The Sine Rule
Date: 2025-06-11 • Author: Trafalgar Tuition
Tags: GCSE, Maths, FTSE, Trigonometry, Proof
Introduction
\[ \displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
This formula often overwhelms students — too many fractions, too many letters. But it's really just an elegant extension of triangle logic. Let’s explore where it comes from and why it works.
Motivation
If we don’t know the height of a triangle, we can’t use the familiar formula:
$$\text{Area} = \frac{1}{2} b h$$
But if we know two sides and an angle, trigonometry gives us another way — and that’s where the sine rule comes in.
Housekeeping
We need to label triangles properly with using the same letter for opposite sides and angles, like so:
where
- Use capital letters (A, B, C) for the angles.
- Label the side side opposite the angles using the same letter in lowercase (a, b, c).
For consistency, we can define point D as where the line **AC** meets the perpendicular line from point B.
Proof
Start with triangle ABC. Draw a perpendicular from point B to the base AC, and call the foot of the perpendicular D. This gives us a height BD.
We now have two right-angled triangles: **ABD** and **CBD**.
In triangle **ABC**, let’s first apply the sine definition to angle **A**:
$$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BD}{c} \quad \Rightarrow \quad BD = c \sin A$$
Now apply the same to angle C:
$$\sin C = \frac{BD}{a} \quad \Rightarrow \quad BD = a \sin C$$
Since both expressions equal BD, we can equate them together:
$$c \sin A = a \sin C$$
Now rearrange:
$$\frac{a}{\sin A} = \frac{c}{\sin C} \;\;\;\square$$
This is the sine rule — and you can derive similar expressions by choosing different pairs of angles and sides.
Conclusion
The sine rule allows us to solve any triangle — not just right-angled ones. It works because we can always draw a height, no matter which side we treat as the base. In the next post, we’ll use this rule in a worked example.
🔗 🔗 Next up: Before the introduction of the cosine rule, introductions must be made to one of the most common and well-known equations in all of maths: Pythagoras' theorem!