Law of Sines

The Law of Sines is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function.


In triangle $\triangle ABC$, where $a$ is the side opposite to $A$, $b$ opposite to $B$, $c$ opposite to $C$, and where $R$ is the circumradius:

\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\]


Method 1

[asy] real r = 1; pair A=(-1,5), B=(-4,-1), C=(4,-1), D, O; O = circumcenter(A,B,C); D = (B+C)/2;  draw(C--A--B--C--O--B); draw(O--D); draw(circumcircle(A,B,C));  label("$A$",A,(-1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1));label("$O$",O,(0,1)); label("$a/2$",(B+D)/2,(0,-1));label("$R$",(3*O+2*B)/5,(-1,1));label("$\theta$",O,(-0.8,-1.2));label("$\theta$",A,(0,-1.5));  dot(A^^B^^C^^D^^O); [/asy]

In the diagram above, point $O$ is the circumcenter of $\triangle ABC$. Point $D$ is on $BC$ such that $OD$ is perpendicular to $BC$. Since $\triangle ODB \cong \triangle ODC$, $BD = CD = \frac a2$ and $\angle BOD = \angle COD$. But $2\angle BAC = \angle BOC$ making $\angle BOD = \angle COD = \theta$. We can use simple trigonometry in right triangle $\triangle BOD$ to find that

$\sin \theta = \frac{\frac a2}R \iff \frac a{\sin\theta} = 2R.$

The same holds for $b$ and $c$, thus establishing the identity.

Method 2

This method only works to prove the regular (and not extended) Law of Sines.

The formula for the area of a triangle is $[ABC] = \frac{1}{2}ab\sin C$.

Since it doesn't matter which sides are chosen as $a$, $b$, and $c$, the following equality holds:

\[\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C\]

Assuming the triangle in question is nondegenerate, $abc \ne 0$. Multiplying the equation by $\frac{2}{abc}$ yields:

\[\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\]



  • If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
$\mathrm{(A) \  2}  \qquad \mathrm{(B) \ } 8/\sqrt{15}  \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ }  (\sqrt{6} + 1)/2$





Let $ABCD$ be a convex quadrilateral with $AB=BC=CD$, $AC \neq BD$, and let $E$ be the intersection point of its diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120^{\circ}$.


See Also

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