Angles made by a transversal: solve for x
Grade 10 Math Worksheet

Circle the best answer.
Ari knows m∠UGF = 110°, m∠FGH = (3x + 10)° and m∠GHQ = (5x − 30)°. Which conclusion about DF and QS is justified?
- Setting 3x + 10 = 5x − 30 gives x = 20, so the angles are equal and DF ∥ QS
- The linear pair gives m∠FGH = 70°, so x = 20, m∠GHQ = 70°, and DF ∥ QS by the alternate interior converse
- Vertical angles give m∠FGH = 110°, so x = 33⅓ and m∠GHQ ≈ 136.7°: not parallel
- 3x + 10 + 5x − 30 = 180 gives x = 25, so DF ∥ QS by the same-side converse
Bo knows m∠UGD = 110°, m∠DGH = (2x + 18)° and m∠QHV = (3x − 7)°. Which conclusion about DF and QS is justified?
Use the picture in question 1.
- Parallel: 2x + 18 = 3x − 7 gives x = 25, so the corresponding angles match
- Not parallel: x = 26 gives 70° and 71°, and corresponding angles must be equal
- Parallel: 70° and 71° are close enough on a sketch
- Not parallel: m∠DGH = 110° as a vertical angle, so x = 46 and m∠QHV = 131°
Above p right of t: 68°. At q, below-right: (5x + 7)°; below-left: (2x + 33)°. Above r left of t: (7x − 28)°. Which lines can Esme prove parallel?
Sketch, not to scale: line t crosses lines p, q and r at G, H and M, from top to bottom. Each angle is named by its side of the crossed line and its side of t. Sketch, not to scale: line t crosses lines p, q and r at G, H and M, from top to bottom. Each angle is named by its side of the crossed line and its side of t. p q r t G H M line t crosses lines p, q and r at G, H and M, from top to bottom. Each angle is named by its side of the crossed line and its side of t. Not to scale. - All three: x = 20 solves every one of the expressions
- p ∥ r only: x = 20 gives 180° − 112° = 68° above r, right of t
- p ∥ q: 68° and 107° add to about 180°, close enough
- None: the measured 68° and 112° are not equal angles
Jun knows m∠UGF = (3x)°, m∠FGH = (x + 40)° and m∠GHS = (2x + 20)°. Which conclusion about DF and QS is justified?
Sketch, not to scale: lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Sketch, not to scale: lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. D F G Q S H U V lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Not to scale. - Not parallel: same-side angles of 75° and 90° would have to be equal
- Parallel: 75° + 90° = 165° is near 180° on a hand sketch
- Not parallel: x = 35 gives 75° and 90°, which add to 165°, not 180°
- Parallel: x + 40 + 2x + 20 = 180 gives x = 40
Read this first
Use this
Proof: 1. DF ∥ QS (given). 2. m∠UGD = m∠GHQ (corresponding angles postulate). 3. m∠UGD = m∠FGH (reason ?). 4. m∠FGH = m∠GHQ (substitution).
Kira proves the alternate interior angles theorem from the corresponding angles postulate. Which reason belongs in step 3?
Sketch, not to scale: line DF ∥ line QS (arrows on both). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Proof: 1. DF ∥ QS (given). 2. m∠UGD = m∠GHQ (corresponding angles postulate). 3. m∠UGD = m∠FGH (reason ?). 4. m∠FGH = m∠GHQ (substitution). Sketch, not to scale: line DF ∥ line QS (arrows on both). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Proof: 1. DF ∥ QS (given). 2. m∠UGD = m∠GHQ (corresponding angles postulate). 3. m∠UGD = m∠FGH (reason ?). 4. m∠FGH = m∠GHQ (substitution). D F G Q S H U V Not to scale. - Angles in a linear pair add to 180°
- Vertical angles are equal
- Corresponding angles are equal
- Alternate interior angles are equal
Use this
Proof: 1. m∠FGH = m∠GHQ (given). 2. m∠FGH = m∠UGD (vertical angles). 3. ? (substitution). 4. DF ∥ QS (converse of corresponding angles).
Luz proves the converse of the alternate interior angles theorem. Which statement belongs in step 3?
Sketch, not to scale: lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Proof: 1. m∠FGH = m∠GHQ (given). 2. m∠FGH = m∠UGD (vertical angles). 3. ? (substitution). 4. DF ∥ QS (converse of corresponding angles). Sketch, not to scale: lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Proof: 1. m∠FGH = m∠GHQ (given). 2. m∠FGH = m∠UGD (vertical angles). 3. ? (substitution). 4. DF ∥ QS (converse of corresponding angles). D F G Q S H U V Not to scale. - m∠UGF = m∠GHS
- m∠UGD = m∠FGH
- m∠FGH + m∠GHS = 180°
- m∠UGD = m∠GHQ
Use this
Line l through A is ∥ BC. angle 1 is between l and AB, angle 2 is ∠BAC, angle 3 is between l and AC. Proof: 1. l ∥ BC (construction). 2. mangle 1 = m∠B and mangle 3 = m∠C (reason ?). 3. mangle 1 + mangle 2 + mangle 3 = 180° (angles on a line). 4. m∠B + mangle 2 + m∠C = 180° (substitution).
Milo proves a triangle's angles add to 180°. Which reason belongs in step 2?
Sketch, not to scale: triangle ABC with A at the top, B lower left, C lower right. Line ℓ through A is parallel to BC. ∠1 is between ℓ and AB, ∠2 is ∠BAC, ∠3 is between ℓ and AC. Proof: 1. ℓ ∥ BC (construction). 2. m∠1 = m∠B and m∠3 = m∠C (reason ?). 3. m∠1 + m∠2 + m∠3 = 180° (angles on a line). 4. m∠B + m∠2 + m∠C = 180° (substitution). Sketch, not to scale: triangle ABC with A at the top, B lower left, C lower right. Line ℓ through A is parallel to BC. ∠1 is between ℓ and AB, ∠2 is ∠BAC, ∠3 is between ℓ and AC. Proof: 1. ℓ ∥ BC (construction). 2. m∠1 = m∠B and m∠3 = m∠C (reason ?). 3. m∠1 + m∠2 + m∠3 = 180° (angles on a line). 4. m∠B + m∠2 + m∠C = 180° (substitution). 1 2 3 A B C Not to scale. - Vertical angles are equal at A
- Corresponding angles are equal, with AB and AC as transversals
- The converse of the alternate interior angles theorem
- Alternate interior angles are equal, with AB and AC as transversals
Use this
Proof: 1. m∠DGH + m∠GHQ = 180° (given). 2. m∠DGH + m∠UGD = 180° (linear pair). 3. m∠UGD = m∠GHQ (reason ?). 4. DF ∥ QS (converse of corresponding angles).
Noah proves the converse of the same-side interior angles theorem. Which reason belongs in step 3?
Sketch, not to scale: lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Proof: 1. m∠DGH + m∠GHQ = 180° (given). 2. m∠DGH + m∠UGD = 180° (linear pair). 3. m∠UGD = m∠GHQ (reason ?). 4. DF ∥ QS (converse of corresponding angles). Sketch, not to scale: lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Proof: 1. m∠DGH + m∠GHQ = 180° (given). 2. m∠DGH + m∠UGD = 180° (linear pair). 3. m∠UGD = m∠GHQ (reason ?). 4. DF ∥ QS (converse of corresponding angles). D F G Q S H U V Not to scale. - Corresponding angles are equal, since DF ∥ QS
- Same-side interior angles add to 180°
- Supplements of the same angle are equal
- Vertical angles are equal
Sketch, not to scale: lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Sketch, not to scale: lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. D F G Q S H U V lines DF and QS (parallel is not assumed). Line UV crosses DF at G and QS at H; U is above, V below; D and Q are on the left. Not to scale. Ren hears: if two equal angles lie on opposite sides of the transversal, one at each crossing, the lines are parallel. Which case is a counterexample?
- m∠UGD = m∠DGH = 90°
- m∠FGH = m∠QHV = 70°
- m∠DGH = m∠GHS = 70°
- m∠UGF = m∠QHV = 70°
Tala hears: if a transversal is perpendicular to one of two lines, it is perpendicular to the other. Which case is a counterexample?
Use the picture in question 9.
- UV ⊥ DF, DF ∥ QS, so m∠GHS = 90°
- UV ⊥ DF and m∠GHS = 80°
- UV ⊥ DF and m∠UGD = 90°
- UV ⊥ DF and m∠FGH = 90°