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# Ambiguous Case

## Evaluate triangles given two sides and one angle not between them

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The Ambiguous Case-SSA

A triangle has two sides of lengths 2 (a) and 5 (b). The non-included angle (B) of the triangle measures 45\begin{align*}45^\circ\end{align*}. What are possible measures for the two other angles and the remaining side?

### SSA

Recall that the sine ratios for an angle and its supplement will always be equal. In other words, sinθ=sin(180θ)\begin{align*}\sin \theta=\sin(180 - \theta)\end{align*}. In Geometry you learned that two triangles could not be proven congruent using SSA and you investigated cases in which there could be two triangles. In the first problem below, we will explore how the Law of Sines can be used to find two possible triangles when given two side lengths of a triangle and a non-included angle.

Given ΔABC\begin{align*}\Delta ABC\end{align*} with mA=30\begin{align*}m \angle A=30^\circ\end{align*}, a=5\begin{align*}a=5\end{align*}, and b=8\begin{align*}b=8\end{align*}, let's solve for the other angle and side measures.

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 C\begin{align*}C\end{align*}:

sin305sinCC=sinC8=8sin305=sin1(8sin305)53.1\begin{align*}\frac{\sin 30^\circ}{5}&=\frac{\sin C}{8} \\ \sin C&=\frac{8 \sin 30^\circ}{5} \\ C&=\sin^{-1}\left(\frac{8 \sin 30^\circ}{5}\right) \approx 53.1^\circ\end{align*}

From here we can find mA=96.9\begin{align*}m \angle A=96.9^\circ\end{align*}, since the three angles must add up to 180\begin{align*}180^\circ\end{align*}. We can also find the third side using another Law of Sines ratio:

sin305a=sin96.9a=5sin96.9sin309.9\begin{align*}\frac{\sin 30^\circ}{5}&=\frac{\sin 96.9^\circ}{a} \\ a&=\frac{5 \sin 96.9^\circ}{\sin 30^\circ} \approx 9.9\end{align*}

Putting these measures in the triangle, we get:

But, we know that sinθ=sin(180θ)\begin{align*}\sin \theta=\sin(180 -\theta)\end{align*} so when we solved for C\begin{align*}C\end{align*} we only got one of the two possible angles. The other angle will be 18053.1=126.9\begin{align*}180^\circ-53.1^\circ=126.9^\circ\end{align*}. Next we need to determine the measure of angle A\begin{align*}A\end{align*} for and the length of the third side in this second possible triangle. The sum of the three angles must still be 180\begin{align*}180^\circ\end{align*}, so mA=23.1\begin{align*}m \angle A=23.1^\circ\end{align*}. Now set up a proportion to solve for the third side just as before:

sin305a=sin23.1a=5sin23.1sin303.9\begin{align*}\frac{\sin 30^\circ}{5}&=\frac{\sin 23.1^\circ}{a} \\ a&=\frac{5 \sin 23.1^\circ}{\sin 30^\circ} \approx 3.9\end{align*}

The second triangle would look like this:

In this instance there were two possible triangles.

Now, given ΔABC\begin{align*}\Delta ABC\end{align*} with mB=80\begin{align*}m \angle B=80^\circ\end{align*}, a=5\begin{align*}a=5\end{align*} and b=7\begin{align*}b=7\end{align*}, let's solve for the other angle and side measures.

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

sin807sinAA=sinA5=5sin807=sin1(5sin807)44.7\begin{align*}\frac{\sin 80^\circ}{7}&=\frac{\sin A}{5} \\ \sin A&=\frac{5 \sin 80^\circ}{7} \\ A&=\sin^{-1}\left(\frac{5 \sin 80^\circ}{7}\right) \approx 44.7^\circ\end{align*}

Now find the third angle, 1808044.7=55.3\begin{align*}180^\circ - 80^\circ - 44.7^\circ=55.3^\circ\end{align*} and solve for the third side:

sin807c=sin55.3c=7sin55.3sin805.8\begin{align*}\frac{\sin 80^\circ}{7}&=\frac{\sin 55.3^\circ}{c} \\ c&=\frac{7 \sin 55.3^\circ}{\sin 80^\circ} \approx 5.8\end{align*}

Because we used the inverse sine function to determine the measure of angle A\begin{align*}A\end{align*}, the angle could be the supplement of 44.7\begin{align*}44.7^\circ\end{align*} or 135.3\begin{align*}135.3^\circ\end{align*} so we need to check for a second triangle. If we let mA=135.3\begin{align*}m \angle A=135.3^\circ\end{align*} and then attempt to find the third angle, we will find that the sum of the two angles we have is greater than 180\begin{align*}180^\circ\end{align*} and thus no triangle can be formed.

mA+mB+mC=180135.3+80+mC=180215.3+mC>180\begin{align*}m \angle A + m \angle B + m \angle C=180^\circ \\ 135.3^\circ + 80^\circ+m \angle C=180^\circ \\ 215.3^\circ+m \angle C>180^\circ\end{align*}

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 A\begin{align*}A\end{align*} is obtuse:

If a>b\begin{align*}a>b\end{align*}, then one triangle can be formed.

If ab\begin{align*}a \le b\end{align*}, then no triangle can be formed.

Now, consider the possible scenarios when A\begin{align*}A\end{align*} is acute.

If a>b\begin{align*}a>b\end{align*}, the one triangle can be formed.

For the following cases, where a<b\begin{align*}a, keep in mind that we would be using the proportion:

sinAa=sinBb\begin{align*}\frac{\sin A}{a}=\frac{\sin B}{b}\end{align*} and that \begin{align*}\sin B=\frac{b \sin A}{a}\end{align*}

If \begin{align*}b \sin A>a\end{align*}, no triangle can be formed because \begin{align*}B>1\end{align*}.

If \begin{align*}b \sin A=a\end{align*}, one right triangle can be formed because \begin{align*}\sin B=1\end{align*}.

If \begin{align*}b \sin A (and \begin{align*}a), two triangles can be formed because \begin{align*}\sin B<1\end{align*}.

Finally, given \begin{align*}\Delta ABC\end{align*} with \begin{align*}m \angle A=42^\circ\end{align*}, \begin{align*}b=10\end{align*} and \begin{align*}a=8\end{align*}, let's use the rules to determine how many, if any, triangles can be formed and then solve the possible triangle(s).

In this case, \begin{align*}A\end{align*} is acute and \begin{align*}a, 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 and \begin{align*}b \sin A, 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 and \begin{align*}b \sin A.

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 and \begin{align*}b \sin A 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.

1. \begin{align*}m \angle A=65^\circ, a=10, b=11\end{align*}
2. \begin{align*}m \angle A=25^\circ, a=8, b=15\end{align*}
3. \begin{align*}m \angle A=100^\circ, a=6, b=4\end{align*}
4. \begin{align*}m \angle A=75^\circ, a=25, b=30\end{align*}
5. \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.

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

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