To find the sum of the measures of the interior angles at vertices I and J in triangle HIJ, we can use the properties of triangles and exterior angles.
- The sum of the interior angles in any triangle is always 180 degrees.
- Given that the exterior angle at vertex H is 38 degrees, we know that this exterior angle is equal to the sum of the two opposite interior angles (angles I and J).
From the exterior angle theorem: \[ \text{Exterior angle at H} = \text{Angle at I} + \text{Angle at J} \] Thus: \[ 38^\circ = \text{Angle at I} + \text{Angle at J} \]
To find the sum of angles at I and J, we use the fact that: \[ \text{Angle at I} + \text{Angle at J} = 180^\circ - \text{Angle at H} \]
Since the exterior angle at H is 38 degrees, we find the interior angle at H: \[ \text{Angle at H} = 180^\circ - 38^\circ = 142^\circ \]
Now for angles I and J: \[ \text{Angle at I} + \text{Angle at J} = 180^\circ - \text{Angle at H} = 180^\circ - 142^\circ = 38^\circ \]
However, we aim for the sum: Since the exterior angle at H (38 degrees) equals the sum of angles I and J, we now clarify: If the exterior angle adds to 38°, we calculate the remaining for all angles within the triangle, which confirms our previously computed estimate gives simply: \[ \text{Angle at I} + \text{Angle at J} = 142^\circ. \]
Therefore, the sum of the measures of the interior angles at vertices I and J is: \[ \boxed{142^\circ}. \]