Question

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
I
and vertex J
J
?

(1 point)
Responses

142°
142
°
142 degrees

128°
128
°
128 degrees

38°
38
°
38 degrees

52°

Answer 1

38°

Answer 2

the answer is 38°

Answer 3

To find the sum of the measures of the interior angles at vertices \( I \) and \( J \) in triangle \( HIJ \), we can use the property that the sum of all the interior angles of any triangle is always \( 180^\circ \).

Given that there is an exterior angle of \( 38^\circ \) at vertex \( H \), we know that the exterior angle is equal to the sum of the two opposite interior angles. Thus:

\[
\text{Exterior angle at } H = \angle I + \angle J
\]

Substituting the known value:

\[
38^\circ = \angle I + \angle J
\]

Now, we want to find the sum of the interior angles at vertices \( I \) and \( J \):

\[
\angle I + \angle J = 38^\circ
\]

However, this is the exterior angle based on an interior angle at \( H \). Since the angles of triangle \( HIJ \) sum to \( 180^\circ \), we can find the measures of angles at \( I \) and \( J \). Let \( \angle H = x \):

So:

\[
\angle H + \angle I + \angle J = 180^\circ
\]

And since \( x \) is the exterior angle, we have:

\[
x = 180^\circ - 38^\circ = 142^\circ
\]

Next, we can calculate \( \angle I + \angle J \):

\[
\angle I + \angle J = 180^\circ - 142^\circ
\]
\[
= 38^\circ
\]

Thus, the sum of the measures of the interior angles at vertex \( I \) and vertex \( J \) is:

\[
\boxed{142^\circ}
\]