IGCSE Maths 0580 Non-Right-Angled Triangles Past Paper Questions and Worked Solutions

By /
IGCSE Maths (0580) · Trigonometry · 6.5

Every non-right-angled triangles question from 7 Cambridge IGCSE Mathematics (0580) past papers (Oct/Nov 2025 to May/June 2026), syllabus section 6.5. 10 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/42/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 Trigonometry past paper questions.

0580 Non-Right-Angled Triangles past paper questions: Extended (Papers 2 and 4)

19 marksExtended

May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q20(b,c)

The diagram shows a field ABCD with a straight footpath from A to C. AB = 50.3 m, AC = 86.8 m, AD = 38.2 m, angle ABC = 112° and angle DAC = 79°.

0580/42 May/June 2026 Q20 – field ABCD with AB = 50.3 m, AC = 86.8 m, AD = 38.2 m, angles 103°, 112° and 79°86.8 m50.3 m38.2 mNorth103°112°79°ABCDNOT TOSCALE

(b) Calculate the distance CD. (c) Calculate the shortest distance from B to AC.

Show solution
Part (b)
  1. In triangle ACD you know two sides and the angle between them, so use the cosine rule.
  2. CD² = AD² + AC² − 2 × AD × AC × cos 79°.
  3. CD² = 38.2² + 86.8² − 2 × 38.2 × 86.8 × cos 79° = 1459.24 + 7534.24 − 1265.3… = 7728.1…
  4. CD = √7728.1… = 87.909… = 87.9 m
Answer 87.9 m
Part (c)
  1. The shortest distance from B to AC is the perpendicular distance. First find angle BAC.
  2. Sine rule in triangle ABC: sin ACB⁄50.3 = sin 112°⁄86.8.
  3. sin ACB = 50.3 × sin 112°⁄86.8 = 0.5372…, so angle ACB = 32.50…°
  4. Angle BAC = 180 − 112 − 32.50… = 35.50…°
  5. The perpendicular from B makes a right-angled triangle with hypotenuse AB, so distance = 50.3 × sin 35.50…°
  6. 29.2 m
Answer 29.2 m
22 marksExtended

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

In triangle CXD, CX = 5 cm, XD = 8 cm and angle CXD = 80.5°. Work out the area of triangle CXD.

Show solution
  1. You know two sides and the angle between them, so use Area = 1⁄2ab sin C.
  2. Area = 1⁄2 × 5 × 8 × sin 80.5° = 20 × 0.9862… = 19.72…
  3. 19.7 cm²
Answer 19.7 cm²
37 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q13

In triangle ABC, BC = 15 cm, AC = 15√3 cm and angle ABC = 120°.

0580/43 May/June 2026 Q13 – triangle ABC with angle 120° at B, BC = 15 cm and AC = 15√3 cm120°15 cm15√3 cmACBNOT TOSCALE

(a) Show that triangle ABC is an isosceles triangle. (b) Calculate the shortest distance from B to AC.

Show solution
Part (a)
  1. You know two sides and the angle opposite one of them (AC is opposite the 120° angle at B), so use the sine rule.
  2. sin BAC⁄BC = sin ABC⁄AC, so sin BAC⁄15 = sin 120°⁄15√3.
  3. sin 120° = √3⁄2. So sin BAC = 15 × √3⁄2/15√3 = 1⁄2.
  4. Angle BAC = sin⁻¹(1⁄2) = 30°. (It cannot be 150°, because the triangle already has a 120° angle.)
  5. Angle BCA = 180 − 120 − 30 = 30°.
  6. Angle BAC = angle BCA = 30°, so two angles are equal and triangle ABC is isosceles ✓
Answer Angle BAC = angle BCA = 30°
Part (b)
  1. The shortest distance from B to AC is the perpendicular from B to AC.
  2. That makes a right-angled triangle with hypotenuse BC = 15 cm and the 30° angle at C.
  3. The perpendicular is opposite the 30° angle: distance = 15 × sin 30° = 15 × 0.5 = 7.5 cm
Answer 7.5 cm
43 marksExtended

May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q15

The sides of a triangle are 5 cm, 7 cm and 11 cm. Find the size of the largest angle in this triangle.

Show solution
  1. The largest angle is always opposite the longest side, which is 11 cm.
  2. You know all three sides, so use the cosine rule: cos C = a² + b² − c²⁄2ab.
  3. cos C = 5² + 7² − 11²⁄2 × 5 × 7 = 25 + 49 − 121⁄70 = −47⁄70 = −0.6714…
  4. C = cos⁻¹(−0.6714…) = 132.17…
  5. 132.2°
  6. Tip: A negative cosine means the angle is obtuse (between 90° and 180°).
Answer 132.2°
512 marksExtended

Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q20

Points A, B, C and D lie on horizontal ground. AC = 450 m, AB = 600 m, AD = 290 m, angle CAB = 57° and angle ACD = 34°.

0580/42 February/March 2026 Q20 – quadrilateral ABCD with AC = 450 m, AB = 600 m, AD = 290 m, angles 57° and 34°450 m600 m290 m57°34°ACBDNOT TOSCALE

(a) Calculate the area of triangle ABC.

(b) Show that angle CAD = 85.8°, correct to 1 decimal place.

(c) There is a vertical mast of height 250 m at B and a vertical mast of height 45 m at D. Calculate the angle of depression of the top of the mast at D from the top of the mast at B.

Show solution
Part (a)
  1. Two sides and the angle between them are known, so use Area = 1⁄2ab sin C.
  2. Area = 1⁄2 × 450 × 600 × sin 57° = 135 000 × 0.8386… = 113 220.5…
  3. 113 000 m² (3 s.f.)
Answer 113 000 m²
Part (b)
  1. In triangle ACD, the side 450 m (AC) is opposite angle ADC and the side 290 m (AD) is opposite angle ACD = 34°.
  2. Sine rule: sin ADC⁄450 = sin 34°⁄290.
  3. sin ADC = 450 × sin 34°⁄290 = 0.8677…, so angle ADC = 60.19…°
  4. Angle CAD = 180 − 34 − 60.19… = 85.80… = 85.8° ✓
Answer 85.8°
Part (c)
  1. First find BD on the ground. Angle BAD = angle CAD + angle CAB = 85.8 + 57 = 142.8°.
  2. Cosine rule: BD² = 290² + 600² − 2 × 290 × 600 × cos 142.8°.
  3. BD² = 84 100 + 360 000 + 277 191… = 721 291…, so BD = 849.29… m.
  4. The top of mast B is 250 − 45 = 205 m higher than the top of mast D.
  5. The angle of depression θ is in a right-angled triangle with opposite 205 and adjacent 849.29…
  6. tan θ = 205⁄849.29… = 0.2413…, so θ = 13.57…
  7. 13.6°
  8. Tip: cos 142.8° is negative, so −2 × 290 × 600 × cos 142.8° is added.
Answer 13.6°
63 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q23

Triangle ABC has AB = 6 cm, BC = 10 cm and angle ABC = 30°. Work out the area of triangle ABC.

0580/23 October/November 2025 Q23 – triangle ABC with AB = 6 cm, BC = 10 cm and angle 30° at B6 cm10 cm30°BACNOT TOSCALE
Show solution
  1. Two sides and the angle between them are known, so use Area = 1⁄2ab sin C.
  2. The angle ABC = 30° is between the sides AB = 6 cm and BC = 10 cm.
  3. Area = 1⁄2 × 6 × 10 × sin 30° = 30 × 1⁄2 = 15 cm²
Answer 15 cm²
72 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q21

In triangle STU, ST = 8 cm, SU = 9 cm and angle TSU = 50°. Calculate the area of triangle STU.

Show solution
  1. Two sides and the angle between them are known, so use Area = 1⁄2ab sin C.
  2. a = 8, b = 9 and C = 50° (the angle between ST and SU).
  3. Area = 1⁄2 × 8 × 9 × sin 50° = 36 × 0.7660… = 27.57…
  4. Area = 27.6 cm²
Answer 27.6 cm²
88 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q29

In the diagram, AB = 10 cm, BC = 13 cm and AC = 14 cm. Angle BAC = y°. In triangle ACD, angle ACD = 38° and angle ADC = 97°.

0580/41 October/November 2025 Q29 – quadrilateral ABCD with diagonal AC, sides 10 cm, 13 cm and 14 cm, angles 38° and 97°14 cm10 cm13 cmy°38°97°ACBDNOT TOSCALE

(a) Calculate the value of y. (b) Calculate BD.

Show solution
Part (a)
  1. In triangle ABC all three sides are known, so use the cosine rule for angle A.
  2. cos y = AB² + AC² − BC²⁄2 × AB × AC = 10² + 14² − 13²⁄2 × 10 × 14
  3. = 100 + 196 − 169⁄280 = 127⁄280 = 0.4535…
  4. y = cos⁻¹(0.4535…) = 63.03… = 63.0
Answer 63.0
Part (b)
  1. First find AD using the sine rule in triangle ACD.
  2. AD is opposite the 38° angle and AC = 14 cm is opposite the 97° angle.
  3. AD⁄sin 38° = 14⁄sin 97°, so AD = 14 × sin 38°⁄sin 97° = 8.684… cm.
  4. Angle CAD = 180 − 97 − 38 = 45° (angles in a triangle).
  5. Angle BAD = angle BAC + angle CAD = 63.03… + 45 = 108.03…°.
  6. Now use the cosine rule in triangle ABD: BD² = 10² + 8.684…² − 2 × 10 × 8.684… × cos 108.03…°
  7. BD² = 100 + 75.41… + 53.75… = 229.16…
  8. BD = √229.16… = 15.13… = 15.1 cm
Answer 15.1 cm
910 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q15

A field ABCD is a quadrilateral. BD is a straight path across the field. AB = 180 m, BD = 300 m, DC = 112 m and angle BDC = 140°.

0580/42 October/November 2025 Q15 – field ABCD with AB = 180 m, BD = 300 m, DC = 112 m and angle BDC = 140°300 m180 m112 m140°DBCANOT TOSCALE

(a) Calculate BC. (b) Calculate angle DBC.

(c) The total area of the field, ABCD, is 35 900 m². Work out the length of the shortest distance from D to AB.

Show solution
Part (a)
  1. In triangle BCD you know two sides and the angle between them, so use the cosine rule.
  2. BC² = 300² + 112² − 2 × 300 × 112 × cos 140°.
  3. cos 140° is negative, so the last term is added: BC² = 90 000 + 12 544 + 51 478.2… = 154 022.2…
  4. BC = √154 022.2… = 392.4… = 392 m
Answer 392 m
Part (b)
  1. Sine rule in triangle BCD: sin DBC⁄112 = sin 140°⁄392.4….
  2. sin DBC = 112 × sin 140°⁄392.4… = 0.1834…
  3. Angle DBC = sin⁻¹(0.1834…) = 10.6°
Answer 10.6°
Part (c)
  1. Area of triangle BCD = 1⁄2 × 300 × 112 × sin 140° = 10 798.9… m².
  2. Area of triangle ABD = 35 900 − 10 798.9… = 25 101.1… m².
  3. The shortest distance from D to AB is the perpendicular height h of triangle ABD, with AB as the base.
  4. 1⁄2 × 180 × h = 25 101.1…, so 90h = 25 101.1… and h = 278.9…
  5. 279 m
Answer 279 m
106 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q18

In the upper triangle, a side of 64.5 cm is opposite an angle of 102° and a side of 41.3 cm is opposite angle x°. In the lower triangle, sides 64.5 cm and 52.1 cm meet at angle y° and the third side is 70.2 cm.

0580/43 October/November 2025 Q18 – two triangles sharing a 64.5 cm side, with angles 102°, x° and y°

(a) Calculate the value of x. (b) Calculate the value of y.

Show solution
Part (a)
  1. In the upper triangle you know two sides and an angle opposite one of them, so use the sine rule.
  2. sin x⁄41.3 = sin 102°⁄64.5
  3. sin x = 41.3 × sin 102°⁄64.5 = 41.3 × 0.9781…⁄64.5 = 0.6263…
  4. x = sin⁻¹(0.6263…) = 38.77… = 38.8
Answer 38.8
Part (b)
  1. In the lower triangle you know all three sides, so use the cosine rule. y is opposite the 70.2 cm side.
  2. cos y = 64.5² + 52.1² − 70.2²⁄2 × 64.5 × 52.1 = 4160.25 + 2714.41 − 4928.04⁄6720.9 = 1946.62⁄6720.9 = 0.2896…
  3. y = cos⁻¹(0.2896…) = 73.16… = 73.2
Answer 73.2

Question papers used on this page

All non-right-angled triangles questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/42/M/J/260580_s26_qp_42May/June 2026 · Paper 4 (Extended)Q20(b,c)
0580/43/M/J/260580_s26_qp_43May/June 2026 · Paper 4 (Extended)Q8(b)(ii), Q13, Q15
0580/42/F/M/260580_m26_qp_42Feb/Mar 2026 · Paper 4 (Extended)Q20
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q23
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q21, Q29
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q15
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q18

Related 0580 topics to practise next

More IGCSE Maths 0580 Trigonometry questions

Browse IGCSE Maths questions by topic

See every topic and subtopic on one page →

Book a free trial class

Tell us a little about your requirement.