IB & A-Level Maths
The sine rule connects each side of a triangle to the angle opposite it. It works in any triangle, not just right-angled ones, which is exactly why it exists.
The sine rule
a ÷ sin A = b ÷ sin B = c ÷ sin C — use this form to find a side. Flip it to sin A ÷ a = sin B ÷ b = sin C ÷ c to find an angle. Each letter pairs with the angle directly opposite it.
Most pages stop there. Two things decide whether you actually get the marks: knowing when the sine rule is the wrong tool, and knowing that one particular set of information gives two valid triangles — and that examiners set that case deliberately.
In any triangle, label each side with the lower-case letter matching the angle opposite it: side a is opposite angle A, side b opposite B, side c opposite C. That pairing is the whole idea.
Turning every fraction upside down gives an equally true statement, and it is the one you want when the unknown is an angle:
Which version to write down
This is the decision that decides the question, and it takes about three seconds once you know what to look for. Count what you have been given.
Scroll the diagram sideways →
Two sides with the angle between them is the classic cosine-rule setup, because the angle you are given sits opposite the side you do not know. There is no complete pair, so the sine rule has nothing to work with.
A useful follow-on: once the cosine rule has given you that third side, you now do have a matching pair — so the rest of the triangle can be finished with the sine rule. Many exam questions are built exactly that way.
In triangle ABC, angle A = 40°, angle B = 75° and side a = 9 cm. Find side b.
Sanity check it before moving on: angle B is larger than angle A, so side b must be longer than side a. 13.5 is longer than 9. If your answer had come out smaller, something went wrong.
In triangle ABC, a = 12 cm, b = 8 cm and angle A = 65°. Find angle B.
Step 6 is not optional
Whenever you use the sine rule to find an angle, you must ask whether a second answer exists. Your calculator only ever gives you one. Skipping that check is the single most common way to lose marks on this topic, and it is why the next section exists.
Here is the situation examiners use to separate students who understand the sine rule from students who can operate a calculator.
You are given two sides and an angle that is not between them — a side, another side, and an angle opposite one of them. Sometimes that information describes exactly one triangle. Sometimes it describes two completely different triangles, both perfectly valid. And sometimes no triangle at all.
In triangle ABC, a = 7 cm, b = 10 cm and angle A = 40°. Find angle B.
Scroll the diagram sideways →
You do not have to guess. Given side a, side b and angle A, compare a with b and with b sin A:
| Condition | Triangles | Why |
|---|---|---|
| a < b sin A | None | Side a is too short to reach the base. sin B would come out greater than 1, which your calculator rejects. |
| a = b sin A | Exactly one | Side a just reaches, meeting the base at a right angle. B = 90° and the two possible answers collapse into one. |
| b sin A < a < b | Two | This is the ambiguous case. Side a can swing to either of two positions, both valid. |
| a ≥ b | Exactly one | The obtuse option would make the angles add to more than 180°, so it is impossible. |
If the side you are given is shorter than the other side, stop and check for a second answer.
In the worked example above, a = 7 and b = 10, so a is shorter — and b sin A = 10 × sin 40° = 6.43, which is less than 7. That puts it squarely in the two-triangle band.
Sometimes the question itself rules one out. Look for wording like “the obtuse triangle”, “where angle B is acute”, or a diagram showing the shape. If a diagram is given, measure it by eye — an obviously wide triangle rules out the acute answer.
If nothing in the question narrows it down and the maths allows two, give both. A correct pair of answers scores full marks. One answer out of two scores partial marks at best.
When there is no matching pair, the cosine rule takes over. It relates all three sides to one angle.
Rearranged to find an angle when you know all three sides:
Worth noticing: put A = 90° into the first version and cos 90° = 0, so the whole last term vanishes and you are left with a² = b² + c². Pythagoras is just the cosine rule for a right-angled triangle.
The cosine rule has no ambiguous case
The third formula in this family. You will not get far with ½ × base × height in a triangle with no right angle, so use two sides and the angle between them:
The angle must be the one between the two sides you use. Any pair of sides works as long as you take the angle they enclose.
A triangle has sides of 9 cm and 12 cm with an angle of 58° between them. Find its area.
| The mistake | What goes wrong | Do this instead |
|---|---|---|
| Missing the second answer | The most common error in the topic. Your calculator gives one angle; the triangle may allow two. | Every time you find an angle, check whether 180° minus it also works. |
| Pairing the wrong side and angle | Using side b with angle A. The sine rule only works with a side and the angle opposite it. | Label the diagram first. Side a faces angle A, always. |
| Reaching for the sine rule with no pair | Two sides and the angle between them has no matching pair, so the sine rule cannot start. | No pair means cosine rule. Check before you write anything. |
| Calculator in radians | Every answer is wrong while every method is right. Nothing on screen warns you. | Check DEG before question 1. |
| Rounding during the working | Rounding an angle to 1 d.p. mid-question and feeding it into the next step shifts the final answer. | Keep full accuracy in the calculator and round only at the end. |
Work each one on paper, then tap to check. Every answer below was verified numerically.
Pair available: a with A. Sides on top.
b / sin 75° = 9 / sin 40°
b = 9 × sin 75° / sin 40°
b = 13.5 cm (3 s.f.)
Finding an angle, so sines on top
sin B / 6 = sin 50° / 9
sin B = 6 × sin 50° / 9 = 0.5107
B = 30.7° (1 d.p.)
a is longer than b, so there is no second answer
a is shorter than b, so check for two triangles
b sin A = 9 × sin 35° = 5.16
5.16 < 6 < 9, so this is the ambiguous case
sin B = 9 × sin 35° / 6 = 0.8604
B = 59.4° or 180° − 59.4° = 120.6°
Check both: C = 85.6° and C = 24.4°, both positive
B = 59.4° or 120.6°
The given angle sits opposite the unknown side
cosine rule
There is no side paired with its opposite angle, so the sine rule has nothing to start from
Two sides and the angle between them, so cosine rule
a² = 8² + 11² − 2(8)(11) cos 37°
a² = 64 + 121 − 176 × 0.7986
a² = 185 − 140.6 = 44.4
a = 6.67 cm (3 s.f.)
Largest angle is opposite the longest side, 9
cos C = (5² + 7² − 9²) / (2 × 5 × 7)
cos C = (25 + 49 − 81) / 70 = −0.1
Negative cosine means the angle is obtuse
C = 95.7° (1 d.p.)
Area = ½ × 9 × 12 × sin 58°
= 54 × 0.8480
Area = 45.8 cm² (3 s.f.)
No pair yet, so start with the cosine rule
a² = 7² + 10² − 2(7)(10) cos 62°
a = 9.125 cm
Now a pairs with A, so switch to the sine rule
sin B = 7 × sin 62° / 9.125 = 0.6773
a = 9.13 cm, B = 42.6°
a is longer than b, so no second answer for B
The sine rule states that in any triangle, a ÷ sin A = b ÷ sin B = c ÷ sin C, where each lower-case letter is the side opposite the angle with the matching capital letter. Flipped over, sin A ÷ a = sin B ÷ b = sin C ÷ c, which is the version to use when you are looking for an angle. It works in every triangle, not only right-angled ones.
Use the sine rule when you have a side together with the angle opposite it — a matching pair. If you have three sides, or two sides and the angle between them, there is no matching pair and you need the cosine rule instead.
Because sine is positive for both acute and obtuse angles. sin 66.7° and sin 113.3° are the same number, so when you use inverse sine your calculator returns only the acute one. If the obtuse option still leaves a positive third angle, it is an equally valid triangle and the question has two answers.
It is the situation where the information given describes two different valid triangles. It happens when you are given two sides and an angle that is not between them, and the side opposite the known angle is the shorter of the two. Specifically, with sides a and b and angle A, you get two triangles when b sin A < a < b and A is acute.
Compare side a with b sin A and with b. If a < b sin A there is no triangle. If a = b sin A there is exactly one, and it is right-angled. If b sin A < a < b with A acute there are two. If a is greater than or equal to b there is exactly one.
Only when both are genuinely possible and the question does not rule one out. Read the wording for phrases like ‘the obtuse triangle’ or ‘where B is acute’, and look at any diagram. If nothing narrows it down and the maths allows two, give both — one answer out of two usually scores partial marks at best.
No. Cosine is negative for obtuse angles and positive for acute ones, so the sign of the answer tells you unambiguously which it is. That makes the cosine rule the safer choice whenever you have a genuine option between the two.
Area = ½ ab sin C, where a and b are two sides and C is the angle between them. The angle must be the one enclosed by the two sides you use. Any pair of sides works as long as you take the angle between them.
Yes — Pythagoras is the cosine rule for a right-angled triangle. Put A = 90° into a² = b² + c² − 2bc cos A. Since cos 90° = 0, the last term disappears and you are left with a² = b² + c².
Yes, it works perfectly well. But SOH-CAH-TOA is usually quicker for a right-angled triangle, so most questions expect that instead. The sine rule earns its place when there is no right angle to work with.
The three usual causes are pairing a side with the wrong angle instead of the one opposite it, leaving the calculator in radians rather than degrees, and rounding part-way through and feeding the rounded value into the next step. Keep full accuracy until the final line.
Nothing has gone wrong with your method — it means no triangle exists with that combination of sides and angle. Side a is too short to reach. If this happens in an exam, say so: ‘no triangle is possible’ is the correct answer, not an error.
Everything above, condensed
The sine rule sits alongside the rest of your trigonometry. If the exact values are not automatic yet, our trig exact value chart covers those, and the double angle formulae are the next step once triangles give way to identities.
The ambiguous case is a question type, not an accident — examiners choose those numbers deliberately. Our tutors teach IB and A‑level maths and work through your own past papers, so the habits we fix are the ones actually costing you marks.
Book a free 30-minute meetingRevising the whole subject? Our IB revision notes cover maths topic by topic at SL and HL, or read about our IB maths tutors.
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