Question

Two parallel lines are crossed by a transversal.

Horizontal and parallel lines y and z are cut by transversal x. At the intersection of lines y and x, the top right angle is (2 k + 11) degrees. At the intersection of lines z and x, the bottom left angle is 131 degrees.
What is the value of k?

k = 9
k = 20
k = 60
k = 71

Answer 1

To solve for \( k \), we need to understand the relationship between the angles formed by the transversal crossing two parallel lines.

Given:
- The top right angle at the intersection of lines \( y \) and \( x \) is \( (2k + 11) \) degrees.
- The bottom left angle at the intersection of lines \( z \) and \( x \) is \( 131 \) degrees.

Since lines \( y \) and \( z \) are parallel, the angles formed by the transversal \( x \) have specific relationships. The angle at the top right where line \( y \) intersects line \( x \) and the angle at the bottom left where line \( z \) intersects line \( x \) are corresponding angles, which means they are equal.

Thus, we can set up the equation:

\[
(2k + 11) = 131
\]

Next, we solve for \( k \):

1. Subtract \( 11 \) from both sides:
\[
2k = 131 - 11
\]
\[
2k = 120
\]

2. Divide by \( 2 \):
\[
k = \frac{120}{2} = 60
\]

Therefore, the value of \( k \) is:

\[
\boxed{60}
\]