Why does 120^ make the side longer?

120^ is negative, so subtracting 2bc A adds a positive amount — obtuse angles are opposite long sides.

1MA1 · G22–G23Geometry & MeasuresHigherGrades 7–9

Sine & Cosine Rules

SOH CAH TOA only works in right-angled triangles. For every other triangle there are two rules — the $\textbf{sine rule}$ and the $\textbf{cosine rule}$ — plus a formula for area that needs no perpendicular height. Together they solve any triangle, given enough information.

55 min Pythagoras & Trigonometry

The big picture

Label a triangle so that each side is opposite the angle with the same letter: side $a$ opposite angle $A$ , and so on. The sine rule links pairs of opposite sides and angles; the cosine rule links all three sides with one angle. Deciding which to use comes down to what you are given.

Edexcel Higher questions are usually multi-step: cosine rule to find a side, sine rule to find an angle, then area or a bearing. The cosine rule is actually Pythagoras with a correction term — when the angle is $90^{\circ}$ , $\cos 90^{\circ} = 0$ and it becomes $a^{2} = b^{2} + c^{2}$ .

What you'll learn

Use the sine rule to find sides and angles
Use the cosine rule to find a side or an angle
Choose the correct rule from the information given
Calculate the area of any triangle using $\tfrac12 ab\sin C$

The sine rule

In any triangle, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ .

Use it when you know a side and its opposite angle, plus one more side or angle.

To find an angle, use the flipped form $\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$ , which puts the unknown on top.

\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

Sine rule.

Example 1In triangle

ABC

A = 42^{\circ}

B = 71^{\circ}

and

a = 9

cm. Find

b

\dfrac{b}{\sin 71^{\circ}} = \dfrac{9}{\sin 42^{\circ}}

b = \dfrac{9\sin 71^{\circ}}{\sin 42^{\circ}}

Answer

b = 12.7

Your turnSame method, new numbers — try it before moving on.

In triangle $ABC$ , $A = 48^\circ$ , $B = 65^\circ$ and $a = 10$ cm. Find $b$ , to 1 decimal place.

Your answer

To 1 decimal place.

Show worked solution

Solution

1 $b = \dfrac{10 \sin 65^\circ}{\sin 48^\circ}$ .
2 $= 12.19\ldots$ cm.

Answer: $12.2$ cm

The cosine rule

To find a side when you know the other two sides and the angle between them: $a^{2} = b^{2} + c^{2} - 2bc\cos A$ .

To find an angle when you know all three sides: $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$ .

A negative cosine means the angle is obtuse — the calculator handles this automatically.

a^{2} = b^{2} + c^{2} - 2bc\cos A

Cosine rule for a side.

\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}

Cosine rule for an angle.

Example 1Two sides of a triangle are 8 cm and 11 cm, with an included angle of

50^{\circ}

. Find the third side.

a^{2} = 8^{2} + 11^{2} - 2(8)(11)\cos 50^{\circ}

= 64 + 121 - 113.13 = 71.87

a = \sqrt{71.87}

Answer8.48 cm

Your turnSame method, new numbers — try it before moving on.

Two sides of a triangle are 7 cm and 10 cm with an included angle of $60^\circ$ . Find the third side, to 1 decimal place.

Your answer

To 1 decimal place.

Show worked solution

Solution

1 $c^2 = 49 + 100 - 140\cos 60^\circ = 79$ .
2 $c = \sqrt{79} = 8.89$ cm.

Answer: $8.9$ cm

Example 2A triangle has sides 7 cm, 9 cm and 12 cm. Find the largest angle.

1The largest angle is opposite the longest side, 12 cm.

\cos\theta = \dfrac{49 + 81 - 144}{2 \times 7 \times 9} = \dfrac{-14}{126}

Answer

96.4^{\circ}

Your turnSame method, new numbers — try it before moving on.

A triangle has sides 6 cm, 8 cm and 11 cm. Find the largest angle, to 1 decimal place.

Your answer

To 1 decimal place.

Show worked solution

Solution

1 $\cos\theta = \frac{36 + 64 - 121}{96} = -\frac{21}{96}$ .
2 $\theta = 102.6^\circ$ .

Answer: $102.6^\circ$

Why this works

Setting $A = 90^{\circ}$ makes $2bc\cos A$ vanish, leaving $a^{2} = b^{2} + c^{2}$ . The cosine rule is Pythagoras, corrected for triangles that are not right-angled.

Which rule?

Two sides and the included angle, finding the third side: $\textbf{cosine rule}$ .

Three sides, finding an angle: $\textbf{cosine rule}$ .

A side and its opposite angle are both known: $\textbf{sine rule}$ .

Tip — Look for an opposite pair — a side and its opposite angle both known. If there is one, use the sine rule; if not, use the cosine rule.

Area of a triangle

Area $= \tfrac12 ab\sin C$ , where $C$ is the angle between sides $a$ and $b$ .

This replaces "half base times height" when the perpendicular height is not known.

\text{Area} = \tfrac12 ab\sin C

Two sides and the included angle.

Example 1Find the area of a triangle with sides 8 cm and 11 cm and included angle

50^{\circ}

\tfrac12 \times 8 \times 11 \times \sin 50^{\circ}

Answer33.7 cm²

Your turnSame method, new numbers — try it before moving on.

Find the area of a triangle with sides 9 cm and 12 cm and included angle $40^\circ$ , to 1 decimal place.

Your answer

To 1 decimal place.

Show worked solution

Solution

1Area $= \frac{1}{2} \times 9 \times 12 \times \sin 40^\circ$ .
2 $= 34.71$ cm².

Answer: $34.7$ cm²

Think like an examiner

Label the triangle with matching sides and angles.
Write the rule with values substituted before calculating.
Keep unrounded values between steps of multi-step problems.

Non-right-angled triangles

\tfrac{a}{\sin A} = \tfrac{b}{\sin B}

Sine rule.

a^{2} = b^{2} + c^{2} - 2bc\cos A

Cosine rule.

\tfrac12 ab\sin C

Area.

Watch out for these

Using the cosine rule with an angle that is not between the two known sides.

The angle must be the included angle.

Working out

b^{2} + c^{2} - 2bc

first, then multiplying by

\cos A

Calculate

2bc\cos A

as one term, then subtract.

Using

\tfrac12 ab\sin C

with an angle not between

a

and

b

The angle must be between the two sides.

Rounding early in multi-step problems.

Use calculator memory to keep full values.

Stretch yourself

Triangle $PQR$ has $PQ = 6$ cm, $PR = 10$ cm and angle $QPR = 120^{\circ}$ . Find the length $QR$ , the area of the triangle, and angle $PQR$ .

Hint — Cosine rule for $QR$ (the included angle is known), then the area formula, then the sine rule with the new opposite pair.

Reveal solution

Solution

QR^{2} = 36 + 100 - 2(6)(10)\cos 120^{\circ} = 136 + 60 = 196

, so

QR = 14

cm. Area

= \tfrac12 \times 6 \times 10 \times \sin 120^{\circ} = 30 \times \tfrac{\sqrt3}{2} = 15\sqrt3 \approx 26.0

cm². Sine rule:

\dfrac{\sin PQR}{10} = \dfrac{\sin 120^{\circ}}{14}

, so

\sin PQR = 0.6186

and

PQR = 38.2^{\circ}

(it must be acute, since the triangle already has an obtuse angle).

\blacksquare

Questions students ask

Are these formulas given in the exam?

No — Edexcel expects you to know the sine rule, cosine rule and area formula.

Can I use these rules on right-angled triangles?

Yes, they still work, but SOH CAH TOA and Pythagoras are quicker.

Why does

\cos 120^{\circ}

make the side longer?

\cos 120^{\circ}

is negative, so subtracting

2bc\cos A

adds a positive amount — obtuse angles are opposite long sides.

Key takeaways

Label sides opposite their matching angles.
Sine rule when an opposite side–angle pair is known.
Cosine rule for the third side from two sides and the included angle, or an angle from three sides.
Area $= \tfrac12 ab\sin C$ with the included angle.
Keep full accuracy through multi-step problems.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 72 marks

In triangle $ABC$ , $a = 10$ cm, $c = 8$ cm and angle $A = 55^\circ$ .

Work out angle $C$ , to 1 decimal place.[2]
Show mark scheme
- M1 $\frac{\sin C}{8} = \frac{\sin 55}{10}$
- A1 $40.9^\circ$
Answer: $40.9^\circ$

Question 2Grade 82 marks

Two sides of a triangle are 12 cm and 15 cm with an included angle of $70^\circ$ . Work out the third side, to 3 significant figures.[2]
Show mark scheme
- M1 $144 + 225 - 360\cos 70$
- A1 $15.7$ cm
Answer: $15.7$ cm

Question 3Grade 93 marks

Without a calculator, work out the area of a triangle with sides 6 cm and 10 cm and an included angle of $150^\circ$ .[3]
Show mark scheme
- M1 $\frac{1}{2} \times 6 \times 10 \times \sin 150^\circ$
- M1 $\sin 150^\circ = \frac{1}{2}$
- A1 $15$ cm²
Answer: $15$ cm²

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