11.1: The Law of Sines

The Law of sines is based on right triangle relationships that can be created with the height of a triangle. Often, in this type of a problem, the angles are labeled with capital letters and their corresponding sides are labeled with lower case letters.

If we drop a perpendicular to the base of the triangle from the vertex point at \(\angle C\) this creates two right triangles with which we can make use of the right triangle trigonometry covered in Chapter 1. This perpendicular would be the height of the triangle.

The Law of sines is derived from this configuration and allows us to calculate the value of sides and angles in a triangle without a right angle, based on information about known sides and angles. Given the right triangles in the diagram above, we can see that:
\[ \sin B=\frac{h}{a}
\] and
\[ \sin A=\frac{h}{b}
\]

Clearing the denominator in each fraction, we can see that:
\[ a \sin B=h
\] and
\[ \begin{array}{c}
b \sin A=h \\
\text{so} \\
a \sin B=b \sin A
\end{array}
\]

To put this in the form in which the Law of sines is normally stated, we can divide on both sides of the previous expression by \(a b\) :
\[ \begin{array}{c}
a \sin B=b \sin A \\
\frac{a \sin B}{a b}=\frac{b \sin A}{a b} \\
\frac{\sin B}{b}=\frac{\sin A}{a}
\end{array}
\] A similar process will show that \(\frac{\sin C}{c}\) is equivalent to \(\frac{\sin B}{b}\) and \(\frac{\sin A}{a} .\) The diagram we derived this from used an acute triangle in which all the angles were less than
\(90^{\circ} .\) The process to show that this is true for an obtuse triangle (which has one angle larger than \(90^{\circ}\) ) is relatively simple and is left to the reader to discover or look up in another resource.
The Law of sines
\[ \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}
\]

Sometimes it is handy to set up a problem with the side lengths in the numerator:
The Law of sines
\[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
\]\

Solve the triangle. Round side lengths to the nearest \(100^{\text {th }}\).

In this problem we're given two angles and one side. It's important that the side we're given corresponds to one of the known angles, otherwise we wouldn't be able to use the Law of sines.

since we know two of the angles, then the third will just be \(180^{\circ}-\left(45^{\circ}+95^{\circ}\right)=\) \(180^{\circ}-140^{\circ}=40^{\circ}=\angle A .\) To find the lengths of the unknown sides, we'll use the Law of sines. We should start by choosing a side-angle pair for which we know both the side and the angle. In this case, we know that \(\angle C=95^{\circ}\) and side \(c=5\)
\[ \begin{array}{c}
\frac{c}{\sin C}=\frac{b}{\sin B} \\
\frac{5}{\sin 95^{\circ}}=\frac{b}{\sin 45^{\circ}}
\end{array}
\] If we multiply on both sides by \(\sin 45^{\circ},\) then
\[ \sin 45^{\circ} * \frac{5}{\sin 95^{\circ}}=b
\]

To arrive at an approximate value for \(\sin 45^{\circ} * \frac{5}{\sin 95^{\circ}},\) we can say:
\[ \begin{array}{c}
0.7071 * \frac{5}{0.9962} \approx b \\
3.55 \approx b
\end{array}
\] To find the length of side \(a,\) I would recommend that we use the exact side-angle pair that was given in the problem, rather than using the approximate value of side \(b\) that we just solved for.

This will make our value for side \(a\) more accurate:
\[ \begin{array}{c}
\frac{c}{\sin C}=\frac{a}{\sin A} \\
\frac{5}{\sin 95^{\circ}}=\frac{a}{\sin 40^{\circ}}
\end{array}
\] Multiplying on both sides by \(\sin 40^{\circ},\) then
\[ \sin 40^{\circ} * \frac{5}{\sin 95^{\circ}}=a
\] To arrive at an approximate value for \(\sin 40^{\circ} * \frac{5}{\sin 95^{\circ}},\) we can say:
\[ \begin{array}{c}
0.6428 * \frac{5}{0.9962} \approx a \\
3.23 \approx a
\end{array}
\]

\begin{array}{lll}
\angle A=40^{\circ} & a \approx 3.23 \\
\angle B=45^{\circ} & b \approx 3.55 \\
\angle C=95^{\circ} & c=5
\end{array}

Example 2
Some problems don't come with diagrams:
Solve the triangle if: \(\quad \angle A=40^{\circ}, \quad \angle B=20^{\circ}, \quad a=2\)
Round side lengths to the nearest \(100^{\text {th }}\).
Just as in the previous example, we can begin by finding the measure of the third angle \(\angle C .\) This would be \(180^{\circ}-\left(40^{\circ}+20^{\circ}\right)=180^{\circ}-60^{\circ}=120^{\circ}=\angle C\)
To find the missing sides, we should use the complete side-angle pair that is given in the problem: \(\angle A=40^{\circ}\) and \(a=2\)

We can find side \(b\) first or side \(c\) first, it doesn't matter which:
\[ \begin{array}{c}
\frac{a}{\sin A}=\frac{b}{\sin B} \\
\frac{2}{\sin 40^{\circ}}=\frac{b}{\sin 20^{\circ}} \\
\sin 20^{\circ} * \frac{2}{\sin 40^{\circ}}=b
\end{array}
\] Then,
\[ 0.3420 * \frac{2}{0.6428} \approx b
\] \[ 1.06 \approx b
\]

For side \(c:\)
\[ \begin{array}{c}
\frac{a}{\sin A}=\frac{c}{\sin C} \\
\frac{2}{\sin 40^{\circ}}=\frac{c}{\sin 120^{\circ}} \\
\sin 120^{\circ} * \frac{2}{\sin 40^{\circ}}=c
\end{array}
\] Then,
\[ 0.8660 * \frac{2}{0.6428} \approx c
\] \[ 2.69 \approx c
\]

\begin{array}{ll}
\angle A=40^{\circ} & a=2 \\
\angle B=20^{\circ} & b \approx 1.06 \\
\angle C=120^{\circ} & c \approx 2.69
\end{array}

Exercises 4.1
In each problem below, solve the triangle. Round side lengths to the nearest \(100^{t h}\)

7. \(\quad \angle A=50^{\circ}, \quad \angle C=27^{\circ}, \quad a=3\)
8. \(\quad \angle B=70^{\circ}, \quad \angle C=10^{\circ}, \quad b=5\)
9. \(\quad \angle A=110^{\circ}, \quad \angle C=30^{\circ}, \quad c=3\)
10. \(\quad \angle A=50^{\circ}, \quad \angle B=68^{\circ}, \quad a=230\)
11. \(\quad \angle A=23^{\circ}, \quad \angle B=110^{\circ}, \quad c=50\)
12. \(\quad \angle A=22^{\circ}, \quad \angle B=95^{\circ}, \quad a=420\)
13. \(\quad \angle B=10^{\circ}, \quad \angle C=100^{\circ}, \quad c=11\)
14. \(\quad \angle A=30^{\circ}, \quad \angle C=65^{\circ}, \quad b=10\)
15. \(\quad \angle A=82^{\circ}, \quad \angle B=65.4^{\circ}, \quad b=36.5\)
16. \(\quad \angle B=28^{\circ}, \quad \angle C=78^{\circ}, \quad c=44\)
17. \(\quad \angle A=42^{\circ}, \quad \angle B=61^{\circ}, \quad a=12\)
18. \(\quad \angle A=42.5^{\circ}, \quad \angle B=71.4^{\circ}, \quad a=215\)