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.

Theorem

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:

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

Proof

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:

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

Problems

Introductory

If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

$\mathrm{(A) \ 2} \qquad \mathrm{(B) \ } 8/\sqrt{15} \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2$

(Source)

Intermediate

Triangle $ABC$ has sides $\overline{AB}$ , $\overline{BC}$ , and $\overline{CA}$ of length 43, 13, and 48, respectively. Let $\omega$ be the circle circumscribed around $\triangle ABC$ and let $D$ be the intersection of $\omega$ and the perpendicular bisector of $\overline{AC}$ that is not on the same side of $\overline{AC}$ as $B$ . The length of $\overline{AD}$ can be expressed as $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find the greatest integer less than or equal to $m + \sqrt{n}$ .

(Source)

Olympiad

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 $\angle BAD+\angle ADC = 120^{\circ}$ .

(Source)

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Law of Sines

Contents

Theorem

Proof

Method 1

Method 2

Problems

Introductory

Intermediate

Olympiad

See Also