IGCSE Maths · Trigonometry · Extended only

Non-Right-Angled Triangles and 3D Trigonometry Extended only

This entire note is Extended-only, and it closes out the Trigonometry unit. SOHCAHTOA only works when there’s a right angle — the sine rule and cosine rule work on any triangle. Both are given on the exam formula sheet, so the real skill is recognising which one fits the information you’ve been given.

1. The Sine Rule

a/sin A = b/sin B = c/sin C

Use the sine rule when you know a matching angle-side pair (an angle and the side directly opposite it), plus one more piece of information.

Worked example — finding a side A B C a b Triangle ABC has angle A = 40°, angle B = 60°, and side a = 8 cm (opposite A). Find side b.
a/sin A = b/sin B → b = a × sin B / sin A = 8 × sin(60°) / sin(40°) = 10.8 cm (3 s.f.)
Worked example — finding an angle Triangle has a = 10 cm, b = 7 cm, and angle B = 35°. Find angle A.
sin A / a = sin B / b → sin A = a × sin B / b = 10 × sin(35°) / 7 = 0.819
A = sin⁻¹(0.819) = 55.0° (1 d.p.)
Try it — Sine Rule
  1. Triangle has angle A = 50°, angle B = 70°, side a = 9 cm. Find side b.
  2. Triangle has a = 12 cm, b = 9 cm, angle A = 65°. Find angle B.
Show answers
  1. b = 9×sin70°/sin50° = 11.0 cm
  2. sinB = 9×sin65°/12 = 0.680. B = 42.8°

2. The Ambiguous Case

When you’re given two sides and an angle that isn’t between them (SSA), the sine rule can sometimes give two valid triangles — because sin θ and sin(180−θ) are always equal. Always check whether the second answer actually makes sense (angles in the triangle must sum to less than 180°).

Worked example Triangle has a = 8 cm, b = 10 cm, angle A = 35°. Find the possible values of angle B.
sin B = b × sin A / a = 10 × sin(35°) / 8 = 0.717
First solution: B = sin⁻¹(0.717) = 45.8°
Second solution: B = 180 − 45.8 = 134.2°
Check: 35° + 134.2° = 169.2°, which is less than 180° — so both solutions are valid triangles.
Try it — The Ambiguous Case
  1. Triangle has a = 6 cm, b = 9 cm, angle A = 30°. Find both possible values of angle B, and check both are valid.
  2. Explain, in one sentence, why the ambiguous case doesn’t happen with ASA or SAS information.
Show answers
  1. sinB = 9×sin30°/6 = 0.75. B = 48.6° or 131.4°. Check second: 30+131.4=161.4° < 180° — both valid.
  2. With ASA or SAS, the given angle sits between (or determines) enough fixed information that only one triangle shape can satisfy it — there’s no leftover freedom for a second solution.

3. The Cosine Rule

c² = a² + b² − 2ab cos C

Use the cosine rule when you have three sides (SSS) and want an angle, or two sides plus the angle between them (SAS) and want the third side.

Worked example — finding a side (SAS) Triangle has a = 7 cm, b = 9 cm, angle C = 50° (between them). Find side c.
c² = 7² + 9² − 2(7)(9)cos(50°) = 49 + 81 − 81.0 = 49.0
c = √49.0 = 7.00 cm (3 s.f.)
Worked example — finding an angle (SSS) Triangle has a = 8 cm, b = 10 cm, c = 13 cm. Find angle C.
Rearranged: cos C = (a² + b² − c²) / (2ab) = (64+100−169) / 160 = −0.0313
C = cos⁻¹(−0.0313) = 91.8° (1 d.p.)
Handling obtuse angles Notice the example above gave a negative value for cos C, and the angle came out just over 90°. This is completely normal — unlike the sine rule, the cosine rule never needs a special “ambiguous case” check, because cos⁻¹ always gives a unique answer between 0° and 180°, correctly identifying whether the angle is acute or obtuse.
Try it — Cosine Rule
  1. Triangle has a = 6 cm, b = 8 cm, angle C = 70° (between them). Find side c.
  2. Triangle has a = 5 cm, b = 7 cm, c = 11 cm. Find angle C, and state whether it’s acute or obtuse.
Show answers
  1. c² = 36+64−2(6)(8)cos70° = 100−32.86 = 67.14. c=8.19 cm
  2. cosC = (25+49−121)/70 = −0.671. C = cos⁻¹(−0.671) = 132.2° — obtuse.

4. Area of a Triangle: ½ab sinC

When you don’t know a base and perpendicular height directly, but you do know two sides and the angle between them, this formula finds the area without needing to work out the height first.

Area = ½ab sin C
Worked example A triangle has sides 8 cm and 6 cm, with an included angle of 50°. Find its area.
Area = ½ × 8 × 6 × sin(50°) = 18.4 cm² (3 s.f.)
Try it — Area Formula
  1. Find the area of a triangle with sides 10 cm and 12 cm, included angle 65°.
  2. A triangle has area 30 cm², with two sides of 9 cm and 10 cm. Find the included angle.
Show answers
  1. ½(10)(12)sin65° = 54.4 cm²
  2. 30 = ½(9)(10)sinC → sinC = 60/90 = 0.667 → C = 41.8°

5. Pythagoras’ Theorem in 3D

Extend Pythagoras to three dimensions by applying it twice — once across a face of the solid, then again using that result as one leg of a second right triangle.

Worked example — space diagonal of a cuboid A cuboid measures 3 cm × 4 cm × 12 cm. Find the length of its longest diagonal (corner to opposite corner).
First, the diagonal across the base: √(3² + 4²) = √25 = 5 cm
Then, combine with the height: √(5² + 12²) = √169 = 13 cm
(This combines into a single formula: diagonal = √(l² + w² + h²) = √(9+16+144) = √169 = 13 cm)
Try it — 3D Pythagoras
  1. Find the space diagonal of a cuboid measuring 6 cm × 8 cm × 10 cm.
  2. A cuboid has a space diagonal of 15 cm, with length 10 cm and width 6 cm. Find its height.
Show answers
  1. √(36+64+100) = √200 = 14.1 cm (3 s.f.)
  2. 15²=10²+6²+h² → h²=225−100−36=89 → h=9.43 cm (3 s.f.)

6. 3D Trigonometry — Angle Between a Line and a Plane

To find the angle between a sloping line and a flat plane, drop a perpendicular from the line down onto the plane — this creates a right-angled triangle you can solve with ordinary SOHCAHTOA.

Worked example bottom corner top corner θ A cuboid has base 6 cm × 8 cm and height 5 cm. Find the angle between the space diagonal (from a bottom corner to the opposite top corner) and the base plane.
First, find the base diagonal: √(6² + 8²) = √100 = 10 cm
This base diagonal and the height (5 cm) form a right-angled triangle with the space diagonal as hypotenuse.
tan θ = height / base diagonal = 5/10 = 0.5
θ = tan⁻¹(0.5) = 26.6° (1 d.p.)
Try it — 3D Trigonometry
  1. A cuboid has base 9 cm × 12 cm and height 8 cm. Find the base diagonal.
  2. Using the same cuboid, find the angle between the space diagonal and the base plane.
Show answers
  1. √(81+144) = √225 = 15 cm
  2. tan⁻¹(8/15) = 28.1°
Common mistakes to watch for
  • Using the sine rule when the given angle isn’t opposite a known side — check you have a genuine angle-side pair before applying it.
  • Forgetting to check the ambiguous case when using the sine rule with SSA information — a second valid triangle might exist.
  • Mixing up the cosine rule’s two forms — one solves for a side (SAS known), the other rearranges to solve for an angle (SSS known); make sure you’re using the version that matches what you’re given.
  • In 3D problems, trying to solve everything in one triangle — usually you need to work through one right-angled triangle first (often across a base or face) before you have enough information for the second one.
  • Forgetting that the angle between a line and a plane is measured to its projection onto the plane, found via the perpendicular drop — not to some other line in the plane.

7. Quick-Fire Challenge Round

True or False?
  1. The sine rule can be used on any triangle, not just right-angled ones.
  2. The cosine rule can give an ambiguous (two-solution) result, just like the sine rule.
  3. The area formula ½ab sinC requires the angle between the two given sides.
  4. 3D Pythagoras problems often require solving two right-angled triangles in sequence.
Show answers
  1. True.
  2. False — the cosine rule always gives one unique angle between 0° and 180°; the ambiguous case is specific to the sine rule.
  3. True — using the formula with an angle that isn’t between the two given sides gives the wrong area.
  4. True.
Spot the Error
  1. “Triangle with a=5, b=7, angle C=60° (between them) — using the sine rule since I have two sides.”
  2. “Area = ½(6)(9)sin(40°), where 40° is an angle NOT between the two given sides.”
  3. “Cosine rule gave cosC = −0.4, so there’s no valid angle.”
Show answers
  1. This is SAS information (two sides and the included angle) — that calls for the cosine rule, not the sine rule (which needs a matching angle-side pair).
  2. The area formula only works with the angle between the two given sides — using a different angle gives an incorrect area.
  3. A negative cosine value is completely valid — it just means the angle is obtuse: C = cos⁻¹(−0.4) = 113.6°.

8. Mixed Practice — Bringing It All Together

  1. Triangle has angle A=45°, angle B=75°, side a=10 cm. Find side b.
  2. Triangle has a=9 cm, b=11 cm, angle C=55° (between them). Find side c.
  3. Triangle has sides 7 cm, 9 cm, 13 cm. Find the largest angle.
  4. Find the area of a triangle with sides 11 cm and 8 cm, included angle 72°.
  5. A cuboid measures 4 cm × 5 cm × 20 cm. Find its space diagonal.
Show answers
  1. b = 10×sin75°/sin45° = 13.7 cm
  2. c² = 81+121−2(9)(11)cos55° = 202−113.6 = 88.4. c=9.40 cm
  3. Largest angle is opposite longest side (13): cosC=(49+81−169)/126=−0.310. C=108.0°
  4. ½(11)(8)sin72° = 41.8 cm²
  5. √(16+25+400) = √441 = 21 cm

Practice at Home

A large set again, to close out the entire Trigonometry unit — take it steadily.

Part A — Sine rule
  1. Triangle has angle A=55°, angle B=65°, side a=14 cm. Find side b.
  2. Triangle has a=11 cm, b=8 cm, angle A=70°. Find angle B.
Part B — The ambiguous case
  1. Triangle has a=7 cm, b=11 cm, angle A=32°. Find both possible values of angle B, and check whether both are valid.
Part C — Cosine rule
  1. Triangle has a=8 cm, b=13 cm, angle C=62° (between them). Find side c.
  2. Triangle has sides 6 cm, 10 cm, 14 cm. Find the largest angle.
Part D — Area formula
  1. Find the area of a triangle with sides 9 cm and 14 cm, included angle 48°.
  2. A triangle has area 45 cm², with sides 12 cm and 10 cm. Find the included angle.
Part E — 3D Pythagoras and trigonometry
  1. Find the space diagonal of a cuboid measuring 5 cm × 12 cm × 15 cm.
  2. A cuboid has base 8 cm × 6 cm and height 9 cm. Find the angle between the space diagonal and the base.
Part F — Applied / challenge
  1. Two ships leave a port at the same time. Ship A sails on a bearing of 040° at 20 km/h; Ship B sails on a bearing of 110° at 15 km/h. Find the distance between them after 2 hours (hint: find the angle between their paths first, then use the cosine rule).
  2. A right pyramid has a square base of side 10 cm and a height of 12 cm. Find the length of a sloping edge (from a base corner to the apex), and the angle it makes with the base.
  3. A triangular field has sides 120 m, 150 m, and 180 m. Find its area using the cosine rule (to find an angle first) followed by the area formula.

Key Vocabulary Recap

Sine rule — a/sinA = b/sinB = c/sinC; needs a matching angle-side pair
Cosine rule — c² = a²+b²−2ab cosC; needs SAS or SSS
Ambiguous case — when SSA data gives two valid triangle solutions
Space diagonal — the longest diagonal through a 3D solid, corner to opposite corner

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) sections E6.5 and E6.6 (Extended subject content) — Non-right-angled triangles (sine rule, cosine rule, area formula, the ambiguous case), and Pythagoras’ theorem and trigonometry in 3D. There is no Core equivalent for either section. The sine rule, cosine rule, and area formula are all given in the exam’s List of Formulas. This closes out the entire Trigonometry unit.

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