Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 3 Geometry and trigonometry · 3.2 Trigonometry in triangles
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Use sine, cosine and tangent to find sides and angles of right-angled triangles | SL, HL | "Find BC" or "Find the angle BÂC" (2 marks) |
| Use inverse sine, cosine and tangent to find an angle from a ratio | SL, HL | the second line of any angle question, linked to inverse functions in 2.2 |
| Use the sine rule a ÷ sin A = b ÷ sin B = c ÷ sin C to find a side or an angle | SL, HL | "Find the length of PR" (3 marks), Paper 2 |
| Use the cosine rule c² = a² + b² − 2ab cos C to find a side | SL, HL | "Find QR" given two sides and the included angle (3 marks) |
| Use cos C = (a² + b² − c²) ÷ 2ab to find an angle from three sides | SL, HL | "Show that cos A = −1/4" or "Find the largest angle" (3 to 4 marks) |
| Find the area of a triangle as ½ab sin C, and use it backwards | SL, HL | "Find the area of triangle ABC" (2 marks); "the area is 30 cm², find θ" (3 marks) |
| Choose the right rule from what is given | SL, HL | every multi-part triangle question |
| Sketch a clearly labelled diagram to support a solution | SL, HL | expected in every question on this topic, and it earns method marks |
Before you start
You need right-angled trigonometry from before the course and the angle sum of a triangle, 180°. You need inverse functions from 2.2, because finding an angle means undoing sin, cos or tan. The formula booklet prints the sine rule, both forms of the cosine rule and the area rule, so the marks are for choosing and using them. It does not print SOH CAH TOA; learn it. Exact values such as sin 60° = √3/2 are taught fully in 3.5; the few used on Paper 1 here are listed in section 2. This page does not include the ambiguous case of the sine rule; that is in 3.5.
1The idea in one paragraph
A triangle has three sides and three angles, and if you know the right three of those six, the other three are fixed. Trigonometry is the machinery for finding them. In a right-angled triangle, SOH CAH TOA is enough. In any other triangle you use one of two rules. The sine rule says each side divided by the sine of the angle facing it gives the same number, so it works whenever you know a side and the angle opposite it. The cosine rule is Pythagoras' theorem with a correction term for the angle not being 90°, and it works when you know two sides and the angle between them, or all three sides. A third formula, ½ab sin C, gives the area without needing the height. The whole skill is looking at what you are given and picking the rule that fits.
2Right-angled triangles, revisited
In a right-angled triangle, the longest side, opposite the right angle, is the hypotenuse. The other two sides are named from the angle you are working with: the opposite faces it and the adjacent runs alongside it to the right angle. Figure 1 shows that opposite and adjacent swap when you change angle, while the hypotenuse stays put.
Finding a side. A right-angled triangle has hypotenuse 14 cm and an angle of 35°. The side opposite the 35° angle is 14 sin 35° = 8.03 cm, and the side adjacent to it is 14 cos 35° = 11.5 cm (3 s.f.).
Finding an angle. The two shorter sides are 5 cm and 9 cm, and θ faces the 5 cm side. Then tan θ = 5/9, so θ = tan⁻¹(5/9) = 29.1°. The inverse function undoes tan: it takes a ratio and returns the angle. On the calculator it is the shift or second function of the tan key.
Exact values for Paper 1. Two special triangles give exact ratios you should know without a calculator: half an equilateral triangle (angles 30°, 60°, 90°, sides 1, √3, 2) and half a square (45°, 45°, 90°, sides 1, 1, √2).
| θ | 30° | 45° | 60° |
|---|---|---|---|
| sin θ | 1/2 | √2/2 | √3/2 |
| cos θ | √3/2 | √2/2 | 1/2 |
| tan θ | √3/3 | 1 | √3 |
In 3.5 these extend to angles beyond 90°, for example sin 120° = √3/2 and cos 120° = −1/2, which Paper 1 cosine-rule questions use.
3Labelling any triangle
Once a triangle has no right angle there is no hypotenuse, so the sides need a new naming system. Call the angles A, B and C, and give each side the small letter of the angle that faces it, as in Figure 2.
The formula booklet writes every rule in these letters, but a question will label its triangle PQR or XYZ. Before using a rule, match the letters: in triangle PQR, the side opposite angle P is QR. Mislabelling is the root of most wrong answers on this page.
4The sine rule
a ÷ sin A = b ÷ sin B = c ÷ sin C · in the formula booklet. Use it when you know a side and the angle opposite it.
Why it is true. Drop a perpendicular of length h from C to side c, as in Figure 3. It splits the triangle into two right-angled triangles. In the left one, h = b sin A. In the right one, h = a sin B. Both equal h, so b sin A = a sin B, and dividing by sin A sin B gives a ÷ sin A = b ÷ sin B. Dropping a perpendicular from a different vertex brings in c ÷ sin C the same way.
Finding a side. In triangle ABC, A = 40°, B = 75° and a = 12 cm. Find b and c.
Finding an angle. When the unknown is an angle, flip the rule so the sines are on top: sin A ÷ a = sin B ÷ b. In triangle ABC, a = 9 cm, b = 7 cm and A = 65°. Find B.
A check that costs nothing: the largest angle faces the longest side. Here b < a, so B must be smaller than A, and 44.8° < 65° agrees.
A warning about sine and obtuse angles. Sine is positive for every angle between 0° and 180°, so sin B = 0.7049 is also true for B = 180° − 44.8° = 135.2°. The calculator only ever gives the acute one. The guide leaves the "two possible triangles" case, the ambiguous case, to 3.5, and questions on this subtopic are set so that only one triangle exists. Here 135.2° is impossible anyway, since 65° + 135.2° exceeds 180°. The safe habit: use the sine rule to find an angle only when that angle is not the largest in the triangle. For the largest angle, use the cosine rule, which never hides an obtuse answer.
5The cosine rule
c² = a² + b² − 2ab cos C, and rearranged, cos C = (a² + b² − c²) ÷ 2ab · both in the formula booklet.
Use the first form when you know two sides and the angle between them (the included angle) and want the third side. Use the second when you know all three sides and want an angle.
Why it is Pythagoras with a correction. If C = 90°, then cos C = 0, the last term vanishes and c² = a² + b², which is Pythagoras. If C is smaller than 90°, cos C is positive and c² is less than a² + b²: the triangle is squashed and c is shorter. If C is larger than 90°, cos C is negative, the correction is added, and c is longer. Figure 4 keeps a = 5 and b = 8 and opens the angle between them from 60° to 90° to 120°.
Where it comes from (for the curious). Put C at the origin and side b along the x-axis, so A = (b, 0). Then B = (a cos C, a sin C). The distance formula gives c² = (a cos C − b)² + (a sin C)² = a²cos²C − 2ab cos C + b² + a²sin²C, and because cos²C + sin²C = 1 (an identity from 3.6), this is a² + b² − 2ab cos C.
Finding a side (Paper 2). Two sides are 7 cm and 10 cm, with an angle of 52° between them. Find the third side c.
Type the whole right-hand side into the calculator in one go, then square-root. The classic slip is to work out 149 − 140 first and multiply that by cos 52°; subtraction does not come before multiplication.
Finding an angle, exactly (Paper 1). A triangle has sides 5, 7 and 8. Find the angle opposite the side of length 7.
An obtuse angle. A triangle has sides 4, 5 and 8. Its largest angle faces the side of length 8.
The negative cosine is the signal that the angle is obtuse, and cos⁻¹ returns it correctly, which is exactly why the cosine rule is the right tool for a largest angle.
Exact side (Paper 1). In triangle ABC, b = 4, c = 6 and A = 60°. The rule, relabelled for angle A, is a² = b² + c² − 2bc cos A.
6The area of a triangle
Area = ½ab sin C · in the formula booklet: two sides and the angle between them.
Area is ½ × base × height, and Figure 5 shows where the height comes from. Take side a as the base. The height from A down to it is the opposite side of a right-angled triangle with hypotenuse b and angle C, so height = b sin C. Then area = ½ × a × b sin C. The angle must be the one between the two sides you use: two sides and the angle trapped between them.
Exact area (Paper 1). Two sides of 8 cm and 11 cm meet at 30°. Area = ½ × 8 × 11 × sin 30° = 44 × ½ = 22 cm². For the triangle with b = 4, c = 6 and A = 60° in section 5, the area is ½ × 4 × 6 × (√3/2) = 6√3.
Backwards: from area to angle. A triangle has sides 8 cm and 10 cm and area 20 cm². Find the angle between those sides.
Both answers give a genuine triangle with that area: a narrow one and a wide one. Unless the question says the angle is acute or obtuse, give both. This is not the ambiguous case of the sine rule; it simply follows from sin θ = sin(180° − θ).
7Choosing the rule, and solving a whole triangle
Figure 6 sorts every starting position.
- A side and its opposite angle are both known: the sine rule.
- Two sides and the included angle, wanting the third side: the cosine rule.
- Three sides, wanting an angle: the cosine rule rearranged.
- Two sides and the included angle, wanting the area: ½ab sin C.
To "solve a triangle" means to find every side and angle. Work in an order that keeps you safe.
Worked example (Paper 2). In triangle PQR, PQ = 9 cm, PR = 6 cm and QPR = 110°. Find QR, the other two angles, and the area. Figure 7 shows the finished triangle.
Notice two choices. After the cosine rule, the sine rule was used for Q, the angle facing the shortest side, which is certainly acute. And the unrounded QR stayed in the calculator: using 12.4 changes nothing here at 3 s.f., but on a longer chain it can.
Sketch every time. The guide asks you to sketch well-labelled diagrams to support solutions. A quick sketch with the given values on it shows which side faces which angle, and a marker can award method marks from a clear diagram even when the arithmetic slips.
8Where marks are lost
Naming opposite and adjacent from the wrong angle. They depend on the angle in use. Mark the angle first, then name the sides.
Using the sine rule for an obtuse angle. sin⁻¹ only returns angles up to 90°. For the largest angle, or any angle that might be obtuse, use the cosine rule.
Mismatching side and angle. a goes with A, the angle facing it. In PQR, QR goes with P. Write the pairs on your sketch.
Evaluating the cosine rule in the wrong order. 149 − 140 cos 52° is not (149 − 140) cos 52°. Enter the whole expression, or work out 140 cos 52° first.
Using ½ab sin C with the wrong angle. The angle must be between the two sides used. If it is not, find the included angle first.
Calculator in radians. On a degree question, a calculator left in radian mode gives nonsense like sin 30 = −0.988. Check the mode before the first calculation, and remember that radians are the default in the rest of the course (3.4).
Forgetting the second angle from an area. sin θ = 0.5 gives 30° and 150°. Give both unless the question rules one out.
Rounding a side, then reusing it. Keep the calculator value of each length for the next step and round only the final answers.
9Work it right
- Sketch the triangle and put every given value on it. Mark the unknown.
- Match letters: write the pairs side–opposite angle before using a rule.
- Decide: right angle? SOH CAH TOA. Known pair? Sine rule. Two sides and included angle? Cosine rule or area. Three sides? Cosine rule for an angle.
- Write the rule with the numbers substituted before you calculate. That line is the method mark.
- For an angle that could be obtuse, use the cosine rule.
- Check the answer is sensible: the largest angle faces the longest side, and the angles sum to 180°.
- Paper 1: leave surds and use the exact values. Paper 2: 3 significant figures for lengths, 1 decimal place or 3 s.f. for angles, with units.
10Try it
Marks in brackets. Q1 and Q2 are Paper 1 style, no calculator. Q3 to Q5 are Paper 2 style, with a GDC.
Q1. A triangle has sides of length 3 cm, 5 cm and 7 cm.
(a) Show that the largest angle of the triangle is 120°. 3 marks
(b) Find the exact area of the triangle. 3 marks
Q2. In triangle ABC, the angle at B is 90°, AC = 10 cm and the angle at A is 30°.
(a) Find the exact lengths of BC and AB. 3 marks
(b) The point D lies on AB, between A and B, such that the angle BDC is 45°. Find the exact length AD. 3 marks
Q3. In triangle XYZ, XY = 15 cm, the angle at X is 48° and the angle at Y is 63°.
(a) Find the length YZ. 3 marks
(b) Find the area of the triangle. 2 marks
Q4. Two sides of a triangle are 8.4 cm and 11.2 cm, and the angle between them is 38°.
(a) Find the length of the third side. 3 marks
(b) Find the size of the largest angle of the triangle. 3 marks
Q5. A triangle has two sides of 9 cm and 12 cm, with angle θ between them. Its area is 30 cm².
(a) Find the two possible values of θ. 3 marks
(b) Given that θ is obtuse, find the length of the third side. 3 marks
11In one breath
In a right-angled triangle, name the opposite and adjacent from the angle you are using and apply SOH CAH TOA, with the inverse functions to get an angle back. In any other triangle, label side a opposite angle A. If you know a side and the angle facing it, the sine rule, a ÷ sin A = b ÷ sin B = c ÷ sin C, finds another side or an angle, but never trust it for an angle that might be obtuse. If you know two sides and the angle between them, the cosine rule c² = a² + b² − 2ab cos C gives the third side; if you know all three sides, cos C = (a² + b² − c²) ÷ 2ab gives any angle, and a negative cosine means obtuse. The area is ½ab sin C with the angle trapped between the two sides, and working backwards from an area gives two angles, θ and 180° − θ. Sketch, label, pick the rule from what you have, and keep unrounded values.
Answers
Q1. (a) The largest angle faces the longest side, 7. cos θ = (3² + 5² − 7²) ÷ (2 × 3 × 5) = (9 + 25 − 49) ÷ 30 = −15/30 = −1/2, so θ = 120°. R1 for identifying the angle opposite 7 (may be implied by the substitution), M1 for the cosine rule correctly substituted, A1 for −1/2 leading to 120°. This is "show that": the value −1/2 must appear.
(b) Area = ½ × 3 × 5 × sin 120° = (15/2) × (√3/2) = 15√3/4 cm². M1 for ½ab sin C with the two sides that enclose 120°, A1 for sin 120° = √3/2, A1 for 15√3/4.
Q2. (a) BC = 10 sin 30° = 5 cm. AB = 10 cos 30° = 10 × √3/2 = 5√3 cm. M1 for a correct trigonometric ratio with the hypotenuse 10, A1 for BC = 5, A1 for AB = 5√3.
(b) In triangle BDC, the angle at B is 90° and the angle at D is 45°, so tan 45° = BC ÷ BD gives BD = 5. Then AD = AB − BD = 5√3 − 5 = 5(√3 − 1) cm. M1 for using triangle BDC with tan 45° = 1 (or noticing it is isosceles), A1 for BD = 5, A1 for AD. Either form of the answer is accepted.
Q3. (a) The angle at Z is 180° − 48° − 63° = 69°, and it faces XY = 15. YZ faces the angle at X. YZ ÷ sin 48° = 15 ÷ sin 69°, so YZ = 15 sin 48° ÷ sin 69° = 11.9 cm (3 s.f.). A1 for the angle 69°, M1 for the sine rule with correctly matched pairs, A1 for 11.9. Using sin 63° in place of sin 48° is a mismatched pair and scores M0.
(b) Area = ½ × XY × YZ × sin 63° = ½ × 15 × 11.940… × sin 63° = 79.8 cm² (3 s.f.). M1 for ½ab sin C with the angle between the two sides used, A1 for 79.8. Any correct pairing, such as ½ × 15 × XZ × sin 48° with XZ = 14.3, earns full marks.
Q4. (a) c² = 8.4² + 11.2² − 2(8.4)(11.2) cos 38° = 47.727…, so c = 6.91 cm (3 s.f.). M1 for the cosine rule substituted, A1 for c² = 47.7, A1 for 6.91.
(b) The largest angle faces the longest side, 11.2. cos θ = (8.4² + 6.9085…² − 11.2²) ÷ (2 × 8.4 × 6.9085…) = −0.0616…, so θ = 93.5° (3 s.f.). M1 for identifying the angle opposite 11.2, M1 for the cosine rule for an angle with their c, A1 for 93.5°. The sine rule gives 86.5°, the acute partner of the true angle, and scores M1 M0 A0: this is exactly the trap in section 4.
Q5. (a) ½ × 9 × 12 × sin θ = 30, so sin θ = 60 ÷ 108 = 0.5556…, giving θ = 33.7° or θ = 146.3° (1 d.p.). M1 for the area rule set equal to 30, A1 for 33.7°, A1 for 146.3° (180° minus their acute angle).
(b) With θ = 146.25…°: c² = 9² + 12² − 2(9)(12) cos 146.25…° = 404.60…, so c = 20.1 cm (3 s.f.). M1 for the cosine rule with the obtuse angle, A1 for c² ≈ 405, A1 for 20.1. Using the acute angle gives 6.74 cm and scores M1 A0 A0.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 3.2 Trigonometry in triangles. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.