The sum of the measures of the interior angles at vertex \( I \) and vertex \( J \) in triangle \( HIJ \) can be found using the fact that the sum of the interior angles of a triangle is always \( 180^\circ \).
Additionally, we know that the exterior angle at vertex \( H \) is \( 38^\circ \). The exterior angle is equal to the sum of the two opposite interior angles. Thus, we can set up the following equation:
\[ \text{Exterior angle (at H)} = \text{Interior angle (at I)} + \text{Interior angle (at J)} \]
Let's denote the interior angles at vertices \( I \) and \( J \) as \( \angle I \) and \( \angle J \) respectively. Therefore:
\[ 38^\circ = \angle I + \angle J \]
We also know that the sum of all interior angles in triangle \( HIJ \) is:
\[ \angle H + \angle I + \angle J = 180^\circ \]
Now, we can determine \( \angle H \) by rearranging the above equation. The interior angle at \( H \) can be expressed as:
\[ \angle H = 180^\circ - (\angle I + \angle J) \]
Substituting \( \angle I + \angle J = 38^\circ \) into this equation gives:
\[ \angle H = 180^\circ - 38^\circ = 142^\circ \]
Now, we also know that:
\[ \angle I + \angle J = 38^\circ \]
So, the sum of the measures of the interior angles at vertex \( I \) and vertex \( J \) is:
\[ \angle I + \angle J = 38^\circ \]
This means the sum of the measures of the interior angles at vertex \( I \) and vertex \( J \) is
\(\boxed{142^\circ}\)
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