In any triangle, the sum of the interior angles is always 180 degrees. For triangle HIJ, you have an exterior angle at vertex H which measures 38 degrees.
The exterior angle is also equal to the sum of the two opposite interior angles. Therefore:
\[ \text{Exterior angle} = \text{Angle at I} + \text{Angle at J} \]
Given that the measure of the exterior angle at vertex H is 38 degrees, we can write:
\[ \text{Angle at I} + \text{Angle at J} = 38° \]
However, we want the sum of the measures of the interior angles at vertex I and vertex J specifically.
Since we also know that:
\[ \text{Angle at H} + \text{Angle at I} + \text{Angle at J} = 180° \]
Let’s denote the angle at H as \(H\). Therefore we can express this as:
\[ H + \text{Angle at I} + \text{Angle at J} = 180° \]
Since we also know that:
\[ \text{Angle at H} = \text{Exterior angle} = 38° \]
Substituting this value into the equation gives:
\[ 38° + \text{Angle at I} + \text{Angle at J} = 180° \]
Subtracting 38° from both sides gives:
\[ \text{Angle at I} + \text{Angle at J} = 180° - 38° \]
\[ \text{Angle at I} + \text{Angle at J} = 142° \]
So the sum of the measures of the interior angles at vertex I and vertex J is:
142°