A triangle has two angles that measure
Guidance
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
Since both are equal to
If we construct the altitude from a different vertex, say
Example A
Solve the triangle.
Solution: 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
Taking two ratios at a time we can solve the proportions to find
To find
To find
This particular triangle is an example in which we are given two angles and the nonincluded side or AAS (also SAA).
Example B
Solve the triangle.
Solution: In this example 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 Example A. First, find the third angle:
Second, write out the appropriate proportions to solve for the unknown sides,
To find
To find
Notice that
Example C
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
Solution: 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
Now find
The Beverly and the Chester are about 27 km apart.
Concept Problem Revisit The measure of the triangle's third angle is
Guided Practice
Solve the triangles.
1.
2.
3. A surveying team is measuring the distance between point
Answers
1. \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*}
2. \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*}
3.
\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*}
Explore More
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 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&=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&=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*}

\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?