## The geometry of the law of sines

Given three non-collinear points, $$A, B$$ and $$C$$, there is only one circle containing those three points. Three non-collinear points also form the vertices of a triangle. According to a theorem from geometry, the ratio of a side of a triangle to the sine of the opposite angle equals the diameter of the circumscribed circle. Since for a given triangle there is only one circumscribed circle, then one may compute its diameter by computing the ratio between any of its three sides and the sine of the angle opposite that side. That is,

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

This identity is called the Law of Sines.

The Law of Sines, together with the Law of Cosines which we will study shortly, are useful for solving triangles.

There are six pieces of information that must be known before a triangle is 'solved': the size of its three angles and the size of its three sides.

In order to solve a triangle, one must know at least three of the six pieces of information at the start. The one exception to this principle is this: knowing only the three angles is sufficient for finding the ratios of the three sides, but is insufficient for finding the actual lengths of the three sides.

## Exercise 6.1.1

Why can't one determine the length of the three sides of a triangle if one knows all three angles?

See Solution

## The three given pieces of information

There are six different ways to classify the three given pieces of information:

SSS: The lengths of all three sides are given.

SAS: The lengths of two sides and the size of the angle between them are given.

SSA: The lengths of two sides and the size of the angle opposite one of them are given.

ASA: The sizes of two angles and the length of the side between them are given.

SAA: The sizes of two angles and the length of the side opposite one of them are given.

AAA: The sizes of all three angles are given.

Note that some combinations of three pieces of information will not have a solution.

For example, the sum of the three angles of a triangle must be $$180^\circ$$. Thus in the case of ASA, if the two angles add to more than $$180^\circ$$, there will be no solution.

Or in the case of SSS, if the sum of two of the sides is less than the third side, there will be no solution, since the sum of two sides of a triangle is always greater than the third side.

SAS is the only case where all possible lengths of the sides and all possible angles greater than $$0^\circ$$ and less than $$180^\circ$$ will have a solution.

## Exercise 6.1.2

Explain why each of the following 'triangles' has no solution:

1. SSS: $$a = 7, b = 2, c = 4$$
2. ASA: $$A = 85^\circ, b = 10, C = 98^\circ$$
3. SAA: $$a = 12, B = 100^\circ, C = 90^\circ$$
4. AAA: $$A = 65^\circ, B = 82^\circ, C = 35^\circ$$

See Solution

## A side/angle pair

In order to apply the Law of Sines, it is necessary to know the size of at least one side and the size of the angle opposite that side. That is, one must know at least one side-sine pair.

One will have a side-sine pair in the case of SSA or SAA. But if one is given ASA, one can deduce the third angle since the sum all all three angles must be $$180^\circ$$ Thus one will know a side-sine pair given ASA.

So the Law of Sines is used to solve triangles given SSA, SAA or ASA.

SAA and ASA are worked essentially the same: find the third angle, then use the side-sine pair to solve the other two sides.

Example: Given $$a = 10, A = 30^\circ$$ and $$C = 120^\circ$$, solve the triangle.

First find $$B = 180^\circ - 30^\circ - 120^\circ = 30^\circ$$.

Next, use the fact that $$\dfrac{b}{\sin B}=\dfrac{a}{\sin A}$$ to find $$b$$, since $$b = \dfrac{a\sin B}{\sin A}= \dfrac{10 \sin 30^\circ}{\sin 30^\circ} = 10$$.

Likewise, $$\dfrac{c}{\sin C}=\dfrac{a}{\sin A}$$, so $$c = \dfrac{a\sin C}{\sin A}= \dfrac{10 \sin 120^\circ}{\sin 30^\circ} = 10\sqrt{3}$$.

Thus the solution of the triangle is:

 $$B = 120^\circ$$ $$b=10$$ $$c=10\sqrt{3}$$

## Exercise 6.1.3

Given $$A = 24^\circ, B = 80^\circ$$ and $$c = 20$$, solve the triangle.

See Solution

## Exercise 6.1.4

It is possible for the case SSA to have no solution for some combinations of lengths of the sides and angles.

Explain why the following SSA 'triangle' has no solution. Hint: the sine of an angle cannot be greater than 1.

$$A = 60^\circ, a = 2, b = 10$$

See Solution

## Supplementary angles have the same sine

Two angles between $$0^\circ$$ and $$180^\circ$$ are said to be 'supplementary angles' if their sum is $$180^\circ$$.

Supplementary angles have the same sine.

Knowing the sine of one of the angles of a triangle is not sufficient information to uniquely determine the angle.

For example, suppose we want to find $$A$$ and all we know is that $$\sin A = \frac{1}{2}$$.

Then $$A$$ could be $$30^\circ$$ or $$A$$ could be the supplementary angle $$150^\circ$$. We cannot assume that the correct value is $$30^\circ$$ unless we can rule out $$150^\circ$$.

In general for an angle $$A$$ in $$\triangle ABC$$, if $$\sin A = y$$, then either $$A = \arcsin y$$ or $$A = 180^\circ - \arcsin y$$.

$$A = \arcsin y$$ is the acute solution and $$A = 180^\circ - \arcsin y$$ is the obtuse solution.

Sometimes this situation arises when solving SSA triangles. When this happens and when we cannot rule out the larger solution, then we must list two separate solutions to the triangle.

## Exercise 6.1.5

Find two solutions to the triangle:

$$a = 10, b = 4, B = 20^\circ$$

See Solution

## Exercise 6.1.6

A ship sailing due north spots a lighthouse $$30^\circ$$ off the port side (to the left of its line of travel). Two miles later, the lighthouse is $$50^\circ$$ off the port side. How far is the ship from the lighthouse at that point? (Assume the earth is flat.)

See Solution

## Exercise 6.1.7

Scientists wish to measure the diameter of a large circular meteor crater. Two points $$A$$ and $$B$$ on the edge of the pit are 120 feet apart. From a point $$C$$ on the far side of the pit, the angle between the lines $$AC$$ and $$BC$$ is measured to be $$8^\circ$$. Use the Law of Sines to find the diameter of the crater.

See Solution