← Back to the Parallel Lines Transversal Angles worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-CO.C.9 · Level: Grade 10 Geometry
Problem 1 Name the angle pairs
- Between the lines, opposite sides of the transversal: alternate interior.
- Same position at the two intersections: corresponding.
- Between the lines, same side of the transversal: same-side interior.
- Outside the lines, opposite sides: alternate exterior.
(a) alternate interior, (b) corresponding, (c) same-side interior, (d) alternate exterior
Problem 2 Alternate interior and same-side
∠3 and ∠6 are alternate interior angles, so m∠6 = 74°. ∠3 and ∠5 are same-side interior angles, so m∠5 = 180° − 74° = 106°.
m∠6 = 74°, m∠5 = 106°
Problem 3 Same-side interior angles
∠4 and ∠6 are same-side interior angles, so they add to 180°. Then m∠6 = 180° − 127° = 53°.
53°
Problem 4 Alternate exterior angles
∠1 and ∠8 are alternate exterior angles, so m∠8 = 133°. ∠2 forms a linear pair with ∠1, so m∠2 = 47°, and ∠7 is the alternate exterior partner of ∠2, so m∠7 = 47°.
m∠8 = 133°, m∠7 = 47°
Problem 5 Corresponding angles with algebra
Corresponding angles are congruent: 7x − 8 = 4x + 37, so 3x = 45 and x = 15.
Each angle is 7(15) − 8 = 97°, and 4(15) + 37 = 97°.
x = 15; both angles measure 97°
Problem 6 Same-side interior with algebra
Same-side interior angles are supplementary: (4x + 12) + (6x − 2) = 180, so 10x + 10 = 180 and x = 17.
The angles are 4(17) + 12 = 80° and 6(17) − 2 = 100°. Check: 80 + 100 = 180.
x = 17; 80° and 100°
Problem 7 Alternate interior with algebra
Alternate interior angles are congruent: 2x + 31 = 5x − 17, so 48 = 3x and x = 16.
Each angle is 2(16) + 31 = 63°, and 5(16) − 17 = 63°.
x = 16; 63°
Problem 8 All eight angles
∠2 and ∠7 are congruent to ∠6 (corresponding or vertical), and ∠3 is its alternate interior partner, so these measure 49°. The remaining angles each form a linear pair with a 49° angle: 180° − 49° = 131°.
∠2, ∠3, ∠7 = 49°; ∠1, ∠4, ∠5, ∠8 = 131°
Problem 9 Are the lines parallel
- Equal corresponding angles: the lines are parallel.
- 98° + 92° = 190°, not 180°, so the lines are not parallel.
- Equal alternate exterior angles: the lines are parallel.
(a) parallel, (b) not parallel, (c) parallel
Problem 10 Value of x for parallel lines
For parallel lines the corresponding angles must be equal: 5x + 4 = 7x − 26, so 30 = 2x and x = 15.
Each angle is 5(15) + 4 = 79°, and 7(15) − 26 = 79°.
x = 15; 79°
Problem 11 Same-side interior converse
The angles must be supplementary: 6x + 7 + 4x + 13 = 180, so 10x + 20 = 180 and x = 16.
The angles measure 6(16) + 7 = 103° and 4(16) + 13 = 77°, and 103 + 77 = 180.
x = 16
Problem 12 Perpendicular bisector equation
By the Perpendicular Bisector Theorem, PA = PB: 5x − 14 = 3x + 10, so 2x = 24 and x = 12.
PA = 5(12) − 14 = 46, and PB = 3(12) + 10 = 46.
x = 12; PA = 46
Problem 13 Alternate interior proof
| Statement | Reason |
|---|---|
| 1. a ∥ b | Given |
| 2. ∠2 ≅ ∠6 | Corresponding angles are congruent (parallel lines) |
| 3. ∠2 ≅ ∠3 | Vertical angles are congruent |
| 4. ∠3 ≅ ∠6 | Transitive Property of congruence |
∠3 ≅ ∠6
Problem 14 Prove lines parallel
| Statement | Reason |
|---|---|
| 1. ∠4 and ∠6 are supplementary | Given |
| 2. ∠2 and ∠4 form a linear pair | Definition of linear pair |
| 3. ∠2 and ∠4 are supplementary | Linear pairs are supplementary |
| 4. ∠2 ≅ ∠6 | Angles supplementary to the same angle (∠4) are congruent |
| 5. a ∥ b | Converse of the Corresponding Angles Theorem |
a ∥ b
Problem 15 Crossing streets
- This angle is corresponding to the 71° angle, so it measures 71°.
- It forms a linear pair with the angle in part (a): 180° − 71° = 109°.
- It is vertical to the 71° angle, so it measures 71°.
(a) 71°, (b) 109°, (c) 71°
Problem 16 Equidistant lamp post
Equidistance gives 3x + 8 = 5x − 6, so 14 = 2x and x = 7. Then LB = LF = 29 feet.
The midpoint M is 20 feet from each end, and LM ⊥ BF, so the Pythagorean Theorem gives LM = √(29² − 20²) = √441 = 21.
x = 7; LB = 29 ft; the lamp post is 21 ft from the midpoint
Problem 17 Same-side exterior angles
∠1 and ∠3 form a linear pair, so they are supplementary. Because the lines are parallel, ∠3 ≅ ∠7 (corresponding angles). Substituting, ∠1 and ∠7 are supplementary.
(8x + 4) + (4x + 8) = 180 gives 12x + 12 = 180, so x = 14.
∠1 = 8(14) + 4 = 116° and ∠7 = 4(14) + 8 = 64°.
x = 14; 116° and 64°
Problem 18 Bisecting an angle between parallels
Alternate interior angles are equal: 5x + 3 = 7x − 25, so 28 = 2x and x = 14. Then m∠3 = 5(14) + 3 = 73°.
∠4 forms a linear pair with ∠3, so m∠4 = 180° − 73° = 107°. A bisector gives halves of 107° ÷ 2 = 53.5°.
x = 14; m∠4 = 107°; each half is 53.5°
Problem 19 Supplementary angles proof
| Statement | Reason |
|---|---|
| 1. a ∥ b | Given |
| 2. ∠2 ≅ ∠6, so m∠2 = m∠6 | Corresponding angles are congruent |
| 3. ∠1 and ∠2 form a linear pair | Definition of linear pair |
| 4. m∠1 + m∠2 = 180° | Linear pairs are supplementary |
| 5. m∠1 + m∠6 = 180° | Substitution |
| 6. ∠1 and ∠6 are supplementary | Definition of supplementary |
∠1 and ∠6 are supplementary
Problem 20 Prove the bisector theorem
| Statement | Reason |
|---|---|
| 1. ℓ is the perpendicular bisector of AB at M | Given |
| 2. AM = MB | Definition of midpoint |
| 3. ∠PMA and ∠PMB are right angles, so ∠PMA ≅ ∠PMB | Definition of perpendicular; all right angles are congruent |
| 4. PM = PM | Reflexive Property |
| 5. △PMA ≅ △PMB | SAS (two sides and the included angle) |
| 6. PA = PB | Corresponding parts of congruent triangles |
Check: PA = √((0 + 6)² + (8 − 0)²) = √100 = 10 and PB = √((0 − 6)² + 8²) = 10.
PA = PB = 10