# Law of Sines

The Law of Sines establishes the relationship between the lengths of the sides of an arbitrary triangle and the sines of the opposite angles.

## Theorem (Law of Sines)

The Law of Sines (also known as the Sine Rule) says that the ratio of each side of a triangle to the sine of the opposite angle is the same for all three pairs of sides and angles.

So if a triangle has sides of lengths $$a,$$ $$b,$$ and $$c,$$ opposite to the angles $$\alpha,$$ $$\beta,$$ and $$\gamma,$$ respectively, then

$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}$

### Proof

Consider an arbitrary oblique triangle $$\triangle ABC$$ with the sides $$a,b,c$$ and angles $$\alpha, \beta, \gamma.$$ The angle $$\beta$$ can be either acute as in Figure $$2,$$ or obtuse as in Figure $$3.$$ In each case, draw the altitude $$CD = h$$ from the vertex $$C$$ to the base $$AB = c.$$

Applying trigonometric relationships for the right triangles $$\triangle CDA$$ and $$\triangle CDB,$$ we can write:

• If the angle $$\beta$$ is acute (Figure $$2$$):
$h = b\sin \alpha = a\sin \beta .$
• If the angle $$\beta$$ is obtuse (Figure $$3$$):
$h = b\sin \alpha = a\sin \left( {\pi - \beta } \right).$

By reduction identity, $$\sin \left( {\pi - \beta } \right) = \sin \beta .$$ Therefore, in both cases we have

$h = b\sin \alpha = a\sin \beta .$

It follows from the last equation that

$\frac{a}{{\sin \alpha }} = \frac{b}{{\sin \beta }}.$

Similarly, if we draw the altitude from the vertex $$A$$ to the side $$BC = a,$$ we can derive the relationship

$\frac{b}{{\sin \beta }} = \frac{c}{{\sin \gamma }}.$

This proves the Sine Rule:

$\frac{a}{{\sin \alpha }} = \frac{b}{{\sin \beta }} = \frac{c}{{\sin \gamma }}.$

## Extended Law of Sines

For any triangle,

$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R$

where $$R$$ is the radius of the circumcircle of the triangle.

### Proof

It is enough to prove that

$\frac{a}{{\sin \alpha }} = 2R.$

Consider again two cases - depending on whether the center of the circumcircle $$O$$ lies inside the triangle (Figure $$4$$), or outside the triangle (Figure $$5$$). In each case, draw the diameter $$BM$$ for the circumcircle.

Using the properties of angles in a circle, we see that the $$\angle BCM$$ is a right angle, and the angle $$\angle CMB$$ is either equal to $$\alpha,$$ as in Figure $$4,$$ or it is equal to $$\pi - \alpha,$$ as in Figure $$5.$$ Since $$\sin \left( {\pi - \alpha } \right) = \sin \alpha,$$ in each case we get

$a = 2R\sin \alpha .$

By similar reasoning for the other two sides, we can conclude that

$\frac{a}{{\sin \alpha }} = \frac{b}{{\sin \beta }} = \frac{c}{{\sin \gamma }} = 2R.$

See solved problems on Page 2.