Use the figure to answer the question.

An illustration shows a right inclined transversal passing through two parallel lines labeled upper M and upper N. Four angles are formed at the intersections of the transversal with upper M and upper N.

If M and N are parallel and angle 1 is 135° , what is the sum of angle 6 and angle 8?

(1 point)

°

1 answer

Given that angle 1 is 135° and the lines M and N are parallel, we can analyze the relationships between the angles formed by the transversal.

Angle 1 is an exterior angle on the top side of the transversal. The angle directly opposite angle 1 (angle 5) is also 135° since they are vertical angles and are equal.

Since angles on a straight line sum to 180°, we can find angle 2:

\[ \text{Angle 2} = 180° - \text{Angle 1} = 180° - 135° = 45° \]

Now, looking at angle 2 and angle 6 which are corresponding angles (because they are in the same position relative to the parallel lines and the transversal), we can say:

\[ \text{Angle 6} = \text{Angle 2} = 45° \]

Next, angle 6 is supplementary to angle 8 (because they are on the same line created by the transversal where angle 8 is located). Therefore, we find angle 8 as follows:

\[ \text{Angle 8} = 180° - \text{Angle 6} = 180° - 45° = 135° \]

Now we can find the sum of angle 6 and angle 8:

\[ \text{Sum of angle 6 and angle 8} = \text{Angle 6} + \text{Angle 8} = 45° + 135° = 180° \]

Thus, the sum of angle 6 and angle 8 is:

\[ \boxed{180°} \]