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IGCSE / O Level Mathematics · Topic 6 Trigonometry · Chapter 30 · pp. 418–435

Further trigonometry

SOH CAH TOA only works in right-angled triangles. This chapter handles any triangle: the sine rule and the cosine rule find missing sides and angles, ½ab sin C gives the area, and the same ideas, with Pythagoras, solve problems in three dimensions.

Learning objectives

  1. Use the sine rule to find a side or an angle in a non-right-angled triangle.
  2. Spot the ambiguous case, where one set of facts fits two different triangles.
  3. Use the cosine rule to find a side, or an angle.
  4. Find the area of a triangle with ½ab sin C, and the shortest distance from a point to a line.
  5. Solve problems in three dimensions, including the angle between a line and a plane.

Book exercises: 30.1 to 30.6 and Student assessments 1 and 2 (pp. 418–435). The whole chapter is Extended content.

The sine rule Extended

🔑 Key ideas

  • Name the sides with small letters a, b, c. Side a is opposite angle A, b opposite B, c opposite C.
  • Sine rule: = = . Turn it upside down to find an angle: = .
  • Use it when you know a matching pair (a side and its opposite angle) plus one more side or angle.
  • You only ever use two of the three fractions at a time.
C B A a (BC) b (CA) c (AB) a ÷ sin A = b ÷ sin B = c ÷ sin C Each side pairs with the angle facing it.
Side a faces angle A, b faces B and c faces C. The sine rule says every side divided by the sine of its facing angle gives the same number.

📘 Example 1 · Find BC. A = 40°, B = 30°, b = AC = 6 cm

  1. Side a = BC faces angle A (40°). Side b = 6 cm faces angle B (30°). That is a matching pair, so use the sine rule.
  2. Write it with the unknown on top: = .
  3. Multiply both sides by sin 40°: a = .
  4. Work it out: a = 6 × 0.6428 ÷ 0.5 = 7.7135, so BC = 7.7 cm (1 d.p.).

📘 Example 2 · Find angle C. A = 60°, a = 6 cm, c = 6.5 cm

  1. Angle C is unknown, so write the rule with the angles on top: = .
  2. Substitute: = .
  3. Multiply by 6.5: sin C = 6.5 × sin 60° ÷ 6 = 0.938.
  4. Use the inverse: C = sin−1(0.938) = 69.8° (1 d.p.).
  5. Check: 6.5 is a little longer than 6, so C should be a little bigger than 60°. ✔

The ambiguous case Extended

🔑 Key ideas

  • Two sides and an angle that is not between them (SSA) can fit two different triangles.
  • Reason: sin θ = sin (180° − θ). Your calculator gives the acute angle, but the obtuse angle has the same sine.
  • So when you use sin−1 to find an angle, also try 180° − your answer.
  • Keep the second answer only if the three angles still add up to less than 180°.
C A B₁ B₂ 20° AC = 10 cm AB = 8 cm swings and cuts the ray twice: B₁ and B₂
AC = 10 cm, AB = 8 cm and angle C = 20°. Swing AB like a compass arm: it meets the ray from C in two places, so there are two triangles. Angle CBA is acute at B₁ (25.3°) and obtuse at B₂ (154.7°).

📘 Example 3 · A = 30°, a = 6 cm, b = 10 cm. Find the possible values of angle B

  1. Sine rule, angles on top: = .
  2. sin B = 10 × 0.5 ÷ 6 = 0.8333.
  3. First answer: B = sin−1(0.8333) = 56.4°.
  4. Second answer: B = 180° − 56.4° = 123.6°.
  5. Test both: 30° + 56.4° = 86.4° and 30° + 123.6° = 153.6°. Both are under 180°, so both triangles exist.
  6. B = 56.4° or 123.6°.

💡 Not every SSA case gives two triangles. If the second angle makes the total reach 180° or more, only the acute answer is real. The triangle solver tab shows this for you.

✏️ Try it

  1. A = 40°, B = 70° and a = 5 cm. Find b.
  2. A = 50°, a = 12 cm and b = 9 cm. Find angle B.
  3. The sine rule gives sin B = 0.6. What are the two possible angles for B?
Answers

1. b = 5 × sin 70° ÷ sin 40° = 7.3 cm · 2. sin B = 9 × sin 50° ÷ 12 = 0.5745, so B = 35.1° (the second answer 144.9° fails because 50° + 144.9° is more than 180°) · 3. 36.9° and 143.1°

Next: the cosine rule →

The cosine rule Extended

🔑 Key ideas

  • For a side: a² = b² + c² − 2bc cos A. Angle A is the one between sides b and c.
  • For an angle: cos A = .
  • Use it for two sides and the angle between them (SAS), or all three sides (SSS).
  • If A = 90°, cos A = 0 and the rule becomes Pythagoras.
  • A negative cosine means an obtuse angle (over 90°).
SAS: find the third side A c b a = ? SSS: find an angle c b a A = ?
Left: sides b and c with the angle A between them give a². Right: all three sides give cos A.

📘 Example 4 · Find BC. b = 9 cm, c = 7 cm, A = 50°

  1. Two sides (9 cm and 7 cm) and the angle between them (50°): use the cosine rule.
  2. a² = 9² + 7² − 2 × 9 × 7 × cos 50°.
  3. a² = 81 + 49 − 126 × 0.6428 = 130 − 80.99 = 49.0.
  4. a = √49.0 = 7.00, so BC = 7.00 cm (3 s.f.).

📘 Example 5 · Find angle A. a = 20 cm, b = 15 cm, c = 12 cm

  1. All three sides are known: use the angle form. Side a = 20 faces the angle you want.
  2. cos A = = .
  3. cos A = = −0.086.
  4. A = cos−1(−0.086) = 94.9° (1 d.p.).
  5. Check: the longest side faces the biggest angle, and 94.9° is obtuse, so the cosine is negative. ✔

📘 Example 6 · Two walkers leave a gate. One goes 5 km, the other 8 km, and their paths are 70° apart. How far apart are they?

  1. Two sides (5 and 8) with the angle between them (70°): cosine rule.
  2. d² = 5² + 8² − 2 × 5 × 8 × cos 70° = 25 + 64 − 80 × 0.3420.
  3. d² = 89 − 27.36 = 61.64.
  4. d = √61.64 = 7.85 km (3 s.f.).

💡 Do the multiplication before the subtraction: work out 2bc cos A first, then take it away. Typing b² + c² − 2bc cos A in one go on the calculator is fine, but never subtract before you multiply.

✏️ Try it

  1. Sides 6 cm and 10 cm enclose an angle of 60°. Find the third side.
  2. A triangle has sides 7 cm, 8 cm and 9 cm. Find the angle opposite the 9 cm side.
  3. A triangle has sides 4 cm, 5 cm and 6 cm. Find its largest angle.
Answers

1. a² = 36 + 100 − 120 × 0.5 = 76, so a = 8.7 cm · 2. cos = (49 + 64 − 81) ÷ 112 = 0.2857, so 73.4° · 3. The largest angle faces 6 cm: cos = (16 + 25 − 36) ÷ 40 = 0.125, so 82.8°

Next: area and shortest distance →

Area of a triangle Extended

🔑 Key ideas

  • Area = ½ × base × height, and the height is measured perpendicular to the base.
  • In the triangle below, sin C = h ÷ a, so h = a sin C.
  • So Area = ½ab sin C: two sides and the angle between them.
  • Works for obtuse angles too, because sin (180° − θ) = sin θ.
C C A B b (base) a h h ÷ a = sin C h = a sin C Area = ½ × b × h Area = ½ab sin C
Drop the height h from B to the base CA. The right-angled triangle on the left gives h = a sin C, and that goes into ½ × base × height.

📘 Example 7 · Find the area. Sides 9 cm and 7 cm enclose an angle of 50°

  1. Two sides and the angle between them: use ½ab sin C.
  2. Area = ½ × 9 × 7 × sin 50° = 31.5 × 0.7660.
  3. Area = 24.1 cm² (1 d.p.).

📘 Example 8 · Sides 8 cm and 12 cm enclose an acute angle x. The area is 30 cm². Find x

  1. Write the area formula: 30 = ½ × 8 × 12 × sin x = 48 sin x.
  2. Divide by 48: sin x = 30 ÷ 48 = 0.625.
  3. x = sin−1(0.625) = 38.7° (1 d.p.).
  4. Check: ½ × 8 × 12 × sin 38.7° = 48 × 0.625 = 30. ✔ (An obtuse angle of 141.3° would also give 30 cm².)

Shortest distance from a point to a line Extended

🔑 Key ideas

  • The shortest distance from a point to a line is perpendicular to the line, through the point.
  • Make a right-angled triangle: the slanting line is the hypotenuse, so the distance is hypotenuse × sin (angle with the line).
  • If you only know three sides, find an angle with the cosine rule first. Or use area: height = 2 × area ÷ base.
X Y P shortest longer longer O Ship S, 5 km from A angle SAB = 30° distance = 5 × sin 30° = 2.5 km
Any slanting line from P to XY is longer than the perpendicular PO. The shortest distance is OP.

📘 Example 9 · A ship S is 5 km from point A on a straight coast AB. Angle SAB = 30°. How close is S to the coast?

  1. Drop the perpendicular from S to AB, meeting it at O. Triangle SOA has a right angle at O.
  2. SA = 5 km is the hypotenuse, and SO is the side opposite the 30° angle.
  3. sin 30° = , so SO = 5 × sin 30° = 2.5 km.

📘 Example 10 · Area of a quadrilateral split by a diagonal. AB = 8, AC = 10, AD = 5 (cm), angle BAC = 30°, angle CAD = 40°

  1. The line AC splits the quadrilateral into triangles ABC and ACD, both with a corner at A.
  2. Triangle ABC: ½ × 8 × 10 × sin 30° = 40 × 0.5 = 20 cm².
  3. Triangle ACD: ½ × 10 × 5 × sin 40° = 25 × 0.6428 = 16.07 cm².
  4. Add: 20 + 16.07 = 36.1 cm² (1 d.p.).

✏️ Try it

  1. Find the area of a triangle with sides 8 cm and 9 cm enclosing an angle of 40°.
  2. A triangle has area 30 cm² and two sides of 8 cm and 12 cm. Find the acute angle between them.
  3. A point P is 12 cm from A. The line AB makes an angle of 35° with AP. How far is P from the line AB?
Answers

1. ½ × 8 × 9 × sin 40° = 36 × 0.6428 = 23.1 cm² · 2. sin x = 30 ÷ 48 = 0.625, so x = 38.7° · 3. 12 × sin 35° = 6.9 cm (6.88)

Next: three dimensions →

Trigonometry in three dimensions Extended

🔑 Key ideas

  • A solid is a set of flat triangles hiding in space. Find a right-angled triangle that contains the length or angle you want.
  • Pull that triangle out and draw it flat, labelling the sides you know.
  • Use Pythagoras for a length and SOH CAH TOA for an angle.
  • Often you need two steps: a face diagonal first, then a space diagonal.
A B C D E F G H A E G 3 cm √18 cm tan G = 3 ÷ √18 G = 35.3°
Cube of edge 3 cm. The thick lines EG (across the bottom face) and AG (through the cube) form the right-angled triangle AEG, shown flat on the right.
1. Find a right-angled triangle 2. Draw it flat, label what you know 3. Pythagoras for a length; SOH CAH TOA: angle 4. Repeat if it feeds the next triangle
The method for every 3D problem in this chapter.

📘 Example 11 · Cube of edge 3 cm. Find EG, then AG, then angle EGA

  1. EG: triangle EHG is right-angled at H. EG² = 3² + 3² = 18, so EG = √18 = 4.24 cm.
  2. AG: triangle AEG is right-angled at E. AG² = 3² + (√18)² = 9 + 18 = 27.
  3. AG = √27 = 5.20 cm (3 s.f.).
  4. Angle EGA: in triangle AEG, the side opposite G is AE = 3 and the side next to G is EG = √18, so tan G = .
  5. G = 35.3° (1 d.p.).

📘 Example 12 · Cuboid with HG = 5 cm, GF = 12 cm, height 8 cm. Find EG, AG and angle AGE

  1. EG is the diagonal of the 5 by 12 base: EG² = 5² + 12² = 169, so EG = 13 cm.
  2. AG: triangle AEG has a right angle at E with AE = 8 and EG = 13. AG² = 64 + 169 = 233.
  3. AG = √233 = 15.3 cm (3 s.f.).
  4. Angle AGE: tan = opposite ÷ adjacent = 8 ÷ 13 = 0.6154.
  5. Angle AGE = 31.6° (1 d.p.).
  6. Check: sin 31.6° × 15.26 = 8.0, the height. ✔

💡 A cuboid with edges p, q, r has a longest (space) diagonal of √(p² + q² + r²). It is Pythagoras twice.

✏️ Try it

  1. A cuboid is 3 cm by 4 cm by 12 cm. Find the diagonal of the 3 by 4 base.
  2. Find the longest diagonal of that cuboid.
  3. Find the angle the longest diagonal makes with the base.
Answers

1. √(9 + 16) = 5 cm · 2. √(5² + 12²) = 13 cm · 3. tan θ = 12 ÷ 5, so θ = 67.4°

Next: the angle between a line and a plane →

The angle between a line and a plane Extended

🔑 Key ideas

  • A plane is a flat surface, such as a floor, a wall or one face of a solid.
  • From a point B on the line, drop a perpendicular to the plane. It lands at C.
  • Join the point where the line meets the plane, A, to C. AC is the projection of AB on the plane.
  • The angle between AB and the plane is the angle between AB and its projection AC.
the plane A C B angle Line: AB Projection: AC BC is ⟂ the plane tan (angle) = BC ÷ AC
The angle between the line AB and the plane is the angle at A between AB and AC.
A B C D E F G H C D E 5 cm √20 cm √45 cm E = 48.2°
The book's cuboid (HG = 5, GF = 2, height 4). The plane ADHE is the left face. C drops straight onto D, so the angle between CE and the plane is angle CED.

📘 Example 13 · Cuboid with HG = 5 cm, GF = 2 cm and height 4 cm. Find CE, then the angle between CE and the plane ADHE

  1. EG lies across the bottom face: EG² = 2² + 5² = 29.
  2. CE: triangle CGE has a right angle at G. CE² = EG² + CG² = 29 + 16 = 45.
  3. CE = √45 = 6.71 cm (3 s.f.).
  4. Projection: CD is perpendicular to the face ADHE, so the projection of CE onto that plane is DE. The angle is CED.
  5. In the right-angled triangle CDE: CD = 5 (opposite) and CE = √45 (hypotenuse). sin E = .
  6. E = 48.2° (1 d.p.).
A D E B C X 12 cm 5 cm 8 cm 6 cm Right pyramid base DE = 8, EB = 6 edge AB = 13 DB = √(8² + 6²) = 10 XB = 5, AX = 12 cos ∠ABX = 5 ÷ 13 = 67.4°
A right pyramid has its apex A straight above the centre X of the base. The angle between the edge AB and the base is angle ABX, in the vertical triangle AXB. (AX is the height, 12 cm.)

📘 Example 14 · Right pyramid, rectangular base 8 cm by 6 cm, sloping edges 13 cm. Find the height and the angle between an edge and the base

  1. Base diagonal: DB² = 8² + 6² = 100, so DB = 10 cm.
  2. X is the centre of the base, so XB = 10 ÷ 2 = 5 cm.
  3. Triangle AXB has a right angle at X and AB = 13 cm: AX² = 13² − 5² = 169 − 25 = 144.
  4. Height AX = 12 cm.
  5. The projection of AB on the base is XB, so the angle is ABX: cos B = .
  6. Angle ABX = 67.4° (1 d.p.). Check: sin 67.4° × 13 = 12.0. ✔

💡 Naming the projection is half the job. Ask: where does the top end of the line land when I drop it straight down onto the plane? Join that landing point to the bottom end.

✏️ Try it

  1. A cuboid is 8 cm long, 6 cm wide and 5 cm high. Find the angle between its longest diagonal and the base.
  2. A right pyramid has a square base of side 6 cm and a height of 4 cm. Find the length of a sloping edge, and the angle it makes with the base.
  3. A line AB rises 8 m above a plane. Its projection AC on the plane is 6 m. Find the angle between AB and the plane.
Answers

1. Base diagonal 10 cm, tan θ = 5 ÷ 10, so θ = 26.6° · 2. Half the base diagonal is √18 = 4.243 cm. Edge = √(18 + 16) = √34 = 5.83 cm. tan θ = 4 ÷ 4.243, so θ = 43.3° · 3. tan θ = 8 ÷ 6, so θ = 53.1°

Next: the triangle solver →

Which rule? Triangle solver

🔑 Key ideas

  • Look at what you are given. That decides the rule.
  • A side and the angle opposite it, plus one more fact: sine rule.
  • Two sides and the angle between them, or all three sides: cosine rule.
  • Always check the answer: the longest side faces the largest angle, and the angles add up to 180°.
What are you given? SSS three sides Cosine rule Cosine rule for each angle SAS two sides and the angle between Cosine rule for the third side AAS / ASA two angles and any side Sine rule after angle sum SSA two sides and an angle not between Sine rule after angle sum check for 2 answers
Choosing the rule. S means a side, A means an angle, in the order they sit around the triangle.

🧪 Try it: non-right-angled triangle solver

Fill in exactly three values and leave the others blank. Sides are a, b, c; angle A faces side a, B faces b, C faces c (angles in degrees).

Examples:

✏️ Try it

  1. Which rule do you use for a triangle with A = 35°, B = 80° and c = 9 cm?
  2. Which rule for sides a = 7, b = 9 and C = 110°?
  3. Which rule for sides a = 7, b = 9 and c = 12?
Answers

1. Two angles and a side: find C = 65°, then the sine rule · 2. Two sides and the angle between them: cosine rule for c · 3. All three sides: cosine rule for an angle. Check each one in the solver.

Next: summary and quiz →

Chapter summary

RuleFormulaUse it when you know
Sine rule = A side and its opposite angle, plus one more side or angle
Cosine rule (side)a² = b² + c² − 2bc cos ATwo sides and the angle between them
Cosine rule (angle)cos A = All three sides
Area½ab sin CTwo sides and the angle between them
  • Side a faces angle A. Always match sides with the angle opposite.
  • SSA can give two triangles: also try 180° − (your angle) and check the angles still total under 180°.
  • A negative cosine means an obtuse angle.
  • The shortest distance from a point to a line is the perpendicular distance.
  • 3D: find a right-angled triangle, draw it flat, use Pythagoras for lengths and SOH CAH TOA for angles.
  • The angle between a line and a plane is the angle between the line and its projection on the plane.
  • Cuboid space diagonal = √(p² + q² + r²).

📄 Practice questions

The book's Exercises 30.1 to 30.6 and Student assessments 1 and 2 are in the printable assignment for this chapter. Diagrams are described in words there.

Mini-quiz

1. Which is the sine rule for triangle ABC (a is opposite A, b is opposite B)?
2. In triangle ABC, A = 30°, B = 45° and a = 8 cm. What is b to 1 d.p.?
3. You know two sides and the angle between them. Which rule finds the third side?
4. Two sides of 5 cm and 8 cm enclose an angle of 60°. How long is the third side?
5. A triangle has sides 3 cm, 4 cm and 5 cm. What does the cosine rule give for the largest angle?
6. The sine rule gives sin B = 0.8 with B possibly obtuse. What are the two possible values of B?
7. Two sides of 10 cm and 12 cm enclose an angle of 30°. What is the area of the triangle?
8. What is the shortest distance from a point P to a straight line?
9. A cuboid is 3 cm by 4 cm by 12 cm. How long is its longest diagonal?
10. In a 3D problem, what is the angle between a line and a plane?
11. A cube has edges of 4 cm. What angle does its longest diagonal make with the base?
12. Angles of 40° and 60° are known, with a side of 10 cm opposite the 40° angle. What is the third angle?

Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 30 (Further trigonometry). Explanations, the Try it answers, the tool and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.