# Sine Rule (Law of Sines)

The **law of sines** is a relationship linking the sides of a triangle with the sine of their corresponding angles.

## Sine Rule

Given the following triangle $ABC$ with corresponding side lengths $a, b$ and $c$:

the **sine rule** or **law of sines** is the following identity:

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

We will prove the first identity

$\frac { a}{ \sin (A)} = \frac {b}{\sin (B)}.$

The second equality can be proved similarly.

By drawing the height $h$ of the triangle from vertex $C$ to the opposite side, we can express the height $h$ in two different ways:

- First, we have $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{b}$, which implies $h = b\sin(A).$
- Also, $\sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{a}$, which implies $h = a\sin(B).$
By equating these values of $h$, we have

$\begin{aligned} b \sin(A) &= a \sin(B) \\ \frac{a}{\sin(A)} &= \frac{b}{\sin(B)}. \end{aligned}$

By drawing the height $h$ from the other two vertices, we can similarly show the second equality. $_\square$

See the extended sine rule for another proof.

## Examples

One real-life application of the sine rule is the sine bar, which is used to measure the angle of a tilt in engineering. Other common examples include measurement of distances in navigation and measurement of the distance between two stars in astronomy.

In the following triangle, suppose that $a= 1, \angle A = 30^\circ,$ and $\angle B = 45^\circ$. $$ What is the side length $b?$

From the sine rule, we have

$\begin{aligned} \frac{a}{\sin(A)} &= \frac{b}{\sin(B)} \\ \frac{1}{\sin(30^\circ)} &= \frac{b}{\sin(45^\circ)} \\ \frac{1}{\ \frac{1}{2}\ } &= \frac{b}{\ \frac{\sqrt{2}}{2}\ } \\ \frac{2 \cdot \sqrt{2}}{2}&= b. \end{aligned}$

Therefore, $b= \sqrt{2}$. $_\square$

## Ambiguous Case

A common application of the sine rule is to determine the triangle $ABC$ given some of its sides and angles. The *ambiguous case* refers to scenarios where there are 2 distinct triangles that satisfy such a configuration. This occurs when we are given the angle-side-side, as shown in the diagram below:

If the side lengths of $\triangle ABC$ are $a = 10$ and $b=9$ with $\angle A$ opposite to $a$ measuring $26$ degrees, what is the measure of $\angle B$ opposite to $b?$

By the sine rule, we have $\frac a{\sin A} = \frac b{\sin B}$ or $\frac{10}{\sin 26^\circ} = \frac9{\sin B}$. Solving for $B$ yields $\sin B = \frac9{10} \sin 26^\circ$ or $B \approx 23.237^\circ$.

However, note that $\sin x = \sin(180^\circ - x)$. Since $A + B < 180^\circ$ and $A + (180^\circ - B) < 180^\circ ,$ another possible measure of $B$ is approximately $180^\circ - 23.237^\circ = 156.763^\circ$. $_\square$

Using the same example given above, find the measure of $\angle B$ if the measure of $\angle A$ is 46 degrees instead.

By the sine rule, we have $\frac a{\sin A} = \frac b{\sin B}$ or $\frac {10}{\sin 46^\circ} = \frac9{\sin B}$. Solving for $B$ yields $\sin B = \frac9{10} \sin 46^\circ$ or $B \approx 40.346^\circ$.

However, note that $\sin x = \sin(180^\circ - x)$. Since $A + B < 180^\circ$ but it is not true that $A + (180^\circ - B) < 180^\circ,$ there is only one possible measure of $B$, which is approximately $40.346$ degrees. $_\square$

Using the very first example given above, if you are further given the angle of $B$ to be 13 degrees, what is the total possible number of distinct measures of $\angle C?$

By the sine rule we have $\frac a{\sin A} = \frac b{\sin B}$ or $\frac {10}{\sin 46^\circ} = \frac9{\sin 13^\circ}$, which is clearly false, implying there is no such triangle. Hence there is no possible triangle that fits these criteria. $_\square$

## Extended Sine Rule

The **extended sine rule** is a relationship linking the sides of a triangle with the sine of their corresponding angles and the radius of the circumscribed circle. The statement is as follows:

Given triangle $ABC$, with corresponding side lengths $a, b$ and $c$ and $R$ as the radius of the circumcircle of triangle $ABC$, we have the following:

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

**Note:** The statement without the third equality is often referred to as the sine rule. The relationship between the sine rule and the radius of the circumcircle of triangle $ABC$ is what extends this to the extended sine rule.

Extended Sine Rule

Let $O$ be the center of the circumcircle, and $D$ the midpoint of $\overline{BC}.$ Then $\overline{OD}$ is perpendicular to $\overline{BC}$. Now, observe that $\angle BOC$ is equal to $2\alpha$ or $360^\circ - 2\alpha,$ where $\angle A=\alpha,$ depending on whether $O$ is in the triangle or not. Then $\angle BOD = \alpha$ or $180^\circ - \alpha$, and thus $\sin BOD = \sin \alpha$. As such,

$\frac {\ \lvert\overline{BD}\rvert\ }{\lvert\overline{OB}\rvert} = \sin \alpha \implies \frac {a} {\sin \alpha } = 2R.\, _\square$

Show that the area of triangle $ABC$ is equal to $\frac {abc} {4R}.$

Let $D$ be the foot of the perpendicular from $A$ to $\overline{BC}$. Using $\overline{BC}$ as the base and $\overline{AD}$ as the height, the area of the triangle is $\frac {1}{2} a \cdot \lvert\overline{AD}\rvert$. From right triangle $CAD$, $\sin \gamma = \frac {\lvert\overline{AD}\rvert} {b}$. Thus, the area of the triangle is $\frac {1}{2} a \cdot \lvert\overline{AD}\rvert = \frac {1}{2} a b \sin \gamma$, which is often quoted. Now, using the extended sine rule, we have, $\frac {c}{\sin \gamma} = 2R$, and thus the area of the triangle is

$\frac {1}{2} a b \sin \gamma = \frac {1}{2} a b \frac {c}{2R} = \frac {abc} {4R}. \ _\square$

**Cite as:**Sine Rule (Law of Sines).

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/sine-rule/