Sine Rule

10.1 The Sine Rule

Theorem. In any $\triangle ABC$ with sides $a,b,c$ opposite angles $A,B,C$,

where $R$ is the circumradius of the triangle.

Proof. Let $O$ be the circumcentre and $R$ the circumradius of $\triangle ABC$. Draw the diameter $BD$ through $B$, so $BD=2R$. Since $BD$ is a diameter, $\angle BCD=90^\circ$ (angle in a semicircle). Also $\angle BDC=\angle BAC=A$, because both angles stand on the same arc $BC$. In the right triangle $BCD$,

Repeating the argument with the diameter through $A$ gives $\dfrac{b}{\sin B}=2R$, and with the diameter through $C$ gives $\dfrac{c}{\sin C}=2R$. Hence all three ratios are equal to $2R$. $\blacksquare$

Remark. The proof used an acute-triangle picture, but the equal-angles-in-the-same-segment fact and the semicircle argument continue to hold with the usual sign conventions for an obtuse triangle, so the result is completely general.

Solving a triangle. The sine rule solves a triangle immediately whenever two angles and one side (AAS/ASA) are known — the third angle follows from $A+B+C=180^\circ$, and the rule then gives both remaining sides directly. It is also used for the SSA ("ambiguous case") data, where care is needed since two different triangles can sometimes satisfy the same data.

Worked example. Take $A=30^\circ$, $B=45^\circ$, $a=10$. By the sine rule,

The sine rule is the starting point for the projection rule (§10.3), the tangent rule (§10.4), and two of the four area formulas (§10.6), since each substitutes $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$ into an algebraic identity and simplifies.

A second worked example (the ambiguous case). Suppose $a=8,b=10,A=30^\circ$ are given (SSA — two sides and a non-included angle). From the sine rule, $\sin B=\dfrac{b\sin A}{a}=\dfrac{10\times0.5}{8}=0.625$, so $B\approx38.68^\circ$ or $B\approx141.32^\circ$ (the supplementary angle). Both satisfy $\sin B=0.625$, and here both remain consistent with a positive third angle once $A=30^\circ$ is added, so this data genuinely admits two distinct triangles — the hallmark of the SSA ambiguous case, and a reminder to always check that $A+B<180^\circ$ before accepting either solution.

Circumradius via area. Combining the sine rule with the SAS area formula $\Delta=\frac12bc\sin A$ (§10.6) gives a direct formula for $R$ itself: since $\sin A=\frac{a}{2R}$, substituting gives $\Delta=\frac12bc\cdot\frac a{2R}=\frac{abc}{4R}$, so $R=\dfrac{abc}{4\Delta}$ — useful whenever all three sides and the area (via Heron's formula) are already known, without ever needing to locate the circumcentre geometrically.