Triangles (Sine/Cosine Rules) (AA SL)
Right-angled trigonometry only gets you so far - most real triangles don't have a 90° angle to work with. The sine rule and cosine rule let you find any missing side or angle in any triangle, as long as you're given enough information to start. This topic covers both rules, the area formula that doesn't need a height, and the tricky "ambiguous case" where two different triangles can fit the same data.
What the syllabus says
This topic maps onto two points in the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| SL3.2 | The sine rule \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\) (not including the ambiguous case). The cosine rule \(c^2=a^2+b^2-2ab\cos C\); \(\cos C=\dfrac{a^2+b^2-c^2}{2ab}\). Area of a triangle as \(\tfrac12ab\sin C\). |
| SL3.5 | Extension of the sine rule to the ambiguous case, where two different triangles can satisfy the same given information. |
Both are core AA SL syllabus points, examined on Paper 1 and Paper 2.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is the sine rule?
The sine rule links every side of a triangle to the sine of its opposite angle: \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\). It's the tool of choice whenever you know an angle and the side opposite it, plus one more piece of information.
e.g. If \(a=10\), \(A=40^\circ\) and \(B=60^\circ\), then \(b=\dfrac{10\sin60^\circ}{\sin40^\circ}\approx13.5\).
What is the cosine rule?
The cosine rule generalises Pythagoras' theorem to any triangle: \(c^2=a^2+b^2-2ab\cos C\). Use it when you know two sides and the angle between them (SAS), or all three sides (SSS) and need an angle.
e.g. If \(a=5\), \(b=7\) and \(C=60^\circ\), then \(c^2=25+49-2(5)(7)(0.5)=39\), so \(c\approx6.24\).
What is the area formula for a triangle?
\(\text{Area}=\tfrac12ab\sin C\) finds a triangle's area from two sides and the angle between them, without ever needing a perpendicular height. It works for right-angled and non-right-angled triangles alike.
e.g. If \(a=8\), \(b=10\) and \(C=30^\circ\), then Area \(=\tfrac12(8)(10)\sin30^\circ=40(0.5)=20\) cm\(^2\).
What is the ambiguous case?
When you're given two sides and a non-included angle (SSA), the sine rule can give two different valid angles for the same data - one acute, one its obtuse supplement. Both may form a genuine triangle, so you need to check which fits.
e.g. If \(a=8\), \(c=10\) and \(A=30^\circ\), then \(\sin C=\dfrac{10\sin30^\circ}{8}=0.625\), so \(C\approx38.7^\circ\) or \(C\approx141.3^\circ\).
What does it mean to "solve a triangle"?
Solving a triangle means finding every missing side and angle. You generally need three pieces of information to start (SSS, SAS, ASA/AAS, or the ambiguous SSA), including at least one side length.
e.g. Given \(a=6\), \(b=8\), \(C=90^\circ\): \(c^2=36+64-2(6)(8)\cos90^\circ=100-0=100\), so \(c=10\) - the cosine rule reduces to Pythagoras when \(C=90^\circ\).
Key formulas
Three formulas cover every question on this topic. The tables below summarise them at a glance and show which rule to reach for - the explanations underneath go into more depth on each one.
Formula reference
All three formulas below are printed in the official IB formula booklet, under the geometry and trigonometry section.
| Formula | Used for | Booklet? |
|---|---|---|
| \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\) | Sine rule | ✓ Yes |
| \(c^2=a^2+b^2-2ab\cos C\) | Cosine rule (find a side) | ✓ Yes |
| \(\cos C=\dfrac{a^2+b^2-c^2}{2ab}\) | Cosine rule (find an angle) | ✓ Yes |
| \(\text{Area}=\tfrac12ab\sin C\) | Area without a height | ✓ Yes |
Sine rule vs cosine rule
Choosing the right rule comes down to what information you're given. This table lines the two up side by side.
| Feature | Sine rule | Cosine rule |
|---|---|---|
| You know | An angle and its opposite side, plus one more piece (ASA, AAS, or SSA) | Two sides and the included angle (SAS), or all three sides (SSS) |
| Finds | A side or an angle opposite a known pair | The third side (SAS) or any angle (SSS) |
| Ambiguous case? | Yes - SSA data can give 0, 1, or 2 triangles | No - SAS and SSS always give a unique triangle |
| Special case | No simplification for right angles | Reduces to Pythagoras' theorem when \(C=90^\circ\) |
Using the sine rule
The sine rule only works when you can pair up an angle with the side directly opposite it.
Finding a side
\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\]
Use when you know one full angle-side pair and one more angle.
Finding an angle
\[\sin B=\dfrac{b\sin A}{a}\]
Rearrange the sine rule to isolate the unknown angle's sine.
The ambiguous case
\[\sin B=k \Rightarrow B=\sin^{-1}(k) \text{ or } 180^\circ-\sin^{-1}(k)\]
Given SSA data, always check whether the obtuse solution also fits.
Using the cosine rule
The cosine rule handles the two cases the sine rule can't reach directly: SAS and SSS.
Finding a side (SAS)
\[c^2=a^2+b^2-2ab\cos C\]
Substitute the two known sides and the included angle directly.
Finding an angle (SSS)
\[\cos C=\dfrac{a^2+b^2-c^2}{2ab}\]
Rearranged form of the cosine rule - use when all three sides are known.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
A triangle has sides \(8\) cm and \(11\) cm with included angle \(40^\circ.\)
(a) Find its area.
(b) Find the third side.
Worked solution
(a) \(\tfrac12(8)(11)\sin40^\circ\) M1
\(\approx28.3\text{ cm}^2.\) A1
(b) \(c^2=64+121-176\cos40^\circ\approx50.2,\ c\) M1
\(\approx7.08\) cm. A1
In triangle \(ABC,\ AB=11\) cm, \(BC=8\) cm and angle \(BAC=32^\circ.\)
(a)(i) Find the size of angle \(ACB\) with \(ACB<90\).
(a)(ii) Find the size with \(ACB>90\).
(b) For the case where angle \(ABC\) is acute, find the area of triangle \(ABC.\)
Worked solution
(a) Sine rule: \(\dfrac{\sin C}{11}=\dfrac{\sin32^\circ}{8}.\) M1
\(\sin C=\dfrac{11\sin32^\circ}{8}=0.7286\Rightarrow C\approx46.8^\circ\) (acute case). A1
or \(C\approx133.2^\circ\) (obtuse case, since \(180-46.8=133.2\)). A1
(b) For angle \(ABC\) to be acute, take the obtuse solution \(C=133.2^\circ\) (from (a)(ii)): \(B = 180-32-133.2=14.8^\circ.\) M1
\(B=14.8^\circ.\) A1
Area \(=\tfrac12\cdot11\cdot8\sin(14.8^\circ)\) M1
\(\approx11.2\) cm². A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Reaching for the sine rule with SAS data. If you only know two sides and the angle between them, there's no angle-side pair to plug into the sine rule - you need the cosine rule instead.
- Missing the ambiguous case. When given SSA data, taking only the acute answer from \(\sin^{-1}\) can lose marks if the question (or the diagram) needs the obtuse solution too.
- Sign errors in the cosine rule. Writing \(c^2=a^2+b^2+2ab\cos C\) instead of \(-2ab\cos C\) - the minus sign is what makes the formula reduce to Pythagoras when \(C=90^\circ\).
- Forgetting to sanity-check the angle sum. A triangle's angles must add to \(180^\circ\) - if your three angles don't, a mistake happened earlier (often a wrong degree/radian mode on the GDC).
Using your GDC
No calculator model needs a special sequence for this topic - the sine and cosine rules just need ordinary trig evaluation.
Once you've set up the sine rule or cosine rule equation on paper, your GDC does the arithmetic - there's no special calculator mode for "triangle solving" itself.
Make sure your calculator's angle setting (degrees or radians) matches the question before you evaluate any \(\sin\), \(\cos\), or their inverses - this is the single most common source of a wrong answer on this topic. Enter the cosine rule or sine rule expression directly on the home/calculator screen once you've rearranged it algebraically, and use the inverse trig keys (\(\sin^{-1}\), \(\cos^{-1}\)) to recover an angle from a ratio. For the ambiguous case, remember your calculator only ever returns the acute (or principal) solution from \(\sin^{-1}\) - you have to work out the obtuse alternative yourself as \(180^\circ\) minus that value.
Tip: If an angle answer looks impossible (e.g. the three angles you've found don't sum to \(180^\circ\)), check the calculator's degree/radian mode first.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Triangle questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
When do I use the sine rule instead of the cosine rule?
Use the sine rule when you know an angle and its opposite side (ASA, AAS, or the ambiguous SSA case). Use the cosine rule when you know two sides and the included angle (SAS), or all three sides (SSS) and need an angle.
What is the ambiguous case of the sine rule?
When you're given two sides and a non-included angle (SSA), the sine rule can return two valid triangles - one with an acute angle and one with its obtuse supplement (180 minus that angle). You have to check which (or both) fit the given information.
Is the area formula (1/2)ab sin C in the formula booklet?
Yes. The sine rule, cosine rule, and the area formula (1/2)ab sin C are all listed in the IB formula booklet under geometry and trigonometry, so you don't need to memorise them - just know how to use them.
Can I use my GDC for this topic?
Yes - your calculator evaluates sin, cos, and their inverses directly, so once you've set up the correct equation from the sine or cosine rule, the calculator does the arithmetic. See the GDC guide for general tips.
Sub-topics
Triangles (Sine/Cosine Rules) broken down into its individual skills, each with its own focused page.
Related topics
More Geometry & Trigonometry topics from the same AA SL syllabus unit, in case you want to keep going.