Law of Sines is the habit of relating side lengths and opposite angles through a triangle circumcircle. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
In any triangle, each side over the sine of its opposite angle equals the same value, which is the diameter of the circumcircle.
Why It Works
1
Inscribe $\triangle ABC$ in its circumcircle of radius $R$. Side $a = BC$ is a chord subtending the inscribed angle $A$.
2
By the inscribed angle theorem, the central angle on arc $BC$ is $2A$; the chord length is $a = 2R\sin A$ (from the isosceles triangle formed with the two radii).
3
Rearranging gives $\dfrac{a}{\sin A} = 2R$, and the same argument applies to sides $b$ and $c$.
4
Hence all three ratios equal the common value $2R$, linking sides, opposite angles, and the circumradius.
Worked Examples
Example 1
In $\triangle ABC$, $\angle A = 30^\circ$, $\angle B = 45^\circ$, and side $a = BC = 10$. Find side $b = CA$.
1
Law of Sines: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$.
2
Substitute: $\dfrac{10}{\sin 30^\circ} = \dfrac{b}{\sin 45^\circ}$, i.e. $\dfrac{10}{1/2} = \dfrac{b}{\sqrt{2}/2}$.
A triangle has a side of length $12$ opposite a $30^\circ$ angle. Find the radius $R$ of its circumcircle.
1
The extended Law of Sines says $\dfrac{a}{\sin A} = 2R$.
2
Here $a = 12$ and $A = 30^\circ$, so $2R = \dfrac{12}{\sin 30^\circ} = \dfrac{12}{1/2} = 24$.
3
Therefore $R = 12$.
Answer:
$R = 12$
Contest level
In $\triangle ABC$, side $a = 7$ is opposite $\angle A$, side $b = 8$ is opposite $\angle B$, and $\angle A = 60^\circ$. Find all possible values of $\angle B$ (the ambiguous SSA case).
1
Law of Sines: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$, so $\sin B = \dfrac{b\sin A}{a} = \dfrac{8\sin 60^\circ}{7} = \dfrac{8\cdot \tfrac{\sqrt3}{2}}{7} = \dfrac{4\sqrt3}{7}$.
2
Numerically $\sin B = \dfrac{4\sqrt3}{7} \approx \dfrac{6.928}{7} \approx 0.9897$, so $\angle B \approx 81.8^\circ$ or its supplement $\approx 98.2^\circ$.
3
Check feasibility: with $A = 60^\circ$, $B \approx 81.8^\circ$ gives $C \approx 38.2^\circ$ (valid), and $B \approx 98.2^\circ$ gives $C \approx 21.8^\circ$ (also valid). Both keep the angle sum at $180^\circ$, so BOTH triangles exist.
4
(The two are distinguished because $b > a$, so the angle opposite $b$ may be acute or obtuse — the textbook ambiguous case.)
Answer:
Two solutions: $\angle B \approx 81.8^\circ$ or $\angle B \approx 98.2^\circ$.
Olympiad / Challenge
Prove the projection-free area form $[\triangle ABC] = \dfrac{abc}{4R}$, where $R$ is the circumradius, using the Law of Sines.
1
Start from the standard area $[\triangle ABC] = \tfrac12 ab\sin C$.
2
By the extended Law of Sines, $\dfrac{c}{\sin C} = 2R$, so $\sin C = \dfrac{c}{2R}$.
Generalization: the value of the common ratio is not just 'some constant' — it is exactly $2R$, the circumcircle diameter. This upgrade ($a = 2R\sin A$) is what links a triangle's metric data to its circumcircle and yields the area form $[\triangle] = \dfrac{abc}{4R}$ and the chord-length law for cyclic polygons.
Where it appears: the Law of Sines is the right tool when the data pairs a side WITH its opposite angle (or asks for the circumradius). On AMC/AIME it cracks 'angle-and-opposite-side' triangles in one line, and it underlies trigonometric Ceva. Use Law of Cosines instead when you have two sides and the INCLUDED angle, or all three sides.
Pitfall: the SSA (side-side-angle) setup is genuinely ambiguous — $\sin B = \sin(180^\circ - B)$, so a given sine yields an acute AND an obtuse candidate. Always test whether each candidate keeps the angle sum below $180^\circ$; discarding the valid second solution (or keeping an impossible one) is the classic Law of Sines mistake.
Spot the Signal
Look for problems where the key step is relating side lengths and opposite angles through a triangle circumcircle.
You can describe the hard part as relating side lengths and opposite angles through a triangle circumcircle, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the diagram relation that calls for law of sines, then rewrite the givens around it.
Name the relation that makes Law of Sines legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Law of Sines just because the surface looks familiar; verify the required condition first.
Applying Law of Sines because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let law of sines reveal the geometric configuration; then compute only what remains.
Try it on:
Practice a contest problem where the key step is relating side lengths and opposite angles through a triangle circumcircle.