The Sine Rule

The Sine Rule. In $\triangle ABC$,

where $R$ is the circumradius (the radius of the circle passing through $A$, $B$, $C$).

Proof (first part — the three ratios are equal). Draw $AD\perp BC$.

In right triangle $ADC$, $AD=b\sin C$. So the area $A(\triangle ABC)=\dfrac12\cdot BC\cdot AD=\dfrac12\cdot a\cdot b\sin C$, i.e. $2\,A(\triangle ABC)=ab\sin C$. By dropping perpendiculars from the other two vertices in the same way, $2\,A(\triangle ABC)=ac\sin B=bc\sin A$ as well. So $bc\sin A=ac\sin B=ab\sin C$; dividing throughout by $abc$: $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$, i.e. $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$. — (1)

Proof (second part — the common ratio is $2R$). Since the three angles of a triangle sum to $180^\circ$, at least one of them is not a right angle; suppose $\angle A$ is not a right angle. Draw the diameter through $A$, meeting the circumcircle again at $P$; then $AP=2R$ and $\triangle ACP$ is right-angled at $C$ (angle in a semicircle). Since $\angle ABC$ and $\angle APC$ are angles inscribed in the same arc $AC$, $m\angle ABC=m\angle APC$, so $\sin B=\sin P=\dfrac{b}{AP}=\dfrac{b}{2R}$, i.e. $\dfrac{b}{\sin B}=2R$. — (2). Combining (1) and (2): $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$. $\blacksquare$

Different equivalent forms of the Sine Rule (all standard, all worth recognising): (i) $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$; (ii) $a=2R\sin A,\ b=2R\sin B,\ c=2R\sin C$; (iii) $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k$ (some constant $k=1/2R$); (iv) $b\sin A=a\sin B,\ c\sin B=b\sin C,\ c\sin A=a\sin C$; (v) $\dfrac{a}{b}=\dfrac{\sin A}{\sin B},\ \dfrac{b}{c}=\dfrac{\sin B}{\sin C}$.

Ex.(1) In $\triangle ABC$ if $A=30^\circ$, $B=60^\circ$ then find the ratio of its sides. Since $A+B+C=180^\circ$, $C=90^\circ$. By the Sine Rule, $\dfrac{a}{\sin30^\circ}=\dfrac{b}{\sin60^\circ}=\dfrac{c}{\sin90^\circ}$, i.e. $\dfrac{a}{1/2}=\dfrac{b}{\sqrt3/2}=\dfrac{c}{1}$. So $a:b:c=\dfrac12:\dfrac{\sqrt3}{2}:1=1:\sqrt3:2$.

Ex.(2) In $\triangle ABC$ if $a=2$, $b=\sqrt3$ and $\sin A=\dfrac{2}{\sqrt3}$ then find $B$. By the Sine Rule, $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\implies\dfrac{2}{2/\sqrt3}=\dfrac{\sqrt3}{\sin B}\implies\sqrt3=\dfrac{\sqrt3}{\sin B}\implies\sin B=1\implies B=\dfrac{\pi}{2}=90^\circ$. …