Law of Sines

When to use the Law of Sines. It is the right tool in exactly two situations: (i) to find an unknown angle, when two sides and a non-included angle are known (an SSA-type data set), and (ii) to find an unknown side, when two angles and a side opposite one of them are known (an AAS/ASA-type data set).

Theorem 3.1 (Law of Sines). In any $\triangle ABC$,

where $R$ is the circumradius — the sides of a triangle are proportional to the sines of the angles opposite them, and the common ratio is exactly twice the circumradius.

Proof (via the circumcircle). Let $O$ be the circumcentre and $R$ the circumradius. The angle $A$ is acute, right, or obtuse; every case is handled by the same construction.

• $A$ acute or obtuse: produce $BO$ to meet the circumcircle again at $D$, so $BD=2R$ is a diameter, which forces $\angle BCD=90^\circ$ (angle in a semicircle). If $A$ is acute, $\angle BDC=\angle BAC=A$ (angles subtended by the same arc $BC$), and in the right triangle $BCD$, $\sin A=\sin\angle BDC=\dfrac{BC}{BD}=\dfrac{a}{2R}$. If $A$ is obtuse, $ABDC$ is a cyclic quadrilateral, so $\angle BDC=180^\circ-A$; since $\sin(180^\circ-A)=\sin A$, the same right triangle gives $\sin A=a/(2R)$ again.
• $A=90^\circ$: now $O$ lies on $BC$ itself, so $BC$ is a diameter: $a=2R$ and $\sin A=\sin 90^\circ=1$, again giving $a/\sin A=2R$.

In every case $\dfrac{a}{\sin A}=2R$. Repeating the identical construction from vertex $B$ (respectively $C$) gives $\dfrac{b}{\sin B}=2R$ (respectively $\dfrac{c}{\sin C}=2R$). Hence

Remarks.

1. The law splits into three separate ratio equations: $\dfrac{a}{b}=\dfrac{\sin A}{\sin B}$, $\dfrac{a}{c}=\dfrac{\sin A}{\sin C}$, $\dfrac{b}{c}=\dfrac{\sin B}{\sin C}$.
2. It says the sides of a triangle are proportional to the sines of the angles opposite them.
3. The Law of Sines cannot by itself solve a triangle given two sides and the included angle (SAS) — that data set needs the Law of Cosines (§3.7.2).
4. A geometric consequence: the largest side is opposite the largest angle of a triangle. Napier's Formula (the tangent rule). Theorem 3.2. In $\triangle ABC$,

   $\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{C}{2},\qquad \tan\frac{B-C}{2}=\frac{b-c}{b+c}\cot\frac{A}{2},\qquad \tan\frac{C-A}{2}=\frac{c-a}{c+a}\cot\frac{B}{2}.$

   Proof (of the first identity — the others follow by cycling $A\to B\to C\to A$). Write $a=2R\sin A,\ b=2R\sin B$ from the Law of Sines, so

   $\frac{a-b}{a+b}\cot\frac{C}{2}=\frac{\sin A-\sin B}{\sin A+\sin B}\cot\frac{C}{2}.$

   Apply the sum-to-product identities $\sin A-\sin B=2\cos\frac{A+B}2\sin\frac{A-B}2$ and $\sin A+\sin B=2\sin\frac{A+B}2\cos\frac{A-B}2$: …