circ = 106^\circ \]
Now, you want to find the sum of \( m\angle 1 \) and \( m\angle 2 \):
\[ m\angle 2 = m\angle A = 2^\circ \]
Thus, we calculate the sum:
\[ m\angle 1 + m\angle 2 = 106^\circ + 2^\circ = 108^\circ \]
Therefore, the sum of the measures of \( \angle 1 \) and \( \angle 2 \) is:
\[ m\angle 1 + m\angle 2 = 108^\circ \]
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