Use the figure to answer the question.

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?

Answer 1

To determine the sum of angle 6 and angle 8, we need to analyze their relationships given that $M$ and $N$ are parallel lines and angle 1 is ${135}^{\circ }$.

In this scenario, we can use the properties of angles formed by a transversal intersecting parallel lines:

1. Corresponding Angles: Corresponding angles are equal.
2. Alternate Interior Angles: Alternate interior angles are equal.
3. Consecutive Interior Angles (Same-Side Interior Angles): These angles are supplementary (they add up to ${180}^{\circ }$).

If we designate the angles as follows based on their positions:

• Angle 1 is given as ${135}^{\circ }$.

Here's the relationship between the angles:

• Angle 1 (which is ${135}^{\circ }$) and angle 8 are on the same side of the transversal and are supplementary.
• Therefore, angle 1 and angle 8 satisfy the relationship: $$\text{angle 1}+\text{angle 8}={180}^{\circ }$$ Substituting ${135}^{\circ }$ for angle 1: $${135}^{\circ }+\text{angle 8}={180}^{\circ }$$ Solving for angle 8 gives: $$\text{angle 8}={180}^{\circ }-{135}^{\circ }={45}^{\circ }$$

Since angle 6 is vertically opposite to angle 8 (because of the transversal crossings), then:

• Angle 6 is equal to angle 8: $$\text{angle 6}={45}^{\circ }$$

Now, we can find the sum of angle 6 and angle 8: $$\text{Sum of angle 6 and angle 8}=\text{angle 6}+\text{angle 8}={45}^{\circ }+{45}^{\circ }={90}^{\circ }.$$

Thus, the sum of angle 6 and angle 8 is $\boxed{{\displaystyle {90}^{\circ }}}$.

Answer 2

wrong

Answer 3

I apologize for the confusion in my previous response. Let's reanalyze the problem step-by-step given that angle 1 is ${135}^{\circ }$.

Given:

• Transversal intersects two parallel lines $M$ and $N$.
• Angle 1 is ${135}^{\circ }$.

Understand the angles:

1. Angle 1 is one of the angles formed by the transversal intersecting line $M$.
2. Since lines $M$ and $N$ are parallel and angle 1 is at the intersection of the transversal with line $M$, we know:
   • Angle 2 (which is vertically opposite to angle 1) is also ${135}^{\circ }$.
   • Angle 3 (corresponding angle to angle 1 at line $N$) is also ${135}^{\circ }$.
   • Angle 4 (the angle that is supplementary to angle 1) gives: $$\text{Angle 4}={180}^{\circ }-{135}^{\circ }={45}^{\circ }.$$

Next, let’s place the angles:

• If angle 1 is ${135}^{\circ }$:
  • $\text{Angle 2}$ (vertically opposite to angle 1) is ${135}^{\circ }$.
  • $\text{Angle 3}$ (corresponding to angle 1 at line $N$) is ${135}^{\circ }$.
  • $\text{Angle 4}={45}^{\circ }$ (same side interior angle).

Relating with angle 6 and angle 8:

• Angle 8 (which is on the same side as angle 1, and interior to line $N$) is supplementary to angle 3:

  • Therefore, $$\text{Angle 8}={180}^{\circ }-\text{Angle 3}={180}^{\circ }-{135}^{\circ }={45}^{\circ }.$$

• Angle 6, which is vertically opposite to angle 4 (and therefore also equals angle 4), is again:

  • $\text{Angle 6}={45}^{\circ }$.

Final Calculation:

Now, the sum of angle 6 and angle 8 is: $$\text{Angle 6}+\text{Angle 8}={45}^{\circ }+{45}^{\circ }={90}^{\circ }.$$

So, the correct answer is still $\boxed{{\displaystyle {90}^{\circ }}}$. If I've misunderstood any part of your setup or the diagram, please provide further clarification!