.

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

1 answer

When two parallel lines \( M \) and \( N \) are cut by a transversal, several angles are formed. The angles located in corresponding positions relative to the transversal and the two parallel lines have relationships governed by angle relationships (such as corresponding angles, alternate interior angles, etc.).

Let's assume that we label the angles formed at the intersections of the transversal and the parallel lines as follows:

  • Angle 1 is at the intersection of the transversal with line \( M \) on one side,
  • Angle 2 is adjacent to angle 1 at line \( M \),
  • Angle 3 is adjacent to angle 6 at line \( N \),
  • Angle 4 is adjacent to angle 5 at line \( N \),
  • and so on.

Given that angle 1 measures \( 135° \), we can find angle 5, which is adjacent to angle 1 and a straight line is formed when angle 1 and angle 5 are combined. Thus:

\[ \text{Angle 5} = 180° - \text{Angle 1} = 180° - 135° = 45° \]

Next, angle 6 is an alternate interior angle to angle 1, since angles 1 and 6 are on opposite sides of the transversal and between the two parallel lines.

Thus: \[ \text{Angle 6} = \text{Angle 1} = 135° \]

Angle 8 is corresponding to angle 5, which we already calculated: \[ \text{Angle 8} = \text{Angle 5} = 45° \]

Now, to find the sum of angle 6 and angle 8, we calculate:

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

So, the sum of angle 6 and angle 8 is \( 180° \).