Question

in part (a), prove that h || k. in parts (b) and (c), solve for x given that q || p and s || r, respectively
a. diagram with h, k, angles x, y, ( x + y = 180^circ ) b. diagram with angles ( 31^circ ), ( 28^circ ) c. diagram with ( 142^circ ), ( 141^circ )
a. why is h || k?
alternate interior angles are congruent

• since ( x + y = 180^circ ) and if ( x ) is the corresponding angle to ( x ), then ( x ) is supplementary to ( y ) and ( x + y = 180^circ ). therefore ( x ) and ( x ) both have the same measure, ( 180 - y )
• since ( x + y = 180^circ ) and if ( x ) is the corresponding angle to ( x ), then ( x ) is complementary to ( y ) and ( x + y = 180^circ ). therefore ( x ) and ( x ) both have the same measure, ( 90 - y )
• since ( x ) and ( y ) are supplementary angles, the lines are parallel

b. solve for x
( x = square^circ )

Answer:

Part (a): Prove \( h \parallel k \)

Step 1: Recall Parallel Line Theorems

When two lines are cut by a transversal, if consecutive interior angles are supplementary (sum to \( 180^\circ \)), the lines are parallel. Here, \( x + y = 180^\circ \), and \( x, y \) are consecutive interior angles formed by transversal with \( h, k \).

Step 2: Apply Consecutive Interior Angles Theorem

Since \( x + y = 180^\circ \) (given), by the Consecutive Interior Angles Theorem, \( h \parallel k \).

Part (b): Solve for \( x \) (Given \( q \parallel p \))

Step 1: Identify Angle Relationship

When two parallel lines are cut by a transversal, the sum of the angles in a triangle (or using alternate interior angles) can be used. The angles given are \( 31^\circ \), \( 24^\circ \), and \( x \) forms a triangle-like structure (or using the exterior angle property). Wait, actually, when \( q \parallel p \), the angle \( x \) is the sum of the two given angles (by the property of parallel lines and transversal, forming a "Z" or "triangle" with the angles).

Step 2: Calculate \( x \)

Using the property that the angle \( x \) is equal to the sum of the two non-adjacent angles (since \( q \parallel p \), alternate interior angles and triangle angle sum). So \( x = 31^\circ + 24^\circ \).

\[
x = 31 + 24 = 55
\]

Part (c): Solve for \( x \) (Given \( s \parallel t \))

Step 1: Recall Supplementary Angles and Parallel Lines

When \( s \parallel t \), we can use the property that the sum of angles around the transversal and the parallel lines. The angles given are \( 142^\circ \) and \( 141^\circ \), and \( x \) is related to these by the formula: \( x + 142^\circ + 141^\circ - 180^\circ \times 2 = 180^\circ \)? Wait, better: Draw a line through \( B \) parallel to \( s \) and \( t \), creating two alternate interior angles.

Step 2: Calculate \( x \)

Let the line through \( B \) be parallel to \( s \) and \( t \). Then, the angle adjacent to \( 142^\circ \) is \( 180^\circ - 142^\circ = 38^\circ \), and the angle adjacent to \( 141^\circ \) is \( 180^\circ - 141^\circ = 39^\circ \). Then \( x = 38^\circ + 39^\circ = 77^\circ \)? Wait, no, correct approach: The sum of the angles on one side of a straight line is \( 180^\circ \), and for parallel lines, the total angle around \( B \) and the transversal:

Wait, actually, the formula for \( x \) when two parallel lines are cut by a transversal with a "zig-zag" line (like a polygon) is \( x = 180^\circ - (180^\circ - 142^\circ) - (180^\circ - 141^\circ) \)? No, better:

The sum of the interior angles for the "path" through \( A, B, C \): \( x + (180 - 142) + (180 - 141) = 180 \)? No, wait, when \( s \parallel t \), the angle \( x \) is equal to \( 180^\circ - (180^\circ - 142^\circ) - (180^\circ - 141^\circ) \)? Wait, let's do it step by step:

1. The angle at \( A \) with the parallel line: \( 180^\circ - 142^\circ = 38^\circ \) (supplementary angle).
2. The angle at \( C \) with the parallel line: \( 180^\circ - 141^\circ = 39^\circ \) (supplementary angle).
3. Then, in the triangle (or the path), \( x + 38^\circ + 39^\circ = 180^\circ \)? No, wait, the line through \( B \) is parallel to \( s \) and \( t \), so the two angles (38° and 39°) are alternate interior angles, and \( x = 38^\circ + 39^\circ = 77^\circ \)? Wait, no, actually, the correct formula is \( x = 142^\circ + 141^\circ - 180^\circ \). Let's check:

\( 142 + 141 = 283 \), \( 283 - 180 = 103 \)? Wait, no, I think I made a mistake. Let's use the formula for a polygon with two parallel lines: the sum of the angles on the "outside" is related to the angle inside.

Wait, the correct way: When you have two parallel lines \( s \) and \( t \), and a transversal with a "bend" at \( B \), the angle \( x \) is equal to \( 180^\circ - (180^\circ - 142^\circ) - (180^\circ - 141^\circ) \)? No, let's use the concept of consecutive interior angles.

Alternatively, the sum of the angles \( x \), \( (180 - 142) \), and \( (180 - 141) \) should be \( 180^\circ \) (since it's a triangle-like figure). Wait, \( (180 - 142) = 38 \), \( (180 - 141) = 39 \), so \( x + 38 + 39 = 180 \)? No, that would make \( x = 103 \), but that's not right. Wait, no, the correct formula is \( x = 142 + 141 - 180 \). Let's calculate: \( 142 + 141 = 283 \), \( 283 - 180 = 103 \). Wait, maybe that's correct. Let's verify:

If \( s \parallel t \), then the angle at \( A \) (142°) and the angle adjacent to \( x \) (let's say \( a \)) are supplementary: \( a = 180 - 142 = 38 \). Similarly, the angle at \( C \) (141°) and the angle adjacent to \( x \) (let's say \( b \)) are supplementary: \( b = 180 - 141 = 39 \). Then, in the triangle formed by \( A, B, C \), \( x + a + b = 180 \)? No, that would be \( x + 38 + 39 = 180 \), so \( x = 103 \). Yes, that makes sense. So \( x = 103^\circ \).

Final Answers

• Part (a): \( h \parallel k \) (proven by Consecutive Interior Angles Theorem, since \( x + y = 180^\circ \)).
• Part (b): \( x = \boxed{55} \)
• Part (c): \( x = \boxed{103} \)