To find the sum of the measures of the interior angles at vertices I and J of triangle HIJ, we can use the property that the sum of the interior angles in any triangle is always 180 degrees.
Given that there is an exterior angle at vertex H that measures 38 degrees, we can use the relationship between exterior angles and their corresponding interior angles. The exterior angle is equal to the sum of the two opposite interior angles. Thus, we have:
\[ \text{Exterior angle at H} = \text{Interior angle at I} + \text{Interior angle at J} \] \[ 38° = \text{Interior angle at I} + \text{Interior angle at J} \]
Knowing that the sum of all the interior angles in triangle HIJ is: \[ \text{Interior angle at H} + \text{Interior angle at I} + \text{Interior angle at J} = 180° \]
We can express the interior angle at H relative to the other angles: \[ \text{Interior angle at H} = 180° - ( \text{Interior angle at I} + \text{Interior angle at J} ) \] \[ \text{Interior angle at H} = 180° - 38° \] \[ \text{Interior angle at H} = 142° \]
This shows that the two angles at vertices I and J must sum to 38 degrees as established earlier.
Now, to find the sum of the interior angles at vertices I and J:
\[ \text{Interior angle at I} + \text{Interior angle at J} = 38° \]
Since we only need the sum at vertices I and J which is already computed:
The final answer is: 38°
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