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.