A triangle has two sides of lengths 2 (a) and 5 (b). The non-included angle (B) of the triangle measures

### SSA

Recall that the sine ratios for an angle and its supplement will always be equal. In other words,

Given

First, let’s make a diagram to show the relationship between the given sides and angles. Then we can set up a proportion to solve for angle

From here we can find

Putting these measures in the triangle, we get:

But, we know that

The second triangle would look like this:

In this instance there were two possible triangles.

Now, given

Again we will start with a diagram and use the law of sines proportion to find a second angle measure in the triangle.

Now find the third angle,

Because we used the inverse sine function to determine the measure of angle

This problem shows that two triangles are not always possible. Note that if the given angle is obtuse, there will only be one possible triangle for this reason.

In both problems, we simply tested to see if there would be a second triangle. There are, however, guidelines to follow to determine when a second triangle exists and when it does not. The “check and see” method always works and therefore it is not necessary to memorize the following table. It is interesting, however, to see to pictures and make the connection between the inequalities and what if any triangle can be formed.

First, consider when

If **one** triangle can be formed.

If **no** triangle can be formed.

Now, consider the possible scenarios when

If **one** triangle can be formed.

For the following cases, where

If **no** triangle can be formed because

If **one right** triangle can be formed because

If **two** triangles can be formed because

Finally, given

In this case, \begin{align*}A\end{align*} is acute and \begin{align*}a<b\end{align*}, so we need to look at the value of \begin{align*}b \sin a\end{align*}. Since \begin{align*}b \sin A=10 \sin 42^\circ \approx 6.69 < a\end{align*}, there will be two triangles. To solve for these triangles, use the Law of Sines extended proportion instead of making a diagram. Plugging in what we know, we have:

\begin{align*}\frac{\sin 42^\circ}{8}=\frac{\sin B}{b}=\frac{\sin C}{10}\end{align*}

Take the first and last ratios to solve a proportion to find the measure of angle \begin{align*}A\end{align*}.

\begin{align*}\frac{\sin C}{10}&=\frac{\sin 42^\circ}{8} \\ C&=\sin^{-1}\left(\frac{10 \sin 42^\circ}{8}\right) \approx 56.8^\circ\end{align*}

So, the \begin{align*}m \angle C \approx 56.8^\circ\end{align*} or \begin{align*}123.2^\circ\end{align*} and \begin{align*}m \angle B \approx 81.2^\circ\end{align*} or \begin{align*}14.8^\circ\end{align*} respectively.

Solve for the measure of side \begin{align*}b\end{align*} in each triangle:

\begin{align*}\frac{\sin 42^\circ}{8}&=\frac{\sin 81.2^\circ}{b} \qquad \qquad \qquad and \qquad \frac{\sin 42^\circ}{8}=\frac{\sin 14.8^\circ}{b}\\ b&=\frac{8 \sin 81.2^\circ}{\sin 42^\circ} \approx 11.8 \qquad \qquad \qquad \qquad \ b=\frac{8 \sin 14.8^\circ}{\sin 42^\circ} \approx 3.1\end{align*}

Putting it all together, we have:

Triangle 1: \begin{align*}m \angle A \approx 42^\circ, m \angle B \approx 81.2^\circ, m \angle C = 56.8^\circ, a=8, b \approx 11.8, c=10\end{align*}

Triangle 2: \begin{align*}m \angle A \approx 42^\circ, m \angle B \approx 14.8^\circ, m \angle C = 123.2^\circ, a=8, b \approx 3.1, c=10\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the other two angles and the remaining side given that a triangle has two sides of lengths 2 (a) and 5 (b) and the non-included angle (B) measures \begin{align*}45^\circ\end{align*}.

\begin{align*}\frac{\sin 45^\circ}{5}&=\frac{\sin A}{2} \\ \sin A&=\frac{2 \sin 45^\circ}{5} \\ A&=\sin^{-1}\left(\frac{2 \sin 45^\circ}{5}\right) \approx 16.4^\circ\end{align*}

Now find the third angle, \begin{align*}180^\circ - 45^\circ - 16.4^\circ=118.6^\circ\end{align*} and solve for the third side:

\begin{align*}\frac{\sin 45^\circ}{5}&=\frac{\sin 118.6^\circ}{c} \\ c&=\frac{5 \sin 118.6^\circ}{\sin 45^\circ} \approx 6.2\end{align*}

**Use the given side lengths and angle measure to determine whether zero, one or two triangles exists.**

#### Example 2

\begin{align*}m \angle A=100^\circ, a=3, b=4\end{align*}.

Since \begin{align*}A\end{align*} is obtused and \begin{align*}a \le b\end{align*}, no triangle can be formed.

#### Example 3

\begin{align*}m \angle A=50^\circ, a=8, b=10\end{align*}.

Since \begin{align*}A\end{align*} is acute, \begin{align*}a<b\end{align*} and \begin{align*}b \sin A<a\end{align*}, two triangles can be formed.

#### Example 4

\begin{align*}m \angle A=72^\circ, a=7, b=6\end{align*}.

Since \begin{align*}A\end{align*} is acute and \begin{align*}a>b\end{align*}, there is one possible triangle.

**Solve the following triangles.**

#### Example 5

There will be two triangles in this case because \begin{align*}A\end{align*} is acute, \begin{align*}a<b\end{align*} and \begin{align*}b \sin A<a\end{align*}.

Using the extended proportion: \begin{align*}\frac{\sin 25^\circ}{6}=\frac{\sin B}{8}=\frac{\sin C}{c}\end{align*}, we get:

\begin{align*}m \angle B &\approx 34.3^\circ \qquad or \qquad m \angle B \approx 145.7^\circ\\ m \angle C &\approx 120.7^\circ \qquad \qquad \ \ m \angle C \approx 9.3^\circ\\ c &\approx 12.2 \qquad \qquad \qquad \quad \ c \approx 2.3\end{align*}

#### Example 6

Using the extended proportion: \begin{align*}\frac{\sin 50^\circ}{15}=\frac{\sin B}{14}=\frac{\sin C}{c}\end{align*}, we get:

\begin{align*}m \angle B &\approx 45.6^\circ \\ m \angle C &\approx 84.4^\circ \\ c &\approx 19.5\end{align*}

#### Example 7

Given \begin{align*}m \angle A=30^\circ\end{align*}, \begin{align*}a=80\end{align*} and \begin{align*}b=150\end{align*}, find \begin{align*}m \angle C\end{align*}.

In this instance \begin{align*}A\end{align*} is acute, \begin{align*}a<b\end{align*} and \begin{align*}b \sin A<a\end{align*} so two triangles can be formed. So, once we find the two possible measures of angle \begin{align*}B\end{align*}, we will find the two possible measures of angle \begin{align*}C\end{align*}. First find \begin{align*}m \angle B\end{align*}:

\begin{align*}\frac{\sin 30^\circ}{80}&=\frac{\sin B}{150} \\ \sin B&=\frac{150 \sin 30^\circ}{80} \\ B & \approx 69.6^\circ, 110.4^\circ\end{align*}

Now that we have \begin{align*}B\end{align*}, use the triangle sum to find \begin{align*}m \angle C \approx 80.4^\circ, 39.9^\circ\end{align*}.

### Review

For problems 1-5, use the rules to determine if there will be one, two or no possible triangle with the given measurements.

- \begin{align*}m \angle A=65^\circ, a=10, b=11\end{align*}
- \begin{align*}m \angle A=25^\circ, a=8, b=15\end{align*}
- \begin{align*}m \angle A=100^\circ, a=6, b=4\end{align*}
- \begin{align*}m \angle A=75^\circ, a=25, b=30\end{align*}
- \begin{align*}m \angle A=48^\circ, a=41, b=50\end{align*}

Solve the following triangles, if possible. If there is a second possible triangle, solve it as well.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 13.13.