In Grade 10, SOH CAH TOA only worked for right-angled triangles. These three rules work for ANY triangle — including the weird, lopsided ones that appear in exams.
Which Rule Do I Use? — The Decision Flowchart#
This is the most important skill. Before you calculate anything, ask:
| What do I have? | Which rule? |
|---|---|
| Two angles + one side (AAS or ASA) | Sine Rule (find the missing angle first: $180° - A - B$) |
| One matching pair + one extra (side + opposite angle + something else) | Sine Rule |
| Two sides + included angle (SAS — the “sandwich”) | Cosine Rule (to find the third side) |
| Three sides (SSS) | Cosine Rule (to find an angle) |
| Two sides + non-included angle (SSA) | Sine Rule — but watch for the ambiguous case! |
| Need the area (two sides + included angle) | Area Rule |
1. The Sine Rule#
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$Or flipped (useful when finding angles):
$$ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} $$Logic: In any triangle, the ratio of a side to the sine of its opposite angle is constant.
Worked Example: Finding a side#
In $\triangle ABC$: $\hat{A} = 40°$, $\hat{B} = 75°$, $a = 10$ cm. Find $b$.
$\hat{C} = 180° - 40° - 75° = 65°$
$\frac{a}{\sin A} = \frac{b}{\sin B}$
$\frac{10}{\sin 40°} = \frac{b}{\sin 75°}$
$b = \frac{10 \sin 75°}{\sin 40°} = \frac{10 \times 0.9659}{0.6428} = 15.03$ cm
Worked Example: Finding an angle#
In $\triangle PQR$: $p = 8$, $q = 12$, $\hat{P} = 30°$. Find $\hat{Q}$.
$\frac{\sin Q}{q} = \frac{\sin P}{p}$
$\frac{\sin Q}{12} = \frac{\sin 30°}{8} = \frac{0.5}{8}$
$\sin Q = \frac{12 \times 0.5}{8} = 0.75$
$\hat{Q} = \sin^{-1}(0.75) = 48.6°$
⚠️ The Ambiguous Case#
When you use the Sine Rule to find an angle and get $\sin Q = 0.75$, there are actually two possible angles: $48.6°$ AND $180° - 48.6° = 131.4°$.
Check: Does $131.4°$ make sense? Add it to the given angle: $30° + 131.4° = 161.4° < 180°$. Yes, there’s room for a third angle. So there are TWO possible triangles.
When to worry: Only when you have SSA (two sides + non-included angle). If the question gives AAS or ASA, there’s no ambiguity.
2. The Cosine Rule#
Finding a side (SAS):#
$$ a^2 = b^2 + c^2 - 2bc\cos A $$Finding an angle (SSS):#
$$ \cos A = \frac{b^2 + c^2 - a^2}{2bc} $$Logic: This is Pythagoras with an “adjustment term” ($-2bc\cos A$) that accounts for the triangle not being right-angled. When $A = 90°$, $\cos 90° = 0$, and it reduces to Pythagoras exactly.
Worked Example: Finding a side#
In $\triangle ABC$: $b = 7$, $c = 10$, $\hat{A} = 60°$. Find $a$.
$a^2 = 7^2 + 10^2 - 2(7)(10)\cos 60°$
$a^2 = 49 + 100 - 140 \times 0.5$
$a^2 = 149 - 70 = 79$
$a = \sqrt{79} = 8.89$ cm
Worked Example: Finding an angle#
In $\triangle ABC$: $a = 5$, $b = 7$, $c = 9$. Find $\hat{C}$ (the largest angle, opposite the longest side).
$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{25 + 49 - 81}{2(5)(7)} = \frac{-7}{70} = -0.1$
$\hat{C} = \cos^{-1}(-0.1) = 95.7°$
The negative cosine correctly tells us the angle is obtuse. No ambiguity with the cosine rule!
3. The Area Rule#
$$ \text{Area} = \frac{1}{2}ab\sin C $$Where $a$ and $b$ are two sides and $C$ is the included angle (the angle BETWEEN those two sides).
Worked Example#
Find the area of $\triangle PQR$ where $p = 8$ cm, $r = 11$ cm, $\hat{Q} = 50°$.
$\text{Area} = \frac{1}{2}(8)(11)\sin 50° = 44 \times 0.766 = 33.7$ cm²
4. 2D Problems — Combining the Rules#
Exam problems often involve multiple triangles. Strategy:
- Draw and label the diagram with all given info.
- Identify which triangle to work with first.
- Choose the right rule for that triangle.
- Transfer information to the next triangle.
Worked Example: Two Triangles#
A lighthouse is observed from two points A and B, 500 m apart. From A, the angle of elevation to the top of the lighthouse is $25°$. From B, it’s $40°$. Points A, B, and the base of the lighthouse are on the same horizontal level. $A\hat{L}B$ (at the lighthouse base) = $180° - 25° - 40°$… Actually, let’s use a proper setup:
From A: the bearing to the lighthouse L is $N30°E$. From B (which is due east of A, 500 m away): the bearing to L is $N50°W$.
In $\triangle ABL$: $\hat{A} = 60°$ (from bearing), $\hat{B} = 40°$ (from bearing), $AB = 500$ m.
$\hat{L} = 180° - 60° - 40° = 80°$
Using Sine Rule: $\frac{AL}{\sin B} = \frac{AB}{\sin L}$
$AL = \frac{500 \sin 40°}{\sin 80°} = \frac{500 \times 0.6428}{0.9848} = 326.4$ m
🚨 Common Mistakes#
- Choosing the wrong rule: If you have a matching pair (side + opposite angle), use Sine Rule. If you have SAS or SSS, you MUST use Cosine Rule. Using the wrong rule gives wrong answers.
- The included angle for Area Rule: The angle MUST be between the two sides you’re using. $\frac{1}{2}(5)(7)\sin C$ only works if $C$ is between sides 5 and 7.
- Forgetting the ambiguous case: When the Sine Rule gives you an angle, always check if $180° - \theta$ is also valid (only for SSA situations).
- Calculator in DEG mode: If your calculator is in RAD mode, every answer will be wrong. Check before every trig calculation.
- Cosine Rule sign errors: In $a^2 = b^2 + c^2 - 2bc\cos A$, the minus sign is part of the formula. If $A$ is obtuse, $\cos A$ is negative, making $-2bc\cos A$ positive — so $a^2$ becomes LARGER than $b^2 + c^2$. This makes sense (the side opposite an obtuse angle is the longest).
💡 Pro Tip: The “Pairing” Check#
Before you start calculating, mark on the diagram:
- Complete pairs: a side AND its opposite angle ✓✓
- Half pairs: a side OR its opposite angle, but not both ✓
If you have at least one complete pair and one half pair → Sine Rule. If you can’t make any pairs → Cosine Rule.
🔗 Related Grade 11 topics:
- Reduction Formulas & CAST — needed when angles in triangle problems go beyond $90°$
- Trig Identities & Equations — trig equations arise when solving for angles
- Analytical Geometry: Inclination — the angle of inclination uses $\tan\theta = m$
📌 Grade 10 foundation: Trig Ratios & Special Angles — SOH CAH TOA for right-angled triangles.
⏮️ Reduction Formulas | 🏠 Back to Trigonometry | ⏭️ Identities & Equations