Theorem 7.3.1. Law of Sines.
For a triangle with angles and sides as labeled in Figure 7.3.2,
\begin{equation*}
\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.
\end{equation*}
Section 7.3 Non-Right Triangles
In Section 7.1 we learned about relationships between angles of the triangles and their sides. However, most of our work was restricted to right triangles. Here we learn some relationships that do not require any angle to be a right angle.
Note the one relationship that did not require a right angle is the angle sum property Theorem 7.1.3.
Subsection 7.3.1 Law of Sines
In Figure 7.1.8 we saw that there was a relationship between angles of a triangle and the side ratios. More generally there is a relationship between an angle of a triangle and the side opposite it. Note that this property is a proportion.
Example 7.3.3.
A triangle has an angle that is \(50^\circ\) which is opposite a side of length 6. It has another angle that is \(45^\circ\text{.}\)
(a)
What is the length of the side opposite the \(45^\circ\) angle?
Solution.
According to the law of sines
\begin{align*}
\frac{\sin(50^\circ)}{6} \amp = \frac{\sin(45^\circ)}{b}.\\
b \sin(50^\circ) \amp = 6 \sin(45^\circ). \amp \text{ Clearing the denominators}\\
b \amp = 6 \frac{\sin(45^\circ)}{\sin(50^\circ)}.\\
\amp \approx 5.54.
\end{align*}
(b)
What is the third angle and the length of the side opposite it?
Solution.
We can use the angle sum fact to calculate the third angle. \(50+45+\alpha=180\) so \(\alpha=85\text{.}\) The side length can be calculated using the law of sines.
\begin{align*}
\frac{\sin(50^\circ)}{6} \amp = \frac{\sin(85^\circ)}{c}.\\
c \sin(50^\circ) \amp = 6 \sin(85^\circ). \amp \text{ Clearing the denominators}\\
c \amp = 6 \frac{\sin(85^\circ)}{\sin(50^\circ)}.\\
\amp \approx 7.80.
\end{align*}
Example 7.3.4.
A triangle has an angle of measure \(40^\circ\text{.}\) The side opposite it is length 5. The other two sides are length 6.74 and 7.66.
(a)
What is the measure of the angle opposite the side of length 6.74?
Solution.
We can use the law of sines.
\begin{align*}
\frac{\sin(40^\circ)}{5} \amp = \frac{\sin(B)}{6.74}.\\
6.75 \sin(40^\circ) \amp = 5\sin(B). \amp \text{ Clearing the denominators}\\
\frac{6.75 \sin(40^\circ)}{5} \amp = \sin(B)\\
\arcsin\left(\frac{6.75 \sin(40^\circ)}{5}\right) \amp = B.\\
60.20 \approx B.
\end{align*}
(b)
What is the measure of the angle opposite the side of length 7.66?
Solution.
We can now use the angle sum fact. \(40+60.20+\alpha=180\) so \(\alpha \approx 79.80\text{.}\)
Checkpoint 7.3.5.
A triangle has angles of measure \(A={50}\) and \(B={70}\text{.}\) The side opposite angle \(A\) has length 7. What is the length of the side opposite angle \(B\text{?}\)
Answer.
\(\frac{\sin\!\left(70\right)}{\sin\!\left(50\right)}\cdot 7\)
Subsection 7.3.2 Ambiguous Triangles
We have calculated angles and side lengths given partial information about a triangle. Here we look at one case we cannot resolve without additional information.
Example 7.3.6.
A triangle has an angle of measure \(45^\circ\) and the side opposite it is length 4. Another side is length 5. What are the other angles and side lengths?
We can try to use the law of sines.
\begin{align*}
\frac{\sin(45^\circ)}{4} \amp = \frac{\sin(\theta)}{5}.\\
5 \cdot \frac{\sin(45^\circ)}{4} \amp = \sin(\theta).\\
\arcsin\left( \frac{5}{4}\sin(45^\circ) \right) \amp = \theta.\\
62.11^\circ \amp \approx \theta.
\end{align*}
Using the angle sum fact we learn the other angle is \(45+62.11+\alpha = 180\) or \(\alpha \approx 72.89^\circ\text{.}\) We use the law of sines again to find the length of the final side.
\begin{align*}
\frac{\sin(45^\circ)}{4} \amp = \frac{\sin(72.89^\circ)}{c}.\\
c \sin(45^\circ) \amp = 4\sin(72.89^\circ).\\
c \amp = 4 \frac{\sin(72.89^\circ)}{\sin(45^\circ)}.\\
c \amp \approx 5.41.
\end{align*}
This gives us a triangle with angles: \(45^\circ, 62.11^\circ, 72.89^\circ\text{;}\) and with side lengths: 4, 5, and 5.41.
Note \(\sin(117.89^\circ)=\sin(62.11^\circ)\text{,}\) that is, If we use \(117.89^\circ\) as the second angle, the third angle is \(45+117.89+\alpha = 180\) or \(\alpha \approx 17.11^\circ\text{.}\)
\begin{align*}
\frac{\sin(45^\circ)}{4} \amp = \frac{\sin(17.11^\circ)}{c}.\\
c \sin(45^\circ) \amp = 4\sin(17.11^\circ).\\
c \amp = 4 \frac{\sin(17.11^\circ)}{\sin(45^\circ)}.\\
c \amp \approx 1.66.
\end{align*}
Notice we have two, distinct triangles that match the initial angle and side information. They can be seen in Figure 7.3.7. This indicates an ambiguity if what we know is this particular information.
Subsection 7.3.3 Law of Cosines
For right triangles we know the Pythagorean theorem is a relationship between the sides of those triangles. For triangles without a right angle that relationship must be slightly modified.
Theorem 7.3.8. Law of Cosines.
For any triangle with side lengths \(a,b,c\) and angle \(C\) which is opposite the side with length \(c\)
\begin{equation*}
c^2 = a^2+b^2-2ab\cos(C)\text{.}
\end{equation*}
Example 7.3.9.
A triangle has sides of lengths 4, 5.39, and 6.13. What are the angles?
Solution.
We can use the Law of Cosines.
\begin{align*}
4^2 \amp = 5.39^2+6.13^2-2(5.39)(6.13)\cos(A).\\
-50.63 \amp = -66.08 \cos(A).\\
\frac{-50.63}{-66.08} \amp = \cos(A).\\
0.7662 \amp = \cos(A).\\
\arccos(0.7662) \amp = A.\\
39.99^\circ \amp \approx A.
\end{align*}
With an angle, we can now use the Law of Sines, but for practice we will use Law of Cosines again.
\begin{align*}
5.39^2 \amp = 4^2+6.13^2-2(4)(6.13)\cos(B).\\
-24.52 \amp = -49.04\cos(B).\\
0.5 \amp = \cos(B).\\
\arccos(0.5) \amp = B.\\
60^\circ \amp = B.
\end{align*}
Knowing two of the angles we can use the angle sum fact to calculate the third angle measure. \(39.99+60+C=180\) so \(C = 80.01\text{.}\)
Example 7.3.10.
A triangle has side lengths 5 and 7 and the angle between them is \(40^\circ\text{.}\) What are the length of the other side and the measures of the other angles?
Solution.
We can use the Law of Cosines because we know \(a,b\) and \(C\text{.}\)
\begin{align*}
c^2 \amp = 5^2+7^2-2(5)(7)\cos(40^\circ)\\
c^2 \amp \approx 20.38.\\
\sqrt{c^2} \amp \approx \sqrt{20.38}.\\
c \amp \approx 4.51.
\end{align*}
Now that we know a side and the angle opposite it, we can use the Law of Sines to find the remaining two angles.
\begin{align*}
\frac{\sin(40^\circ)}{4.51} \amp = \frac{\sin(A)}{5}.\\
5 \cdot \frac{\sin(40^\circ)}{4.51} \amp = 5 \cdot \frac{\sin(A)}{5}.\\
0.7126 \amp = \sin(A).\\
\arcsin(0.7126) \amp = A.\\
45.45 \amp \approx A.
\end{align*}
Finally we can use the angle sum theorem to calculate the final angle. \(40+45.45+B = 180\) so \(B=94.55\text{.}\)
Checkpoint 7.3.11.
A triangle has sides of length \(a={5}\text{,}\) \(b={5.9155}\text{,}\) and \(c={5.9155}\text{.}\) What is the measure of the angle opposite the side of length \(a\text{?}\)
Answer.
\(50\)
