A triangle has two angles that measure \begin{align*}60^\circ\end{align*}
Law of Sines
Consider the non right triangle below. We can construct an altitude from any one of the vertices to divide the triangle into two right triangles as show below.
from the diagram we can write two trigonometric functions involving \begin{align*}h\end{align*}
\begin{align*}\sin C&=\frac{h}{b} \qquad and \qquad \sin B=\frac{h}{c}\\
b \sin C&=h \qquad \qquad \quad \ c \sin B=h\end{align*}
Since both are equal to \begin{align*}h\end{align*}
\begin{align*}b \sin C=c \sin B\end{align*}
\begin{align*}\frac{\sin C}{c}=\frac{\sin B}{b}\end{align*}
If we construct the altitude from a different vertex, say \begin{align*}B\end{align*}
Let's solve the following triangles using the Law of Sines.
Since we are given two of the three angles in the triangle, we can find the third angle using the fact that the three angles must add up to \begin{align*}180^\circ\end{align*}. So, \begin{align*}m \angle A=180^\circ  45^\circ  70^\circ = 650^\circ\end{align*}. Now we can substitute the known values into the Law of Sines proportion as shown:
\begin{align*}\frac{\sin 65^\circ}{a}=\frac{\sin 70^\circ}{15}=\frac{\sin 45^\circ}{c}\end{align*}
Taking two ratios at a time we can solve the proportions to find \begin{align*}a\end{align*} and \begin{align*}c\end{align*} using cross multiplication.
To find \begin{align*}a\end{align*}:
\begin{align*}\frac{\sin 65^\circ}{a}&=\frac{\sin 70^\circ}{15} \\ a&=\frac{15 \sin65^\circ}{\sin 70^\circ} \approx 14.5\end{align*}
To find \begin{align*}c\end{align*}:
\begin{align*}\frac{\sin 70^\circ}{15}&=\frac{\sin 45^\circ}{c} \\ c&=\frac{15 \sin45^\circ}{\sin 70^\circ} \approx 11.3\end{align*}
This particular triangle is an example in which we are given two angles and the nonincluded side or AAS (also SAA).
In this problem we are given two angles and a side as well but the side is between the angles. We refer to this arrangement as ASA. In practice, in doesn’t really matter whether we are given AAS or ASA. We will follow the same procedure as #1 above. First, find the third angle: \begin{align*}m \angle A=180^\circ  50^\circ  80^\circ = 50^\circ\end{align*}.
Second, write out the appropriate proportions to solve for the unknown sides, \begin{align*}a\end{align*} and \begin{align*}b\end{align*}.
To find \begin{align*}a\end{align*}:
\begin{align*}\frac{\sin 80^\circ}{a}&=\frac{\sin 50^\circ}{20} \\ a&=\frac{20 \sin80^\circ}{\sin 50^\circ} \approx 25.7\end{align*}
To find \begin{align*}b\end{align*}:
\begin{align*}\frac{\sin 50^\circ}{b}&=\frac{\sin 50^\circ}{20} \\ b&=\frac{20 \sin50^\circ}{\sin 50^\circ} = 20\end{align*}
Notice that \begin{align*}c=b\end{align*} and \begin{align*}m \angle C=m \angle B\end{align*}. This illustrates a property of isosceles triangles that states that the base angles (the angles opposite the congruent sides) are also congruent.
Now, let's solve the following problem.
Three fishing ships in a fleet are out on the ocean. The Chester is 32 km from the Angela. An officer on the Chester measures the angle between the Angela and the Beverly to be \begin{align*}25^\circ\end{align*}. An officer on the Beverly measures the angle between the Angela and the Chester to be \begin{align*}100^\circ\end{align*}. How far apart, to the nearest kilometer are the Chester and the Beverly?
First, draw a picture. Keep in mind that when we say that an officer on one of the ships is measuring an angle, the angle she is measuring is at the vertex where her ship is located.
Now that we have a picture, we can determine the angle at the Angela and then use the Law of Sines to find the distance between the Beverly and the Chester.
The angle at the Angela is \begin{align*}180^\circ  100^\circ  25^\circ = 55^\circ\end{align*}.
Now find \begin{align*}x\end{align*},
\begin{align*}\frac{\sin 55^\circ}{x}&=\frac{\sin 100^\circ}{32} \\ x&=\frac{32 \sin 55^\circ}{\sin 100^\circ} \approx 27\end{align*}
The Beverly and the Chester are about 27 km apart.
Examples
Example 1
Earlier, you were asked to find the lengths of the other two sides of a triangle that has two angles that measure \begin{align*}60^\circ\end{align*} and \begin{align*}45^\circ\end{align*} and a side between these two angles with length 10.
The measure of the triangle's third angle is \begin{align*}180^\circ  60^\circ 45^\circ =75^\circ\end{align*}
\begin{align*}\frac{\sin 45^\circ}{x}&=\frac{\sin 75^\circ}{10}, \ \text{so} \ x=\frac{10 \sin 45^\circ}{\sin 75^\circ} \approx 7.29 \\ \frac{\sin 60^\circ}{y}&=\frac{\sin 75^\circ}{10}, \ \text{so} \ y=\frac{10 \sin 60^\circ}{\sin 75^\circ} \approx 8.93\end{align*}
Example 2
Solve the following triangle.
\begin{align*}m \angle A=180^\circ  82^\circ 24^\circ =74^\circ\end{align*}
\begin{align*}\frac{\sin 24^\circ}{b}&=\frac{\sin 74^\circ}{11}, \ \text{so} \ b=\frac{11 \sin 24^\circ}{\sin 74^\circ} \approx 4.7 \\ \frac{\sin 82^\circ}{c}&=\frac{\sin 74^\circ}{11}, \ \text{so} \ c=\frac{11 \sin 82^\circ}{\sin 74^\circ} \approx 11.3\end{align*}
Example 3
Solve the following triangle.
\begin{align*}m \angle C=180^\circ  110^\circ 38^\circ =32^\circ\end{align*}
\begin{align*}\frac{\sin 38^\circ}{a}&=\frac{\sin 110^\circ}{18}, \ \text{so} \ a=\frac{18 \sin 38^\circ}{\sin 110^\circ} \approx 11.8 \\ \frac{\sin 32^\circ}{c}&=\frac{\sin 110^\circ}{18}, \ \text{so} \ c=\frac{18 \sin 32^\circ}{\sin 110^\circ} \approx 10.2\end{align*}
Example 4
A surveying team is measuring the distance between point \begin{align*}A\end{align*} on one side of a river and point \begin{align*}B\end{align*} on the far side of the river. One surveyor is positioned at point \begin{align*}A\end{align*} and the second surveyor is positioned at point \begin{align*}C\end{align*}, 65 m up the riverbank from point \begin{align*}A\end{align*}. The surveyor at point \begin{align*}A\end{align*} measures the angle between points \begin{align*}B\end{align*} and \begin{align*}C\end{align*} to be \begin{align*}103^\circ\end{align*}. The surveyor at point \begin{align*}C\end{align*} measures the angle between points \begin{align*}A\end{align*} and \begin{align*}B\end{align*} to be \begin{align*}42^\circ\end{align*}. Find the distance between points \begin{align*}A\end{align*} and \begin{align*}B\end{align*}.
\begin{align*}m \angle B &= 180^\circ  103^\circ  42^\circ = 35^\circ \\ \frac{\sin 35^\circ}{65} &= \frac{\sin 42^\circ}{c} \\ c &= \frac{65 \sin 42^\circ}{\sin 35^\circ} \approx 75.8 \ m\end{align*}
Review
Solve the triangles. Round your answers to the nearest tenth.
Using the given information, solve \begin{align*}\Delta ABC\end{align*}.
 .

 \begin{align*}m \angle A&=85^\circ \\ m \angle C&=40^\circ \\ a&=12\end{align*}
 .

 \begin{align*}m \angle B&=60^\circ \\ m \angle C&=25^\circ \\ a&=28\end{align*}
 .

 \begin{align*}m \angle B&=42^\circ \\ m \angle A&=36^\circ \\ b&=8\end{align*}
 .

 \begin{align*}m \angle B&=30^\circ \\ m \angle A&=125^\circ \\ c&=45\end{align*}
Use the Law of Sines to solve the following world problems.
 A surveyor is trying to find the distance across a ravine. He measures the angle between a spot on the far side of the ravine, \begin{align*}X\end{align*}, and a spot 200 ft away on his side of the ravine, \begin{align*}Y\end{align*}, to be \begin{align*}100^\circ\end{align*}. He then walks to \begin{align*}Y\end{align*} the angle between \begin{align*}X\end{align*} and his previous location to be \begin{align*}20^\circ\end{align*}. How wide is the ravine?
 A triangular plot of land has angles \begin{align*}46^\circ\end{align*} and \begin{align*}58^\circ\end{align*}. The side opposite the \begin{align*}46^\circ\end{align*} angle is 35 m long. How much fencing, to the nearest half meter, is required to enclose the entire plot of land?
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 13.12.