Sine Rule, Cosine Rule and Vectors

This lesson is Higher tier throughout. SOHCAHTOA only works in right-angled triangles, but many problems involve triangles with no right angle at all. For those you need the sine rule and the cosine rule, together with the formula for the area of any triangle using two sides and the included angle. The second half of the lesson covers vectors: representing journeys as directed quantities, adding and subtracting them, and using them to prove geometric facts. The exam skills are choosing the correct rule from the information given, substituting carefully, and — for vectors — writing a clear chain of vector steps. Selecting and using a rule is AO1; deciding which rule fits an unfamiliar triangle is AO2; and constructing a vector proof is AO3.

Key Vocabulary

Term	Meaning
Sine rule	Relates each side to the sine of its opposite angle
Cosine rule	Relates all three sides and one angle
Included angle	The angle between two given sides
Vector	A quantity with both magnitude and direction
Resultant	The single vector equivalent to two or more combined
Scalar multiple	A vector multiplied by a number, changing its length
Parallel vectors	Vectors that are scalar multiples of each other

Labelling a Non-Right Triangle

Before either rule can be used, the triangle must be labelled in the standard way, and this convention underpins everything that follows. Each side is named with the lower-case letter of the angle that lies opposite it. So the side labelled $a$ is opposite the angle labelled $A$ , the side $b$ is opposite angle $B$ , and the side $c$ is opposite angle $C$ . A side and the angle facing it across the triangle therefore share the same letter in different cases. Getting this pairing right is the foundation of both the sine rule and the cosine rule, because both rules are stated in terms of these opposite pairs; a single mislabelling will send the whole calculation wrong. When a question gives a triangle with its own letters, take a moment to re-label mentally so that each unknown you want and each quantity you know are correctly identified as a side or an angle, and as opposite or included. The figure shows a triangle labelled this way.

The Sine Rule

The sine rule states that $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ , which says that in any triangle the ratio of each side to the sine of its opposite angle is the same. You do not use all three fractions at once; you pick the two that contain the quantities you know and the one you want, giving a simple equation to solve. The decisive thing to recognise is when the sine rule applies: it needs a complete "matching pair" — a side together with the angle directly opposite it — plus one further piece of information. In practice that means it is the right tool when you are given two angles and any side (you can always find the third angle first, then a matching pair exists), or when you are given two sides and an angle opposite one of them. A small but important efficiency tip concerns the way you write the rule. When you are finding a missing side, leave the sides on top as written. When you are finding a missing angle, turn every fraction upside down so the sines are on top, $\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$ ; this puts the unknown sine in the numerator and makes the rearrangement far cleaner.

Worked Example 1

In triangle $ABC$ , $\angle A = 40^{\circ}$ , $\angle B = 75^{\circ}$ and side $a = 8$ cm. Work out side $b$ , to $1$ decimal place.

By the sine rule, $\dfrac{b}{\sin 75^{\circ}} = \dfrac{8}{\sin 40^{\circ}}$ , so $b = \dfrac{8 \sin 75^{\circ}}{\sin 40^{\circ}} \approx \dfrac{8 \times 0.9659}{0.6428} = 12.0$ cm.

Answer: $12.0$ cm.

Worked Example 2

In triangle $ABC$ , $a = 9$ cm, $b = 7$ cm and $\angle A = 65^{\circ}$ . Work out $\angle B$ , to $1$ decimal place.

Flip the rule: $\dfrac{\sin B}{7} = \dfrac{\sin 65^{\circ}}{9}$ , so $\sin B = \dfrac{7 \sin 65^{\circ}}{9} \approx 0.7050$ , giving $\angle B = \sin^{-1}(0.7050) \approx 44.8^{\circ}$ .

Answer: $44.8^{\circ}$ .

Common error: keeping the sines on the bottom when finding an angle, which makes the rearrangement awkward and error-prone.

The Cosine Rule

The cosine rule states that $a^{2} = b^{2} + c^{2} - 2bc\cos A$ . You can think of it as Pythagoras' theorem with a correction term: if angle $A$ were $90^{\circ}$ then $\cos A = 0$ and the rule would collapse to $a^{2} = b^{2} + c^{2}$ , ordinary Pythagoras, but for any other angle the $-2bc\cos A$ term adjusts for the fact that the triangle is not right-angled. The cosine rule is the tool to reach for whenever the sine rule cannot get started — and the sine rule cannot start when you have no matching side–angle pair. Concretely, use the cosine rule in two situations: when you know all three sides and want to find an angle, or when you know two sides and the angle between them (the included angle) and want the third side. For finding a side you substitute directly into $a^{2} = b^{2} + c^{2} - 2bc\cos A$ and square-root at the end. For finding an angle, it is cleaner to use the rearranged form $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$ , then apply $\cos^{-1}$ to the result. A useful sanity check is that a negative value of $\cos A$ tells you the angle is obtuse.

Worked Example 3

In triangle $ABC$ , $b = 6$ cm, $c = 9$ cm and the included angle $\angle A = 50^{\circ}$ . Work out side $a$ , to $1$ decimal place.

$a^{2} = 6^{2} + 9^{2} - 2 \times 6 \times 9 \times \cos 50^{\circ} = 36 + 81 - 108 \times 0.6428 = 117 - 69.42 = 47.58$ . So $a = \sqrt{47.58} \approx 6.9$ cm.

Answer: $6.9$ cm.

Worked Example 4

A triangle has sides $a = 7$ cm, $b = 8$ cm and $c = 10$ cm. Work out the largest angle, to $1$ decimal place.

The largest angle is opposite the longest side ( $c = 10$ ), namely $\angle C$ . Rearranging the cosine rule: $\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2ab} = \dfrac{49 + 64 - 100}{2 \times 7 \times 8} = \dfrac{13}{112} \approx 0.1161$ , so $\angle C = \cos^{-1}(0.1161) \approx 83.3^{\circ}$ .

Answer: $83.3^{\circ}$ .

Common error: evaluating $b^{2} + c^{2} - 2bc\cos A$ as $(b^{2} + c^{2} - 2bc)\cos A$ — the cosine multiplies only the $2bc$ term, so follow the order of operations.

Area of a Triangle

The ordinary area formula $\dfrac{1}{2} \times \text{base} \times \text{height}$ is awkward for a non-right triangle because the perpendicular height is rarely given. Fortunately there is a formula that uses only two sides and the angle between them: the area of any triangle is $\text{Area} = \dfrac{1}{2}ab\sin C$ , where $a$ and $b$ are two sides and $C$ is the included angle, the angle that sits between those two sides. The word "included" is doing real work here, because the formula fails if you use an angle that is not between the two chosen sides — so always check that your angle is sandwiched between the two sides you are multiplying. This single formula handles area, and rearranged it can also find a missing side or a missing angle when the area is given, which makes it surprisingly versatile.

Worked Example 5

A triangle has sides $a = 10$ cm and $b = 7$ cm with an included angle of $48^{\circ}$ . Work out its area, to $1$ decimal place.

Area $= \dfrac{1}{2} \times 10 \times 7 \times \sin 48^{\circ} = 35 \times 0.7431 \approx 26.0$ cm $^{2}$ .

Answer: $26.0$ cm $^{2}$ .

Worked Example 6

A triangle has two sides of $9$ cm and $12$ cm and an area of $40$ cm $^{2}$ . Work out the included angle, to $1$ decimal place.

$40 = \dfrac{1}{2} \times 9 \times 12 \times \sin C = 54\sin C$ , so $\sin C = \dfrac{40}{54} \approx 0.7407$ , giving $C = \sin^{-1}(0.7407) \approx 47.8^{\circ}$ .

Answer: $47.8^{\circ}$ .

Common error: using a side that is not adjacent to the angle; the formula needs the angle between the two sides used.

Sine Rule, Cosine Rule and Vectors (Higher)