Use the image to answer the question.

An illustration shows triangle uppercase H uppercase I uppercase J. Uppercase H is the left vertex, uppercase I is the top vertex, and uppercase J is the right vertex. A line extending outward from uppercase H forms an exterior angle of 38 degrees.

What is the sum of the measures of the interior angles at vertex I and vertex J ?

(1 point)
Responses

38°
38 degrees

142°
142 degrees

52°
52 degrees

128°
128 degrees
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1 answer

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°