Chapter 4: Sine Rule & Cosine Rule
SOHCAHTOA only works in right-angled triangles. For any other triangle — whether it has all acute angles or an obtuse angle — we need two new tools: the Sine Rule and the Cosine Rule. Together they allow us to find any unknown side or angle in any triangle, given sufficient information.
Exam Board Coverage
This entire chapter is assessed at Extended tier only for both Cambridge IGCSE (0580) and Edexcel IGCSE (4MA1). The formulae are provided on the formula sheet in the Cambridge exam but not in the Edexcel exam — check your board's specification.
4.1 Labelling Non-Right-Angled Triangles
For the Sine Rule and Cosine Rule to work, the triangle must be labelled with a consistent convention: lower-case letters for sides and the corresponding upper-case letter for the opposite angle.
Labelling Convention
- Side $a$ is opposite angle $A$
- Side $b$ is opposite angle $B$
- Side $c$ is opposite angle $C$
Always re-label the triangle at the start of each question so the convention is clear.
Triangle $ABC$ with sides $a$, $b$, $c$ opposite to angles $A$, $B$, $C$ respectively.
4.2 Sine Rule — Finding a Side
The Sine Rule links each side to its opposite angle. It is used when you know an angle and its opposite side (giving you a complete ratio), plus one more piece of information.
Sine Rule
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
To find a side: put the unknown side in the numerator of the first fraction and a known side-angle pair in the second.
When to Use the Sine Rule to Find a Side
You need: one complete side-angle pair (e.g., $a$ and $A$) plus the angle opposite the unknown side (e.g., $B$ to find $b$).
Given information pattern: AAS (Angle–Angle–Side) or ASA
Example 4.1 — Sine Rule to find a side
In triangle $ABC$: $A = 48°$, $B = 65°$, $a = 9\,\text{cm}$. Find side $b$. Give your answer to 3 s.f.
Step 1 Write the Sine Rule with $b$ and $a$:
$$\frac{b}{\sin B} = \frac{a}{\sin A}$$
Step 2 Substitute: $\dfrac{b}{\sin 65°} = \dfrac{9}{\sin 48°}$
Step 3 $b = \dfrac{9 \sin 65°}{\sin 48°} = \dfrac{9 \times 0.9063}{0.7431} \approx \mathbf{10.97\ldots \approx 11.0\,\text{cm}}$
4.3 Sine Rule — Finding an Angle
Rearrange the Sine Rule so the unknown angle's sine is in the numerator. This form is sometimes written as:
Sine Rule (Angle Form)
$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$
To find an angle: put the unknown $\sin$ in the first fraction, then use a known pair.
Example 4.2 — Sine Rule to find an angle
In triangle $PQR$: $p = 12\,\text{cm}$, $q = 9\,\text{cm}$, $P = 72°$. Find angle $Q$.
Step 1 $\dfrac{\sin Q}{q} = \dfrac{\sin P}{p}$
Step 2 $\dfrac{\sin Q}{9} = \dfrac{\sin 72°}{12}$
Step 3 $\sin Q = \dfrac{9 \sin 72°}{12} = \dfrac{9 \times 0.9511}{12} \approx 0.7133$
Step 4 $Q = \sin^{-1}(0.7133) \approx \mathbf{45.5°}$
4.4 The Ambiguous Case (SSA)
The ambiguous case arises when two sides and a non-included angle (SSA) are given. There may be two valid triangles, one triangle, or no triangle.
When the Ambiguous Case Arises
Suppose you know sides $a$, $b$ and angle $A$ (where $A$ is opposite $a$).
- If $A \geq 90°$: at most one solution (obtuse angle at $A$ cannot also have obtuse $B$).
- If $A < 90°$ and $a < b \sin A$: no triangle exists.
- If $A < 90°$ and $a = b \sin A$: exactly one right-angled triangle.
- If $A < 90°$ and $b \sin A < a < b$: two triangles exist (angles $B$ and $180° - B$ both work).
- If $A < 90°$ and $a \geq b$: exactly one triangle.
Example 4.3 — Ambiguous case with two solutions
In triangle $ABC$: $a = 10\,\text{cm}$, $b = 14\,\text{cm}$, $A = 40°$. Find all possible values of angle $B$.
Step 1 $\dfrac{\sin B}{14} = \dfrac{\sin 40°}{10}$
Step 2 $\sin B = \dfrac{14 \sin 40°}{10} = \dfrac{14 \times 0.6428}{10} \approx 0.8999$
Step 3 First solution: $B_1 = \sin^{-1}(0.8999) \approx 64.1°$
Step 4 Second solution: $B_2 = 180° - 64.1° = 115.9°$
Check For $B_2 = 115.9°$: $A + B_2 = 40° + 115.9° = 155.9° < 180°$ ✓ — both solutions are valid.
So $B \approx 64.1°$ or $B \approx 115.9°$.
4.5 Cosine Rule — Finding a Side
The Cosine Rule relates all three sides and one angle. It is used when no complete side-angle pair is available — typically when two sides and the included angle are given (SAS), or all three sides are known (SSS).
Cosine Rule (Finding a Side)
$$a^2 = b^2 + c^2 - 2bc\cos A$$
Similarly: $b^2 = a^2 + c^2 - 2ac\cos B$ and $c^2 = a^2 + b^2 - 2ab\cos C$
Memory aid: "Square the two sides, add, subtract twice their product times the cosine of the included angle."
Example 4.4 — Cosine Rule to find a side (SAS)
In triangle $ABC$: $b = 8\,\text{cm}$, $c = 11\,\text{cm}$, $A = 60°$. Find side $a$.
Step 1 $a^2 = b^2 + c^2 - 2bc\cos A = 8^2 + 11^2 - 2(8)(11)\cos 60°$
Step 2 $= 64 + 121 - 176 \times 0.5 = 185 - 88 = 97$
Step 3 $a = \sqrt{97} \approx \mathbf{9.85\,\text{cm}}$ (3 s.f.)
4.6 Cosine Rule — Finding an Angle
Rearrange $a^2 = b^2 + c^2 - 2bc\cos A$ to make $\cos A$ the subject.
Cosine Rule (Finding an Angle)
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$
Used when all three sides are known (SSS). Find the largest angle first (opposite the longest side) to check for obtuse angles.
Example 4.5 — Cosine Rule to find an angle (SSS)
A triangle has sides $a = 7\,\text{cm}$, $b = 9\,\text{cm}$, $c = 5\,\text{cm}$. Find angle $A$ to 1 d.p.
Step 1 $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{81 + 25 - 49}{2 \times 9 \times 5} = \dfrac{57}{90} = 0.6\overline{3}$
Step 2 $A = \cos^{-1}(0.6\overline{3}) \approx \mathbf{50.7°}$
Example 4.6 — Cosine Rule with obtuse angle
A triangle has sides $5\,\text{cm}$, $6\,\text{cm}$, $9\,\text{cm}$. Find the largest angle.
Step 1 Largest angle is opposite the longest side $a = 9$. Let $b = 5$, $c = 6$.
Step 2 $\cos A = \dfrac{25 + 36 - 81}{2 \times 5 \times 6} = \dfrac{-20}{60} = -\dfrac{1}{3}$
Step 3 $A = \cos^{-1}\!\left(-\tfrac{1}{3}\right) \approx \mathbf{109.5°}$
A negative cosine value confirms the angle is obtuse ($> 90°$).
4.7 Area of a Triangle: $\frac{1}{2}ab\sin C$
The standard area formula $\frac{1}{2} \times \text{base} \times \text{height}$ requires a perpendicular height, which may not be given. When two sides and the included angle are known, use this formula instead.
Area Formula
$$\text{Area} = \tfrac{1}{2}ab\sin C$$
where $a$ and $b$ are two sides and $C$ is the angle between them.
Example 4.7 — Area using the sine formula
Triangle $PQR$ has $p = 7\,\text{cm}$, $q = 10\,\text{cm}$ and included angle $R = 115°$. Find the area to 3 s.f.
Step 1 The angle between sides $p$ and $q$ is $R = 115°$.
Step 2 Area $= \tfrac{1}{2} \times 7 \times 10 \times \sin 115°$
Step 3 $= 35 \times 0.9063\ldots \approx \mathbf{31.7\,\text{cm}^2}$
4.8 Choosing the Right Rule
Decision Flowchart
Start: What information do you have?
┃
Right-angled triangle? → Yes → Use SOHCAHTOA (or Pythagoras)
┃ No
Do you have a complete side-angle pair (e.g., $a$ and $A$)?
┃ Yes → Use Sine Rule
┃ No (SAS or SSS)
→ Use Cosine Rule
Summary Table
- AAS or ASA (angle, angle, one side) → Sine Rule for a side; find third angle by subtraction first if needed
- SSA (two sides, non-included angle) → Sine Rule for an angle; check for ambiguous case
- SAS (two sides, included angle) → Cosine Rule for a side, then Sine Rule for remaining angles
- SSS (three sides) → Cosine Rule for an angle, then Sine Rule for the rest
- SAS (area) → $\tfrac{1}{2}ab\sin C$
Exercises
All angles to 1 decimal place; all lengths to 3 significant figures unless otherwise stated.
Exercise 4.1
In triangle $ABC$: $A = 55°$, $C = 70°$, $c = 15\,\text{cm}$. Find side $a$.
Show Solution
$\dfrac{a}{\sin 55°} = \dfrac{15}{\sin 70°}$
$a = \dfrac{15\sin 55°}{\sin 70°} = \dfrac{15 \times 0.8192}{0.9397} \approx \mathbf{13.1\,\text{cm}}$
Exercise 4.2
In triangle $PQR$: $p = 8\,\text{cm}$, $r = 11\,\text{cm}$, $P = 35°$. Find angle $R$.
Show Solution
$\dfrac{\sin R}{11} = \dfrac{\sin 35°}{8}$
$\sin R = \dfrac{11\sin 35°}{8} \approx \dfrac{11\times0.5736}{8} \approx 0.7885$
$R = \sin^{-1}(0.7885) \approx \mathbf{52.0°}$
(Check: $35° + 52° = 87° < 180°$ ✓. Second solution $180°-52°=128°$ gives $35°+128°=163° < 180°$ — also valid, so state both.)
Exercise 4.3
In triangle $ABC$: $a = 6\,\text{cm}$, $c = 9\,\text{cm}$, $B = 75°$. Find side $b$.
Show Solution
$b^2 = 6^2 + 9^2 - 2(6)(9)\cos 75° = 36 + 81 - 108\times0.2588 = 117 - 27.95 = 89.05$
$b = \sqrt{89.05} \approx \mathbf{9.44\,\text{cm}}$
Exercise 4.4
A triangle has sides $6\,\text{cm}$, $8\,\text{cm}$, $10\,\text{cm}$. Find the largest angle.
Show Solution
Largest angle opposite $a=10$: $\cos A = \dfrac{36+64-100}{2\times6\times8} = \dfrac{0}{96} = 0$
$A = \cos^{-1}(0) = \mathbf{90°}$ (confirms it is a 6-8-10 right-angled triangle)
Exercise 4.5
Find the area of a triangle with sides $12\,\text{cm}$ and $15\,\text{cm}$ and included angle $50°$.
Show Solution
Area $= \tfrac{1}{2}\times12\times15\times\sin50° = 90\times0.7660 \approx \mathbf{68.9\,\text{cm}^2}$
Exercise 4.6
In triangle $XYZ$: $x = 5\,\text{cm}$, $y = 7\,\text{cm}$, $X = 40°$. Find both possible values of angle $Y$, if they exist.
Show Solution
$\sin Y = \dfrac{7\sin40°}{5} = \dfrac{7\times0.6428}{5} \approx 0.9000$
$Y_1 = \sin^{-1}(0.9) \approx 64.2°$; check: $40°+64.2° < 180°$ ✓
$Y_2 = 180°-64.2° = 115.8°$; check: $40°+115.8° < 180°$ ✓
Two valid triangles: $Y \approx 64.2°$ or $Y \approx 115.8°$
Exercise 4.7
Two ships leave a port. Ship A travels $18\,\text{km}$ on a bearing of $050°$ and Ship B travels $24\,\text{km}$ on a bearing of $110°$. Find the distance between them to 3 s.f.
Show Solution
Angle between the two bearings at the port: $110° - 50° = 60°$.
By the Cosine Rule: $d^2 = 18^2 + 24^2 - 2(18)(24)\cos60° = 324 + 576 - 864\times0.5 = 900 - 432 = 468$
$d = \sqrt{468} \approx \mathbf{21.6\,\text{km}}$
Exercise 4.8
Find the area of a triangle with sides $4\,\text{cm}$, $7\,\text{cm}$, $9\,\text{cm}$, by first finding an angle using the Cosine Rule.
Show Solution
Find angle opposite the longest side $a=9$: $\cos A = \dfrac{16+49-81}{56} = \dfrac{-16}{56} \approx -0.2857$, so $A \approx 106.6°$
Area $= \tfrac{1}{2}\times4\times7\times\sin106.6° \approx 14\times0.9579 \approx \mathbf{13.4\,\text{cm}^2}$
Exercise 4.9
In triangle $ABC$: $b = 13\,\text{cm}$, $c = 13\,\text{cm}$, $A = 50°$. Find the area and then side $a$.
Show Solution
Area $= \tfrac{1}{2}\times13\times13\times\sin50° = \tfrac{169}{2}\times0.7660 \approx \mathbf{64.7\,\text{cm}^2}$
$a^2 = 13^2+13^2-2(13)(13)\cos50° = 338 - 338\times0.6428 = 338 - 217.3 = 120.7$
$a \approx \mathbf{10.99 \approx 11.0\,\text{cm}}$
Exercise 4.10
A triangle has area $30\,\text{cm}^2$, two sides of $10\,\text{cm}$ and $8\,\text{cm}$. Find the possible values of the angle between those sides.
Show Solution
$30 = \tfrac{1}{2}\times10\times8\times\sin\theta = 40\sin\theta$
$\sin\theta = \dfrac{30}{40} = 0.75$
$\theta_1 = \sin^{-1}(0.75) \approx 48.6°$ or $\theta_2 = 180°-48.6° = \mathbf{131.4°}$
Exam Tips
Tip 1 — Write the formula first, then substitute
Always write the Sine or Cosine Rule in its general form before substituting values. Examiners award a method mark for the correct formula, even if you make an arithmetic error later.
Tip 2 — Identify SSS/SAS vs AAS/SSA before you start
Sine Rule needs a complete pair ($a$ and $A$). If you only have sides (or two sides and an included angle), you need the Cosine Rule. Choosing the wrong rule wastes time and marks.
Tip 3 — Check the ambiguous case explicitly
If SSA is given with an acute angle, always check whether $180° - B$ is also valid by testing $A + B_2 < 180°$. Exam questions that award two marks for two answers will penalise you if you only give one.
Tip 4 — Negative cosine means obtuse angle
If your Cosine Rule calculation gives a negative value for $\cos A$, $A$ is obtuse ($>90°$). Make sure your calculator gives an answer between $90°$ and $180°$ — it will do so automatically when $\cos^{-1}$ is applied to a negative number.