Full lesson notes
Everything covered in this lesson, in one place - useful for revision or printing.
\( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} \qquad a^2 = b^2 + c^2 - 2bc\cos A \)
In Grade 10 every triangle you solved had a right angle in it. That is a big restriction — most triangles in the real world do not.
These three rules remove it completely. By the end of this lesson you will be able to:
• find the area of any triangle without knowing its height
• use the sine rule to find a missing side or angle
• use the cosine rule when the sine rule will not work
• choose the right rule by reading what the question gives you
Set your calculator to DEGREE mode before you begin.
1. Labelling
\( \text{side } a \text{ is OPPOSITE angle } A \)
Everything in this topic depends on one naming rule, so get it right first.
• Corners (vertices) get CAPITAL letters: A, B, C
• Sides get the matching lower-case letter
• Side a is the side opposite angle A
\( a \leftrightarrow A \qquad b \leftrightarrow B \qquad c \leftrightarrow C \)
The most common mistake is calling the side next to angle A 'side a'.
Side a never touches angle A. Look across the triangle from the angle — that is its opposite side.
In the diagram on the previous step, side a joins B and C. Angle A is nowhere near it.
\( \hat{A} + \hat{B} + \hat{C} = 180^\circ \)
Every rule you are about to meet pairs a side with the angle opposite it. If you label the triangle wrongly, the rule is applied to the wrong pair and the answer is wrong — even though the method looked right.
One more fact you will use constantly: the three angles of any triangle add to \( 180^\circ \). That single line often turns an unsolvable problem into a solvable one.
2. Area Rule
\( \text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height} \)
The area formula you already know needs a perpendicular height.
But exam questions rarely give you one. They give you two sides and the angle between them. So we need a formula that works from that information.
\( \sin A = \dfrac{h}{b} \;\Rightarrow\; h = b\sin A \)
Drop a perpendicular from C down to side AB. Call its length h.
In the right-angled triangle that this creates, \( \sin A = \dfrac{h}{b} \), so \( h = b\sin A \).
We have just produced the height without measuring it.
\[ \text{Area} = \tfrac{1}{2}\,ab\sin C \]
Substituting \( h = b\sin A \) into \( \text{Area} = \tfrac{1}{2}\times c \times h \) gives \( \text{Area} = \tfrac{1}{2}\,bc\sin A \).
All three versions are the same rule:
\( \tfrac{1}{2}ab\sin C = \tfrac{1}{2}bc\sin A = \tfrac{1}{2}ac\sin B \)
The angle must always sit BETWEEN the two sides you use.
\( \text{Area} = \tfrac{1}{2}(7)(9)\sin 50^\circ = 24{,}13\text{ cm}^2 \)
In triangle ABC, b = 7 cm, c = 9 cm and A = 50°.
Angle A sits between sides b and c, so the area rule applies directly:
\( \text{Area} = \tfrac{1}{2}bc\sin A = \tfrac{1}{2}(7)(9)\sin 50^\circ \)
\[ \text{Area} = \tfrac{1}{2}\,ab\sin C \]
Use it when: you have two sides and the angle between them, and the question asks for area.
Watch out for: using an angle that is not enclosed by your two sides — that gives a wrong answer every time.
Units: area is always in square units.
3. Sine Rule
\[ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \]
Each fraction pairs a side with the angle opposite it.
You never use all three fractions at once — you pick the two that contain what you know and what you want.
\( \text{Do I have an angle AND its opposite side?} \)
The sine rule only starts if you have a complete pair: an angle together with the side opposite it.
That pair is your anchor. Without it there is nothing to compare the unknown to, and you must use the cosine rule instead.
\( b = \dfrac{12\sin 75^\circ}{\sin 40^\circ} = 18{,}03\text{ cm} \)
In triangle ABC, A = 40°, B = 75° and a = 12 cm. Find b.
A pairs with a — that is the complete pair. Put the unknown on top:
\( \dfrac{b}{\sin 75^\circ} = \dfrac{12}{\sin 40^\circ} \)
\( 75^\circ > 40^\circ \;\Rightarrow\; b > a \)
Before moving on, sanity-check the answer.
The bigger angle always sits opposite the bigger side. Here 75° is bigger than 40°, so b must be longer than a = 12 cm.
We got 18,03 cm. That is longer — the answer is sensible.
\( \sin B = \dfrac{11\sin 62^\circ}{15} = 0{,}6475 \;\Rightarrow\; \hat{B} = 40{,}35^\circ \)
When the unknown is an angle, flip every fraction so the sines are on top:
\( \dfrac{\sin A}{a} = \dfrac{\sin B}{b} \)
With a = 15 cm, A = 62° and b = 11 cm:
\( \dfrac{\sin B}{11} = \dfrac{\sin 62^\circ}{15} \)
\( \sin 40^\circ = \sin 140^\circ \)
This one catches people out. Sine is positive for obtuse angles too, so two different angles share the same sine.
Your calculator only ever returns the acute one. If the diagram clearly shows an obtuse angle, the answer is \( 180^\circ - (\text{calculator value}) \).
The cosine rule does not have this problem — you will see why shortly.
\[ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \]
Use it when: you have an angle and the side opposite it.
Sides on top when finding a side. Sines on top when finding an angle.
Watch out for: the obtuse case, and never write \( \sin(75 \div 40) \) — work each sine out separately.
4. Cosine Rule
\( \text{No complete pair} \;\Rightarrow\; \text{no sine rule} \)
Two situations defeat the sine rule completely:
• you know all three sides but no angle at all
• you know two sides and the angle between them, and want the third side
In both cases there is no angle paired with its opposite side. The cosine rule handles both.
\[ a^2 = b^2 + c^2 - 2bc\cos A \]
The angle on the right is always opposite the side on the left.
The other two versions:
\( b^2 = a^2 + c^2 - 2ac\cos B \)
\( c^2 = a^2 + b^2 - 2ab\cos C \)
\( \text{If } A = 90^\circ,\; \cos A = 0 \;\Rightarrow\; a^2 = b^2 + c^2 \)
Look at what happens when A is a right angle: \( \cos 90^\circ = 0 \), so the last term disappears entirely and you are left with Pythagoras.
So the cosine rule is not a new idea — it is Pythagoras with a correction term for triangles that are not right-angled.
\( a^2 = 89 - 80\cos 65^\circ = 55{,}19 \;\Rightarrow\; a = 7{,}43\text{ cm} \)
With b = 8 cm, c = 5 cm and A = 65°:
\( a^2 = 8^2 + 5^2 - 2(8)(5)\cos 65^\circ \)
Work out the whole right-hand side first, then square-root. Forgetting the square root is the classic lost mark here.
\[ \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \]
Rearranging the rule makes \( \cos A \) the subject. Now three sides are enough to find any angle.
With a = 9, b = 7 and c = 5:
\( \cos A = \dfrac{49 + 25 - 81}{70} = \dfrac{-7}{70} = -0{,}1 \)
\( \cos A = -0{,}1 \;\Rightarrow\; \hat{A} = 95{,}74^\circ \)
A negative cosine is not a mistake. Unlike sine, cosine is negative for obtuse angles and positive for acute ones.
So the sign itself tells you the answer: \( \cos A < 0 \) means A is obtuse. The calculator gives you the correct angle directly — no 180° adjustment needed.
Notice too that A is opposite the longest side (9 cm). The largest angle always sits opposite the longest side.
\[ a^2 = b^2 + c^2 - 2bc\cos A \qquad \cos A = \dfrac{b^2+c^2-a^2}{2bc} \]
Use it when: you have three sides, or two sides and the angle between them.
Watch out for: forgetting the square root, and remember the angle must be enclosed by the two sides.
Bonus: it handles obtuse angles automatically.
5. Choosing & Applying
\( \text{Do I have an angle AND the side opposite it?} \)
Do not memorise problem types. Ask one question:
Do I have an angle together with the side opposite it?
• Yes → sine rule
• No → cosine rule
The area rule is the odd one out — it answers 'how big is this triangle', not 'how long is that side'.
\( \text{SSS} \rightarrow \cos \qquad \text{SAS} \rightarrow \cos \qquad \text{AAS} \rightarrow \sin \)
Try these before reading on:
1. Sides 5, 8, 11 — find the largest angle. → Cosine rule (three sides, no angle)
2. A = 40°, B = 60°, a = 9 — find b. → Sine rule (A pairs with a)
3. b = 6, c = 7, A = 55° — find a. → Cosine rule (no complete pair)
\( \hat{C} = 180^\circ - 65^\circ - 78^\circ = 37^\circ \)
A and B are two points 50 m apart on level ground. A tower stands at C. The angle at A is 65° and the angle at B is 78°. Find AC.
At first there is no complete pair — we know the side AB but not the angle opposite it (angle C). So find angle C first.
\( AC = \dfrac{50\sin 78^\circ}{\sin 37^\circ} = 81{,}27\text{ m} \)
Now angle C = 37° is paired with side AB = 50 m, and we have our anchor.
AC is opposite the 78° angle, so:
\( \dfrac{AC}{\sin 78^\circ} = \dfrac{50}{\sin 37^\circ} \)
One line — the angle sum — turned an unsolvable problem into a one-step sine rule.
\( \text{DEG mode. Square root. Obtuse check.} \)
1. Calculator in radians — check for the small D on the display
2. Wrong rule — ask the one question
3. Forgetting the square root after the cosine rule
4. Missing an obtuse angle when using the sine rule
5. Rounding too early — keep full accuracy until the end
6. Angle not between the sides for the area and cosine rules
\[ \begin{aligned}\text{Area} &= \tfrac{1}{2}ab\sin C \\[4pt]\dfrac{a}{\sin A} &= \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \\[4pt]a^2 &= b^2 + c^2 - 2bc\cos A\end{aligned} \]
Side a is opposite angle A. Everything rests on that.
• Area rule — two sides and the angle between them
• Sine rule — an angle and its opposite side
• Cosine rule — three sides, or two sides and the included angle
• Rearranged: \( \cos A = \dfrac{b^2+c^2-a^2}{2bc} \)
Angles add to 180°. Largest angle opposite longest side. Keep the calculator in DEGREE mode and round only at the end.
You are ready for the quiz.