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The Sine and Cosine Rules

Q: How is the cosine rule related to Pythagoras?

It is Pythagoras plus a correction. When the angle is 90^, A = 0 and a^2 = b^2 + c^2 - 2bc A becomes a^2 = b^2 + c^2. The -2bc A term adjusts for angles that are not right.

Q: Which angle goes in the area formula?

The angle enclosed by the two sides you are multiplying. Area = 12 ab C requires C to be the angle between sides a and b.

SOHCAHTOA only works in right-angled triangles. The sine and cosine rules extend trigonometry to any triangle — the trick is knowing, from what you are given, which rule to reach for.

28 min Video by TLMaths Trigonometry

The big picture

Most triangles in the real world are not right-angled, and these two rules handle all of them. The genuinely useful skill is not memorising the formulas but $diagnosing$ which one a problem needs: the sine rule pairs an angle with its opposite side, while the cosine rule connects all three sides with one angle. Get the diagnosis right and these problems become quick. There is also a subtle trap — the “ambiguous case” — where the sine rule can give two valid triangles, and knowing when to expect it separates a confident answer from a lost mark.

A-Level Maths E1-08: Using the Sine RuleWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Use the sine rule to find a missing side or angle
Use the cosine rule for the SAS and SSS situations
Choose the correct rule from the information given
Find the area of a triangle with $\frac{1}{2} ab sin C$ and handle the ambiguous case

The sine rule — matched angle–side pairs

The sine rule says each side is proportional to the sine of the angle $opposite$ it. Use it when you have a $pair$ : a side and the angle facing it, plus one more piece of matching information.

Put the unknown on top when finding a side, and flip the rule (sines on top) when finding an angle.

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

Each side over the sine of its opposite angle. Label sides with lowercase, opposite angles with uppercase.

1Use

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

b = \frac{a sin B}{sin A} = \frac{8 sin 7 5 ^{\circ}}{sin 4 0 ^{\circ}}

= \frac{8 \times 0.966}{0.643} \approx 12.0

Answer

b \approx 12.0

Tip — The sine rule needs a complete angle–side pair to start. If you only have sides and one enclosed angle, it is a cosine-rule problem instead.

The cosine rule — three sides and one angle

The cosine rule links all three sides to one angle. Reach for it in two situations: you know two sides and the angle $between$ them (SAS) and want the third side; or you know all three sides (SSS) and want an angle. It is essentially Pythagoras with a correction term for the angle.

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

Find a side (two sides + included angle).

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

Rearranged to find an angle (three sides).

1Cosine rule:

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

a^{2} = 49 + 25 - 2 (7) (5) cos 6 0^{\circ} = 74 - 70 (0.5) = 74 - 35 = 39

a = 39 \approx 6.24

Answer

a \approx 6.24

Why this works

Notice that if $A = 9 0^{\circ}$ then $cos A = 0$ and the cosine rule collapses to $a^{2} = b^{2} + c^{2}$ — plain Pythagoras. The $- 2 b c cos A$ term is exactly the adjustment for a non-right angle.

Area, and the ambiguous case

The area of any triangle from two sides and the included angle is $\frac{1}{2} ab sin C$ . And a warning: when you use the sine rule to find an $angle$ , remember that $sin θ = sin (18 0^{\circ} - θ)$ , so there may be $two$ possible angles — the $ambiguous case$ . Check whether the obtuse alternative also fits the triangle.

Area = \frac{1}{2} ab sin C

Two sides and the angle between them.

1Area

= \frac{1}{2} ab sin C = \frac{1}{2} (6) (9) sin 5 0^{\circ}

= 27 \times 0.766 \approx 20.7

AnswerArea

\approx 20.7

square units.

Think like an examiner

Diagnose first: a matched angle–side pair → sine rule; two sides and the included angle, or all three sides → cosine rule.
When the sine rule gives an angle, consider the obtuse alternative $18 0^{\circ} - θ$ and check whether it is valid (the ambiguous case).
Keep answers unrounded until the end; rounding early throws off multi-step triangle calculations.
For area, the angle must be the one $between$ the two sides used.

Common misconceptions

Using the cosine rule when you have an angle–side pair.

A matched pair points to the sine rule, which is simpler here.

Forgetting the second solution when finding an angle with the sine rule.

Check

18 0^{\circ} - θ

too — the ambiguous case can give a valid obtuse triangle.

Using

\frac{1}{2} ab sin C

with an angle not between the two sides.

The angle must be enclosed by sides

a

and

b

Mislabelling: putting a side opposite the wrong angle.

Lowercase side

a

is opposite uppercase angle

A

; label carefully before substituting.

Triangle rules

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

Sine rule.

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

Cosine rule (side).

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

Cosine rule (angle).

Area = \frac{1}{2} ab sin C

Triangle area.

Stretch yourself

In triangle $A B C$ , $a = 10$ , $b = 8$ and $A = 4 0^{\circ}$ . Show that there are two possible values for angle $B$ , and find both.

Hint — Use the sine rule to find $sin B$ . Then remember $sin B = sin (18 0^{\circ} - B)$ and check both angles give a valid triangle.

Questions students ask

How do I decide between the sine and cosine rules?

If you have a side and its opposite angle (a matched pair), use the sine rule. If you have two sides and the angle between them, or all three sides, use the cosine rule. The information you are given points to the rule.

What is the ambiguous case?

When you use the sine rule to find an angle, there can be two answers, because

sin θ = sin (18 0^{\circ} - θ)

. Both an acute and an obtuse angle may satisfy the equation, so you check which (or both) actually fit the triangle.

How is the cosine rule related to Pythagoras?

It is Pythagoras plus a correction. When the angle is

9 0^{\circ}

cos A = 0

and

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

becomes

a^{2} = b^{2} + c^{2}

. The

- 2 b c cos A

term adjusts for angles that are not right.

Which angle goes in the area formula?

The angle enclosed by the two sides you are multiplying.

Area = \frac{1}{2} ab sin C

requires

C

to be the angle between sides

a

and

b

Key takeaways

Sine rule: sides proportional to the sine of the opposite angle — use with a matched pair.
Cosine rule: two sides + included angle (find a side), or three sides (find an angle).
Cosine rule reduces to Pythagoras when the angle is $9 0^{\circ}$ .
Area $= \frac{1}{2} ab sin C$ , with $C$ between the two sides.
Finding an angle by sine rule may give two answers — check the ambiguous case.

How this fits the course

Build on

Trigonometric Ratios and Graphs

Radian Measure

Leads to

Radian Measure

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