Consider a triangle with angle \(A = 40^\circ\text{,}\) side \(a = 7\text{,}\) and side \(b = 10\text{.}\) How many triangles are possible, and what are their dimensions?
Solution.
There is a systematic way to solve problems with the ambiguous case. While it’s not absolutely necessary, it helps to keep things organized. First, sketch the known angle (\(A = 40^\circ\)) in the bottom-left corner, with side \(b = 10\) above it, and the unknown side \(c\) horizontally. You should have something like this,
Notice that the bottom length \(c\) is unknown, so just extend it a bit to the right (we don’t know how long). Then, the other side \(a\) is opposite angle \(A\text{,}\) so it’ll go from the top point down to the right.
You should aim to sketch \(a\) so it’s a bit shorter than \(b\) (because 7 is smaller than 10), however, it doesn’t have to be perfect. Now, attach side \(a\) to point \(B\) and swing it down until it touches the base \(c\text{,}\)
Now, we have one triangle. However, this is not the only possibility! In fact, \(a\) can swing the other way to create a different triangle,
This setup also works with the given measurements. Therefore, there are 2 possible triangles with the given information.
Now, to solve both triangles, we can work on them side-by-side. On your paper, draw a vertical line down the middle. The left column is for Triangle 1 (where \(a\) swings to the right), and the right column is for Triangle 2 (where \(a\) swings to the left). We’ll do the same steps for both, and the only difference is whether we use the angle from inverse sine directly, or its supplement.
First, use the law of sines to find \(\sin{B}\text{.}\) This step is the same for both triangles,
\begin{align*}
\frac{7}{\sin{40^\circ}} \amp= \frac{10}{\sin{B}}\\
\sin{B} \amp= \frac{10 \sin{40^\circ}}{7} \approx 0.9165
\end{align*}
Now, here’s where the two triangles differ. Recall that \(\sin^{-1}(0.9165) \approx 66.7^\circ\text{,}\) but there are two angles with the same sine value: \(66.7^\circ\) and its supplement \(180^\circ - 66.7^\circ = 113.3^\circ\text{.}\) You can organize the 2 triangles nicely in 2 columns,
| Triangle 1 | Triangle 2 |
|---|---|
\(B_1 \approx 66.7^\circ\)
|
\(B_2 \approx 180^\circ - 66.7^\circ = 113.3^\circ\)
|
Find \(C_1\text{:}\)
|
Find \(C_2\text{:}\)
|
\(C_1 = 180^\circ - 40^\circ - 66.7^\circ\)
|
\(C_2 = 180^\circ - 40^\circ - 113.3^\circ\)
|
\(C_1 \approx 73.3^\circ\)
|
\(C_2 \approx 26.7^\circ\)
|
Find \(c_1\) using law of sines:
|
Find \(c_2\) using law of sines:
|
\(c_1 = \dfrac{7 \sin{73.3^\circ}}{\sin{40^\circ}} \approx 10.4\)
|
\(c_2 = \dfrac{7 \sin{26.7^\circ}}{\sin{40^\circ}} \approx 4.9\)
|
In summary,
