Sine Rule
The sine rule links every side of a triangle to the angle facing it — the tool of choice whenever a question hands you a matching side–angle pair.
Extending trigonometry beyond right angles
Label a triangle so side $a$ is opposite angle $A$, side $b$ opposite $B$, and side $c$ opposite $C$ — each side paired with the angle facing it. The sine rule states that these ratios are all equal:
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
Use it whenever you have a matching side–angle pair — one side and its opposite angle both known — together with one more piece of information (another angle, or the side opposite it). If instead you're given two sides and the angle between them, or three sides with no matching pair, the cosine rule from the previous lesson is the right tool.
Proving the sine rule from first principles
The sine rule comes straight from a dropped perpendicular. Take triangle $ABC$ and drop the perpendicular $CD$ from $C$ to the base $AB$, calling its length $h$. This creates two right-angled triangles that share the same height $h$ — and expressing $h$ two different ways is the whole trick.
The entire argument hinges on one shared quantity: the height $h$. Both right-angled triangles are built on it, so writing $h$ two ways — $b\sin A$ from the left, $a\sin B$ from the right — and setting them equal forces the two ratios together. That's why the sine rule pairs each side with the angle opposite it: the side is always the hypotenuse, and the opposite angle is the one whose sine reaches down to the shared height.
Look for a matching pair: a side and the angle directly opposite it, both known. If you have that pair plus one more angle (to find a side) or one more side (to find an angle), the sine rule solves it in one line. No matching pair at all usually means the cosine rule instead.
Worked example 1
Keep the unknown on the top of its fraction — it makes the rearrangement a single multiplication. And the bigger angle ($75°$) faces the bigger side ($12.0 > 8$), which is a quick sanity check: in any triangle, larger angles sit opposite longer sides.
Worked example 2
Here the matching pair is $p$ with $P$; the fourth piece of information is the other side $q$, and we want its opposite angle $Q$ — the sine rule rearranges directly for it.
Whenever the sine rule is used to find an angle, remember that $\sin^{-1}$ only ever returns an acute angle by default — but the true angle could be obtuse instead ($180° - 55.9° = 124.1°$ also has sine $0.8285$). Here $124.1°$ is rejected because $Q$ and the given $P=35°$ would then sum to more than $180°$, which is impossible in a triangle. This "two candidate angles, check which fits" situation is explored fully as the ambiguous case in the next lesson.
Worked example 3
Don't automatically find the third angle first — check whether a usable matching pair is already in front of you. Here $Y$ and $y$ were already paired, so the third angle $Z=180°-52°-81°=47°$ was never needed. Extra steps aren't wrong, but they waste time and introduce more chances for arithmetic slips.
Check yourself
In triangle $DEF$, angle $D = 48°$, angle $E = 63°$, and side $d = 10\,\text{cm}$. Find side $e$ to 3 s.f.
(Hint: d and D already form a matching pair — set up the sine rule directly for e.)