In triangle HIJ, the sum of the interior angles is always 180 degrees. Let's denote the angles at vertices I and J as \( \angle I \) and \( \angle J \), respectively. Given that the exterior angle at vertex H is \( 38^\circ \), we can use the exterior angle theorem, which states that the exterior angle is equal to the sum of the two non-adjacent interior angles.
Thus, we have: \[ \angle H + \angle I = 38^\circ \]
Since the sum of all angles in triangle HIJ is \( 180^\circ \): \[ \angle H + \angle I + \angle J = 180^\circ \]
We can solve for \( \angle J \): \[ \angle J = 180^\circ - \angle H - \angle I \]
By substituting \( \angle H = 38^\circ - \angle I \) into this equation: \[ \angle J = 180^\circ - (38^\circ - \angle I) - \angle I \] \[ \angle J = 180^\circ - 38^\circ + \angle I - \angle I \] \[ \angle J = 142^\circ \]
Now we can calculate the sum of the measures of the interior angles at vertices I and J: \[ \angle I + \angle J = \angle I + 142^\circ \]
Since we don't need the measure of \( \angle I \) individually, we know that: \[ \angle I + \angle J = 180^\circ - \angle H \]
Given that \( \angle H = 38^\circ \), we find: \[ \angle I + \angle J = 180^\circ - 38^\circ = 142^\circ \]
Thus, the sum of the measures of the interior angles at vertex I and vertex J is:
142°
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