IGCSE Maths 0580 Angles Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Geometry · 4.6

Every angles question from 25 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 4.6. 34 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Geometry past paper questions.

0580 Angles past paper questions: Core (Papers 1 and 3)

12 marksCore

May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q10

The diagram shows two straight lines intersecting two parallel lines.

0580/11 May/June 2026 Q10 – two lines crossing two parallel lines, angles x°, 67°, 81° and y° markedx°67°81°y°NOT TOSCALE

Complete each statement with the correct geometrical reason: x = 67 because …; y = 81 because …

Show solution
  1. x and the 67° angle are between the parallel lines on opposite sides of the crossing line (a Z-shape).
  2. x = 67 because alternate angles are equal
  3. y and the 81° angle are in the same position at each parallel line (an F-shape).
  4. y = 81 because corresponding angles are equal
  5. Tip: Write the full reason. “Z-angles” or “F-angles” on their own do not score.
Answer alternate angles are equal corresponding angles are equal
23 marksCore

May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q17

(a) Two of the angles in a triangle are 100° and 30°. Work out the size of the third angle.

(b) Work out the size of an interior angle of a regular 10-sided polygon.

Show solution
Part (a)
  1. The angles in a triangle add up to 180°.
  2. Third angle = 180 − 100 − 30 = 50°
Answer 50°
Part (b)
  1. The exterior angles of any polygon add up to 360°. A regular 10-sided polygon has 10 equal exterior angles.
  2. Exterior angle = 360 ÷ 10 = 36°.
  3. Interior angle and exterior angle make a straight line, so interior angle = 180 − 36 = 144°
Answer 144°
32 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q23

Calculate the interior angle of a regular pentagon.

Show solution
  1. The exterior angles of a polygon add up to 360°. A regular pentagon has 5 equal exterior angles.
  2. Exterior angle = 360 ÷ 5 = 72°.
  3. Interior angle = 180 − 72 = 108°
Answer 108°
42 marksCore

May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q19

Calculate the interior angle of a regular polygon with 12 sides.

Show solution
  1. The exterior angles of any polygon add up to 360°. A regular 12-sided polygon has 12 equal exterior angles.
  2. Exterior angle = 360 ÷ 12 = 30°.
  3. Interior and exterior angles make a straight line, so interior angle = 180 − 30 = 150°
Answer 150°
53 marksCore

Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q10

The diagram shows a triangle on a straight line, with a second line parallel to the left side of the triangle. The angles 60° and 65° are marked at the right-hand vertex. Work out the values of a, b and c.

0580/12 February/March 2026 Q10 – triangle on a straight line with parallel lines, angles a°, b°, c°, 60° and 65°a°b°c°60°65°NOT TOSCALE
Show solution
  1. The two sloping lines are parallel. a is at the top vertex and the 60° angle is at the right-hand vertex, between the parallel lines (a Z-shape).
  2. a = 60 because alternate angles are equal.
  3. b is at the left-hand vertex in the same position as the 65° angle is at the right-hand vertex (an F-shape).
  4. b = 65 because corresponding angles are equal.
  5. c, 60° and 65° lie together on the straight line, and angles on a straight line add up to 180°.
  6. c = 180 − 60 − 65, so c = 55
Answer a = 60 b = 65 c = 55
63 marksCore

Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q15

ABC is a straight line and ABD is an isosceles triangle with DA = DB. Angle DBC = 140° and angle ADB = x°. Work out the value of x.

0580/11 October/November 2025 Q15 – isosceles triangle ABD on line ABC, angle 140° at B and x° at D140°x°ABCDNOT TOSCALE
Show solution
  1. ABC is a straight line, so angles on it add up to 180°: angle ABD = 180 − 140 = 40°.
  2. DA = DB, so the triangle is isosceles and the base angles are equal: angle DAB = 40°.
  3. Angles in a triangle add up to 180°: x = 180 − 40 − 40 = 100
Answer 100
75 marksCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q12

The diagram shows a parallelogram with two adjacent angles (3x + 10)° and (x − 30)°.

0580/12 October/November 2025 Q12 – parallelogram with angles (3x + 10)° and (x − 30)°(x − 30)°(3x + 10)°NOT TOSCALE

(a) Write an equation and solve it to find the value of x. (b) Find the value of the largest angle in this parallelogram.

Show solution
Part (a)
  1. Next-door (adjacent) angles in a parallelogram add up to 180°, because the sides are parallel.
  2. 3x + 10 + x − 30 = 180
  3. Collect like terms: 4x − 20 = 180.
  4. Add 20 to both sides: 4x = 200.
  5. Divide by 4: x = 50
Answer 3x + 10 + x − 30 = 180 x = 50
Part (b)
  1. Substitute x = 50 into each angle: 3 × 50 + 10 = 160° and 50 − 30 = 20°.
  2. Check: 160 + 20 = 180 ✓
  3. The largest angle is 160°
Answer 160°
83 marksCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q14

Triangle DEF has its vertices on the sides of triangle ABC, with D on AB, E on AC and F on BC. Angle ABC = 50°, angle BFD = 45°, angle EFC = 35°, angle ADE = 25° and angle DEF = x°.

0580/12 October/November 2025 Q14 – triangle DEF inside triangle ABC with angles 100°, 25°, 50°, 45°, 35° and 30°50°30°100°45°35°25°x°BCAFDENOT TOSCALE

Find the value of x.

Show solution
  1. Work round the shape one triangle or line at a time.
  2. Triangle BDF: angle BDF = 180 − 50 − 45 = 85° (angles in a triangle add up to 180°).
  3. D is on the straight line ADB, so angle FDE = 180 − 85 − 25 = 70° (angles on a straight line).
  4. F is on the straight line BFC, so angle DFE = 180 − 45 − 35 = 100°.
  5. Triangle DEF: x = 180 − 70 − 100 = 10
Answer 10
92 marksCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q22

Work out the size of one interior angle of a regular hexagon.

Show solution
  1. The exterior angles of any polygon add up to 360°. A regular hexagon has 6 equal exterior angles.
  2. Exterior angle = 360 ÷ 6 = 60°.
  3. Interior angle = 180 − 60 = 120°
Answer 120°
102 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q3

(a) Measure angle a. (b) Write down the mathematical name of this type of angle.

0580/13 October/November 2025 Q3 – angle aa
Show solution
Part (a)
  1. Put the centre of the protractor on the corner of the angle, with 0° along one arm.
  2. Read the scale where the other arm crosses it. Use the scale that starts at 0 on your first arm.
  3. Angle a = 123°
Answer 123°
Part (b)
  1. The angle is between 90° and 180°.
  2. It is obtuse
Answer obtuse
114 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q10

The diagram shows an isosceles triangle with a 40° angle at its top vertex, between a pair of parallel lines. Its base angles are x° and the angle y° is on a straight line next to one of them.

0580/13 October/November 2025 Q10 – isosceles triangle between parallel lines with angles 40°, x° and y°x°40°y°NOT TOSCALE

(a) Find the value of x. (b) Find the value of y. Give a geometrical reason for your answer.

Show solution
Part (a)
  1. The angles in a triangle add up to 180°, so the two base angles together are 180 − 40 = 140°.
  2. The triangle is isosceles, so the base angles are equal: x = 140 ÷ 2 = 70
Answer 70
Part (b)
  1. y and x lie together on a straight line.
  2. y = 180 − 70, so y = 110 because angles on a straight line add up to 180°
Answer y = 110 angles on a straight line add up to 180°
122 marksCore

Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q19

Find the number of sides of a regular polygon with interior angle 160°.

Show solution
  1. An interior angle and an exterior angle make a straight line, so exterior angle = 180 − 160 = 20°.
  2. The exterior angles of a polygon add up to 360°.
  3. Number of sides = 360 ÷ 20 = 18
Answer 18
134 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q9

(a) Write down the mathematical name for the type of angle shown (more than 180°).

(b) Triangle ABC has angle A = 54° and angle C = 102°. Lily says angle B, x, is 34. Give a geometrical reason why Lily is not correct.

(c) ABCD is a quadrilateral with angle A = 48°, angle B = 75° and angle D = 126°. DCE is a straight line. Calculate x, the angle BCE.

0580/31 October/November 2025 Q9 – (a) a reflex angle, (b) triangle ABC, (c) quadrilateral ABCD with DCE a straight line(a)54°102°x°(b)48°126°75°x°(c)ABCADCEBNOT TOSCALE
Show solution
Part (a)
  1. An angle bigger than 180° (but less than 360°) is a reflex angle.
Answer reflex
Part (b)
  1. Add Lily’s angle to the other two: 54 + 102 + 34 = 190°.
  2. The total should be exactly 180°, because Angles in a triangle add up to 180°
  3. So Lily is wrong. The correct value is x = 180 − 54 − 102 = 24.
Answer Angles in a triangle add up to 180°
Part (c)
  1. The angles in a quadrilateral add up to 360°.
  2. Angle BCD = 360 − 48 − 75 − 126 = 111°.
  3. D, C and E lie on a straight line, so x and angle BCD add up to 180°.
  4. x = 180 − 111 = 69
Answer 69
142 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q20(a)

Work out the size of one interior angle of a regular 8-sided polygon.

Show solution
  1. The exterior angles of a polygon add up to 360°. A regular 8-sided polygon has 8 equal exterior angles.
  2. Exterior angle = 360 ÷ 8 = 45°.
  3. Interior angle = 180 − 45 = 135°
Answer 135°
151 markCore

Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q15

Three straight lines meet in the diagram, with angles 68°, 43°, 135° and x° around a point. Find the value of x.

0580/33 October/November 2025 Q15 – angles 68°, x°, 43° and 135° around a point68°x°43°135°NOT TOSCALE
Show solution
  1. The four angles go all the way round one point.
  2. Angles at a point add up to 360°.
  3. x = 360 − 68 − 43 − 135 = 114
Answer 114
166 marksCore

May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q18

The diagram shows two straight lines intersecting two parallel lines. Angles of 110° and 50° are marked at the lower parallel line.

0580/11 May/June 2025 Q18 – two straight lines crossing two parallel lines with angles x°, y°, z°, 110° and 50°y°x°z°110°50°NOT TOSCALE

(a) Find the value of x. Give a geometrical reason for your answer.

(b) Find the value of y. Give a geometrical reason for your answer.

(c) Find the value of z.

Show solution
Part (a)
  1. x and the 50° angle are on opposite sides of a line that crosses the parallel lines, between them (a Z-shape).
  2. x = 50, because alternate angles are equal
Answer 50, because alternate angles are equal
Part (b)
  1. y is in the same position at the top parallel line as the 110° angle is at the bottom one (an F-shape).
  2. y = 110, because corresponding angles are equal
Answer 110, because corresponding angles are equal
Part (c)
  1. z is the third angle of the triangle made by the two straight lines and the lower parallel line.
  2. The angle inside the triangle next to the 110° angle is 180 − 110 = 70° (angles on a straight line).
  3. Angles in a triangle add up to 180°: z = 180 − 70 − 50 = 60
Answer 60
173 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q4

The diagram shows a quadrilateral with sides of equal length. Two neighbouring angles are a° and 2a°.

0580/13 May/June 2025 Q4 – quadrilateral with four equal sides and angles a° and 2a°a°2a°NOT TOSCALE

(a) Write down the mathematical name for this quadrilateral.

(b) Work out the value of a.

Show solution
Part (a)
  1. All four sides are equal, but the angles are not right angles, so it is not a square.
  2. The quadrilateral is a rhombus
Answer rhombus
Part (b)
  1. Opposite sides of a rhombus are parallel, so neighbouring angles are co-interior and add up to 180°.
  2. a + 2a = 180
  3. 3a = 180, so a = 60
Answer 60

0580 Angles past paper questions: Extended (Papers 2 and 4)

182 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q11

Find the number of sides of a regular polygon with interior angle 140°.

Show solution
  1. An interior angle and an exterior angle make a straight line: exterior angle = 180 − 140 = 40°.
  2. The exterior angles of any polygon add up to 360°.
  3. Number of sides = 360 ÷ 40 = 9
Answer 9
192 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q11

The interior angle of a regular polygon is 135°. Find the number of sides of this regular polygon.

Show solution
  1. An interior angle and an exterior angle make a straight line: exterior angle = 180 − 135 = 45°.
  2. The exterior angles of any polygon add up to 360°.
  3. Number of sides = 360 ÷ 45 = 8
Answer 8
202 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q1

Triangle ABC is isosceles with AB = AC. Angle BAC = 30° and angle ACB = x°. Find the value of x.

0580/42 May/June 2026 Q1 – isosceles triangle ABC with 30° at A and x° at C30°x°ABCNOT TOSCALE
Show solution
  1. AB = AC, so the triangle is isosceles and the base angles at B and C are equal.
  2. The angles in a triangle add up to 180°, so the two base angles together = 180 − 30 = 150°.
  3. Each base angle = 150 ÷ 2 = 75°.
  4. x = 75
Answer x = 75
211 markExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q8(a)

AB and CD are parallel lines. AD and BC are straight lines that intersect at X. Angle ABX = y°. Complete the statement with a geometrical reason: angle DCX = y° because …

0580/43 May/June 2026 Q8 – parallel lines AB and CD with AD and BC crossing at Xy°ABCDXNOT TOSCALE
Show solution
  1. BC is a straight line crossing the two parallel lines AB and CD.
  2. Angles ABX and DCX are on opposite sides of BC, between the parallel lines (a Z shape).
  3. Angle DCX = y° because alternate angles are equal
Answer alternate angles are equal
224 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q14

The diagram shows part of a regular polygon ABCDE. Angle ABD = 135° and angle CBD = x°. By finding the value of x, work out the number of sides of this polygon.

0580/22 February/March 2026 Q14 – part of a regular polygon with angle ABD = 135° and angle CBD = x°135°x°ABCDENOT TOSCALE
Show solution
  1. In a regular polygon all sides are equal, so BC = CD and triangle BCD is isosceles.
  2. The base angles of triangle BCD are equal: angle CDB = angle CBD = x°, so angle BCD = 180 − 2x.
  3. All interior angles are equal, so angle ABC = angle BCD.
  4. Angle ABC = angle ABD + angle DBC = 135 + x.
  5. Set up the equation: 135 + x = 180 − 2x, so 3x = 45 and x = 15.
  6. Interior angle = 135 + 15 = 150°, so exterior angle = 180 − 150 = 30°.
  7. Number of sides = 360 ÷ 30 = 12
Answer 12
233 marksExtended

Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q2

ABC is a triangle with a right angle at A. ADB and DCE are straight lines and angle ADC = 74°. Angle BDC = x° and angle ACE = y°. Find the value of x and the value of y.

0580/42 February/March 2026 Q2 – right-angled triangle ABC with D on AB, line DCE, angles 74°, x° and y°74°x°y°ABCDENOT TOSCALE
Show solution
  1. ADB is a straight line, so the angles at D on it add up to 180°.
  2. x = 180 − 74 = 106
  3. Triangle ADC has a right angle at A. Angles in a triangle add up to 180°.
  4. Angle ACD = 180 − 90 − 74 = 16°.
  5. DCE is a straight line, so angle ACE and angle ACD add up to 180°.
  6. y = 180 − 16 = 164
Answer 106 164
243 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q2

In the diagram, AC = BC. The angles 148° and 82° are marked at C, round the point with angle ACB. Work out angle CAB.

0580/21 October/November 2025 Q2 – triangle ABC with AC = BC and angles 148° and 82° marked at C82°148°CABNOT TOSCALE
Show solution
  1. The angles round the point C add up to 360°: angle ACB = 360 − 148 − 82 = 130°.
  2. AC = BC, so triangle ABC is isosceles and its base angles CAB and CBA are equal.
  3. The two base angles together are 180 − 130 = 50°.
  4. Angle CAB = 50 ÷ 2 = 25°
Answer 25°
252 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q3

Find the interior angle of a regular 20-sided polygon.

Show solution
  1. The exterior angles of a polygon add up to 360°. A regular 20-sided polygon has 20 equal exterior angles.
  2. Exterior angle = 360 ÷ 20 = 18°.
  3. Interior angle = 180 − 18 = 162°
Answer 162°
264 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q6

ABCD is a quadrilateral. E lies on CD and AE is parallel to BC. EA = ED. Angle BAE = 75° and angle BCD = 80°.

0580/22 October/November 2025 Q6 – quadrilateral ABCD with E on CD, AE parallel to BC, angles 75° and 80°75°80°ABCDENOT TOSCALE

Find (a) angle ABC (b) angle AED (c) angle DAB.

Show solution
Part (a)
  1. AE is parallel to BC, and AB crosses both lines. Angles BAE and ABC are on the same side, between the parallels (co-interior).
  2. Co-interior angles add up to 180°: angle ABC = 180 − 75 = 105°
Answer 105°
Part (b)
  1. AE is parallel to BC, and the straight line DC crosses both. Angle AED and angle BCD are in the same position (corresponding).
  2. Angle AED = 80°
Answer 80°
Part (c)
  1. EA = ED, so triangle AED is isosceles and angle EAD = angle EDA.
  2. The two equal angles together are 180 − 80 = 100°, so angle EAD = 50°.
  3. Angle DAB = angle DAE + angle EAB = 50 + 75 = 125°
Answer 125°
274 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q16

AB and BD are two sides of a regular 15-sided polygon. AB and BC are two sides of a regular n-sided polygon. Angle DBC = 14°. Work out the value of n.

0580/22 October/November 2025 Q16 – sides AB, BD and BC from two regular polygons, angle DBC = 14°14°ABDCNOT TOSCALE
Show solution
  1. Regular 15-sided polygon: exterior angle = 360 ÷ 15 = 24°, so interior angle ABD = 180 − 24 = 156°.
  2. Angle ABC is an interior angle of the n-sided polygon: ABC = ABD + DBC = 156 + 14 = 170°.
  3. Its exterior angle = 180 − 170 = 10°.
  4. n = 360 ÷ 10 = 36
Answer 36
282 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q1

The diagram shows an isosceles triangle with base angles of 25° and top angle x°. Find the value of x.

0580/23 October/November 2025 Q1 – isosceles triangle with base angle 25° and top angle x°25°x°NOT TOSCALE
Show solution
  1. The triangle is isosceles, so both base angles are 25°.
  2. Angles in a triangle add up to 180°: x = 180 − 25 − 25 = 130
Answer 130
292 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q5

ABCD is a rectangle. EFG is an equilateral triangle that touches the rectangle at F (on BC) and G (on AD). Angle BFE = 12°, angle GFC = x° and angle FGD = y°.

0580/23 October/November 2025 Q5 – rectangle ABCD with equilateral triangle EFG touching it at F and G, angles 12°, x° and y°12°x°y°ADCBFGENOT TOSCALE

Find the value of x and the value of y.

Show solution
  1. EFG is equilateral, so each of its angles is 60°. Angle EFG = 60°.
  2. B, F and C lie on a straight line, so the angles at F along it add up to 180°.
  3. x = 180 − 12 − 60 = 108
  4. BC is parallel to AD, and the line FG crosses both. Angles GFC and FGD are co-interior (between the parallels on the same side).
  5. Co-interior angles add up to 180°: y = 180 − 108 = 72
Answer 108 72
302 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q3

The diagram shows two parallel lines and a straight line crossing them, with angles 55° and x° marked.

0580/41 October/November 2025 Q3 – two parallel lines crossed by a straight line, angles x° and 55°x°55°NOT TOSCALE
Show solution
  1. The 55° angle and the angle next to x on the top line are alternate angles (Z-shape between the parallel lines), so that angle is also 55°.
  2. That angle and x lie on the same straight line, and angles on a straight line add up to 180°.
  3. x = 180 − 55 = 125
Answer 125
314 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q5

(a) The diagram shows a regular decagon. AB is a line of symmetry through vertex A, and angle d° is half an interior angle at A. Work out the value of d.

0580/43 October/November 2025 Q5(a) – regular decagon with line of symmetry AB and angle d° at Ad°ABNOT TOSCALE

(b) The exterior angle of a regular polygon with n sides is 45°. Work out the value of n.

Show solution
Part (a)
  1. Exterior angle of a regular decagon = 360 ÷ 10 = 36°.
  2. Interior angle = 180 − 36 = 144°.
  3. The line of symmetry through A cuts this angle exactly in half: d = 144 ÷ 2 = 72
Answer 72
Part (b)
  1. The exterior angles of any polygon add up to 360°, and in a regular polygon they are all equal.
  2. n = 360 ÷ 45 = 8
Answer 8
324 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q2

The diagram shows two parallel lines intersecting two straight lines. One straight line is steep and the other is shallow. Angles of 76° and 158° are marked, along with w°, x° and y°. Find the values of w, x and y.

0580/21 May/June 2025 Q2 – two parallel lines crossed by two straight lines with angles w°, x°, y°, 76° and 158°w°158°76°x°y°NOT TOSCALE
Show solution
  1. w is in the same position at the upper parallel line as the 158° angle is at the lower one, on the shallow line (an F-shape).
  2. w = 158 (corresponding angles are equal).
  3. x is in the same position at the upper parallel line as the 76° angle is at the lower one, on the steep line.
  4. x = 76 (corresponding angles are equal).
  5. y is inside the triangle made by the steep line, the shallow line and a parallel line.
  6. The angle in the triangle next to 158° is 180 − 158 = 22° (angles on a straight line). The angle on the steep line is 76°.
  7. Angles in a triangle add up to 180°: y = 180 − 76 − 22 = 82.
  8. w = 158, x = 76, y = 82
Answer w = 158, x = 76, y = 82
332 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q5

The interior angle of a regular polygon is 150°. Find the number of sides of this polygon.

Show solution
  1. An interior angle and an exterior angle make a straight line: exterior angle = 180 − 150 = 30°.
  2. The exterior angles of any polygon add up to 360°.
  3. Number of sides = 360 ÷ 30 = 12
Answer 12
343 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q3

The diagram shows two straight lines intersecting two parallel lines. Angles of 45° and 65° are marked at the upper parallel line. The two straight lines cross at the lower parallel line, where x° and y° are marked.

0580/22 May/June 2025 Q3 – two straight lines crossing two parallel lines with angles 45°, 65°, x° and y°45°65°x°y°NOT TOSCALE

Find the value of x and the value of y.

Show solution
  1. The two straight lines and the upper parallel line make a triangle. Its angles at the upper line are 45° and 65°.
  2. x is the third angle of that triangle. Angles in a triangle add up to 180°.
  3. x = 180 − 45 − 65 = 70.
  4. y is between the lower parallel line and the line that makes 65° with the upper parallel line, on the opposite side (a Z-shape).
  5. y = 65 because alternate angles are equal.
  6. x = 70, y = 65
Answer x = 70, y = 65

Question papers used on this page

All angles questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/11/M/J/260580_s26_qp_11May/June 2026 · Paper 1 (Core)Q10
0580/12/M/J/260580_s26_qp_12May/June 2026 · Paper 1 (Core)Q17
0580/22/M/J/260580_s26_qp_22May/June 2026 · Paper 2 (Extended)Q11
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q11
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q23
0580/33/M/J/260580_s26_qp_33May/June 2026 · Paper 3 (Core)Q19
0580/42/M/J/260580_s26_qp_42May/June 2026 · Paper 4 (Extended)Q1
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q8(a)
0580/12/F/M/260580_m26_qp_12Feb/Mar 2026 · Paper 1 (Core)Q10
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q14
0580/42/F/M/260580_m26_qp_42Feb/Mar 2026 · Paper 4 (Extended)Q2
0580/11/O/N/250580_w25_qp_11Oct/Nov 2025 · Paper 1 (Core)Q15
0580/12/O/N/250580_w25_qp_12Oct/Nov 2025 · Paper 1 (Core)Q12, Q14, Q22
0580/13/O/N/250580_w25_qp_13Oct/Nov 2025 · Paper 1 (Core)Q3, Q10, Q19
0580/21/O/N/250580_w25_qp_21Oct/Nov 2025 · Paper 2 (Extended)Q2, Q3
0580/22/O/N/250580_w25_qp_22Oct/Nov 2025 · Paper 2 (Extended)Q6, Q16
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q1, Q5
0580/31/O/N/250580_w25_qp_31Oct/Nov 2025 · Paper 3 (Core)Q9, Q20(a)
0580/33/O/N/250580_w25_qp_33Oct/Nov 2025 · Paper 3 (Core)Q15
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q3
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q5
0580/11/M/J/250580_s25_qp_11May/June 2025 · Paper 1 (Core)Q18
0580/13/M/J/250580_s25_qp_13May/June 2025 · Paper 1 (Core)Q4
0580/21/M/J/250580_s25_qp_21May/June 2025 · Paper 2 (Extended)Q2, Q5
0580/22/M/J/250580_s25_qp_22May/June 2025 · Paper 2 (Extended)Q3

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