Question

two parallel lines are cut by a transversal as shown below. suppose m ∠1 = 101°. find m ∠6 and m ∠8.

Explanation:

Step1: Identify relationship for ∠6

∠1 and ∠5 are corresponding angles (parallel lines, transversal), so \( m\angle5 = m\angle1 = 101^\circ \). ∠5 and ∠6 are supplementary (linear pair), so \( m\angle6 = 180^\circ - m\angle5 \).
\( m\angle6 = 180^\circ - 101^\circ = 79^\circ \)

Step2: Identify relationship for ∠8

∠1 and ∠8 are alternate exterior angles (parallel lines, transversal), so \( m\angle8 = m\angle1 \)? Wait, no—wait, ∠1 and ∠4 are supplementary, ∠4 and ∠8 are corresponding? Wait, better: ∠1 and ∠3 are vertical? No, let's re-express. ∠1 and ∠5 are corresponding (so ∠5=101°), ∠5 and ∠8 are vertical angles? Wait, no. Wait, ∠6 and ∠8: ∠6 and ∠8 are vertical angles? Wait, no, ∠5 and ∠8 are vertical? Wait, the diagram: top line has ∠1,2,3,4; bottom line ∠5,6,7,8. Transversal crosses them. So ∠1 and ∠5: corresponding (same position), so equal. ∠5 and ∠6: linear pair (supplementary). ∠1 and ∠8: let's see, ∠1 and ∠4 are supplementary (linear pair), ∠4 and ∠8 are corresponding (since top line ∠4, bottom line ∠8, same position relative to transversal), so ∠4 = ∠8. Since ∠1 + ∠4 = 180°, ∠4 = 180 - 101 = 79°, so ∠8 = 79°? Wait, no, earlier for ∠6: ∠5 is 101°, ∠5 and ∠6 are supplementary, so ∠6=79°. Then ∠6 and ∠8: are they vertical angles? Wait, ∠6 and ∠8: ∠6 is adjacent to ∠7, ∠8 is adjacent to ∠7? Wait, no, ∠5 and ∠8: ∠5 is above bottom line, left; ∠8 is below bottom line, left. So ∠5 and ∠8 are vertical angles? Wait, no, vertical angles are opposite each other when two lines intersect. The transversal intersects the bottom line, creating ∠5,6,7,8. So ∠5 and ∠7 are vertical, ∠6 and ∠8 are vertical. Wait, no: when two lines intersect, vertical angles are opposite. So transversal and bottom line: intersection creates ∠5 (top left), ∠6 (top right), ∠7 (bottom right), ∠8 (bottom left). So ∠5 and ∠8 are not vertical. ∠5 and ∠7 are vertical (opposite), ∠6 and ∠8 are vertical (opposite). So if ∠6 is 79°, then ∠8 is also 79°? Wait, but earlier, ∠1 is 101°, ∠8: let's use alternate exterior angles. ∠1 is top left, ∠8 is bottom left (exterior? Wait, top line: ∠1 is above top line, left; bottom line: ∠8 is below bottom line, left. So they are alternate exterior angles. Wait, alternate exterior angles are equal when lines are parallel. Wait, but ∠1 is 101°, alternate exterior angle would be ∠8? Wait, no, alternate exterior angles: ∠1 and ∠8—wait, top line, above, left; bottom line, below, left. So yes, alternate exterior angles. But if lines are parallel, alternate exterior angles are equal? Wait, no, alternate interior angles are equal, alternate exterior angles are equal. Wait, but ∠1 and ∠5 are corresponding (equal), ∠5 and ∠8: are they something? Wait, maybe I made a mistake earlier. Let's start over.

Given two parallel lines, cut by transversal.

- Corresponding angles: equal. So ∠1 and ∠5 (corresponding) → \( m\angle5 = 101^\circ \).
- ∠5 and ∠6: linear pair (supplementary) → \( m\angle6 = 180 - 101 = 79^\circ \).
- ∠1 and ∠8: let's see, ∠1 and ∠4 are supplementary (linear pair: \( m\angle1 + m\angle4 = 180 \)) → \( m\angle4 = 79^\circ \). ∠4 and ∠8 are corresponding angles (since ∠4 is below top line, left; ∠8 is below bottom line, left) → so \( m\angle8 = m\angle4 = 79^\circ \). Alternatively, ∠6 and ∠8 are vertical angles (since they are opposite each other when transversal intersects bottom line: ∠6 is top right, ∠8 is bottom left? No, wait, no. Wait, the bottom line: angles around intersection: ∠5 (top left), ∠6 (top right), ∠7 (bottom right), ∠8 (bottom left). So ∠5 and ∠7 are vertical (opposite), ∠6 and ∠8 are vertical (opposite). So if ∠6 is 79°, then ∠8 is also 79°, because vertical angles are equal. Ah, that's simpler. So ∠6 and ∠8 are vertical angles, so they are equal. And ∠6 is supplementary to ∠5 (which is equal to ∠1, 101°), so ∠6=79°, hence ∠8=79°.

Answer:

\( m\angle6 = 79^\circ \)
\( m\angle8 = 79^\circ \)