Ambiguous Case of the Sine Law


Slide Notes

Glossary

All Slides

Navigation

Obtuse Angles and Inverse Trigonometric Operations

Using a calculator, we find

 

Example 2

Determine the obtuse angle \(\theta\) with \(\sin(\theta)=\dfrac{1}{2}\).

 

Paused Finished
Slide /

Check Your Understanding 2


Part 1: null
How Did I Do?Try Another

Explore This 2


Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/m7wnxsad#


Explore This 2 Summary

In the activity above, you may have observed the following:

  • There are two distinct triangles, \(\triangle ABC\) and \(\triangle ABC'\), which can be constructed provided \(\angle C\) is not a right angle.

    Triangle ABC with C prime lying on AC such that BC prime is equal to BC. AB is equal to 10. BC is equal to 6. Angle A is equal to 30 degrees.

  • In \(\triangle ABC\), the angle opposite to \(AB\) is acute. In \(\triangle ABC'\), the angle opposite to \(AB\) is obtuse. In fact, these angles turn out to be supplementary.

Let's look now at what can happen when we overlook these observations while solving a triangle.

The Ambiguous Case of the Sine Law

Consider the triangle given in the diagram above with \(AB=10\), \(BC=6\), and \(\angle A = 30^{\circ}\). The first step in solving this triangle would be to use the sine law to get an equation for the sine of \(\angle C\).

\[ \frac{\sin(A)}{a} = \frac{\sin(C)}{c} \]

Rearranging and substituting known values, we get

\[ \begin{align*} \sin(C) &= \frac{c}{a} \cdot \sin(A) \\ &= \frac{10}{6} \cdot \sin(30^{\circ}) \\&= \frac{5}{6} \end{align*} \]

Next, we would use the inverse sine operation on a calculator to determine \(\angle C\).

\[ \sin(C) = \frac{5}{6} \quad \implies \quad \angle C = \sin^{-1}\left(\frac{5}{6}\right) \approx 56^{\circ}\]

This is where we must be careful. Our calculator ignores the possibility that \(\angle C\) is the obtuse angle supplementary to this value (i.e., \(180^{\circ} - 56^{\circ} = 124^{\circ}\)). There are actually two triangles that can be constructed using the given information. As such, we call this the ambiguous case of the sine law.

It is possible that two distinct oblique triangles can be constructed given two side lengths and one opposing angle. In this case, we say that the solution is ambiguous.

Triangle ABC with another line equal to BC shown. This line is connected from B to a point along AC.

Triangle ABC with another line equal to BC shown. This line is connected from B to a point along the extended line AC.

Let's return now to our Try This problem with the possibility of encountering ambiguous cases in mind.

Try This Revisited

In \(\triangle ABC\), determine obtuse \(\angle B\)  to the nearest degree.

Triangle ABC with angle A equal to 40 degrees, side AC equal to 8, and side BC equal to 6.

Solution

Observe that we are given two side lengths and one opposing angle.

  • \(BC = 6.0\)
  • \(AC = 8.0\)
  • \(\angle A = 40^{\circ}\)

Therefore, we should keep in mind the possibility that we will be able to construct two triangles consistent with these parameters.

To begin, let us apply the sine law to get an equation for \(\sin(B)\).

\[ \frac{\sin(A)}{BC} = \frac{\sin(B)}{AC} \]

Rearranging and substituting known values, we get

\[ \begin{align*} \sin(B) &= \frac{AC}{BC} \cdot \sin(A) \\ &= \frac{8.0}{6.0} \cdot \sin(40^{\circ}) \\ & \approx 0.857 \end{align*} \]

Using the inverse sine operation, we find

\[ \begin{align*} \angle B &= \sin^{-1}\left(0.857\right) \\ & \approx 59^{\circ} \end{align*} \]

At this point that we note that the supplementary angle \(180^{\circ} - 59^{\circ} = 121^{\circ}\) will also give a sine ratio of approximately \(0.857\). We have encountered the ambiguous case of the sine law. We illustrate the two cases below.

Triangle ABC with angle B equal to 121 degrees.

Triangle ABC with angle B equal to 59 degrees.

Fortunately, in this problem, we can remove the ambiguity. Recall that the question specifically states that \(\angle B\) is obtuse. This means that we can eliminate \(\angle B \approx 59^{\circ}\) as a possibility leaving us with \(\angle B \approx 121^{\circ}\).

In conclusion, \(\angle B\) is approximately equal to \(121^{\circ}\).


Check Your Understanding 3


Part 1: null
How Did I Do?Try Another

Recognizing the Ambiguous Case

If we are given side lengths and angles of a triangle, then only one triangle with those properties can be constructed if the given information includes

  • three side lengths, or
  • two side lengths and their contained angle, or
  • one side length and any two angles.

On the other hand, if the information we are given includes two side lengths and an angle that is not the angle contained by the two sides, then it is possible that two triangles with those properties can be constructed. How can we quickly determine if this is the case for a given triangle?

As an example, consider again the triangle given in the Try This problem with \(A = 40^{\circ}\), \(a = 6\), and \(b=8\) but suppose we are not told that \(\angle B\) is obtuse.

If we reorient this triangle so that the side whose length is not known acts as the base, then we can use the known angle, \(\angle A\), and the adjacent known side length, \(b\), to compute the height, \(h\).

\[\sin(A) = \frac{h}{b} \quad \implies \quad h = b \cdot \sin(A) = 8 \sin(40^{\circ}) \simeq 5.1\]

The length of \(BC\) is equal to \(6\) which is greater than this height. As such, \(BC\) is long enough to reach \(AB\) which means that at least one triangle could be formed. However, since the length of \(BC\) is also less than the length of \(AC\), it is possible for \(BC\) to meet up with \(AB\) in two different locations. In other words, two triangles can be formed with the given measurements.

Triangle 1

Note that angle b is obtuse and side AB is short compared to triangle 2.

Triangle 2

Note that angle b is acute and side AB is long compared to triangle 2.

Therefore, when solving this triangle, we should expect to encounter the ambiguous case of the sine law.

This procedure can be generalized to quickly recognize if one is dealing with the ambiguous case of the sine law or otherwise.


Explore This 3


Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/qvtwnmk8#


Explore This 3 Summary

From this activity, we observe that one can determine the number of triangles that can be formed with the vertices \(A\), \(B\), and \(C\) given lengths \(a\) and \(b\) and \(\angle A\) by completing the following steps:

  1. Reorient the triangle so that the unknown side, \(AB\), is the base. Observe that the known angle will be adjacent to this base.
  2. Use the sine ratio to compute the height of \(\triangle ABC\) in this orientation.\[\sin(A) = \frac{h}{b} \quad \implies \quad h = b \cdot \sin(A)\]
  3. Compare the length of the side opposite to the known angle, \(a\), to both the height of the triangle, \(h\), and the side length, \(b\).
    1. If \(0 \lt a \lt h\), then it is not possible to form a triangle at all.

    2. If \(a = h\), then \(BC\) would coincide with the height of the triangle in this orientation making \(\triangle ABC\) a right-angled triangle.

    3. If \(h \lt a \lt b\), then \(BC\) could meet \(AB\) in two places. In one case, \(\angle B\) would be acute and in the other case, it would be obtuse. This is the ambiguous case.

      \(\angle B\) Obtuse

      \(\angle B\) Acute

    4. If \(a \geq b\), then a single triangle would be formed.


Check Your Understanding 4


There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:
null

How Did I Do?Try Another