In this explainer, we will learn how to use the rule of sines to solve SSA (side-side-angle) ambiguous triangles.
Letβs begin by recalling the law of sines.
Law: The Law of Sines
The law of sines applies to any triangle, equating the ratios of the sines of angles to corresponding side lengths.
That is,
We can use the law of sines to find unknown angle measures and lengths in triangles where we know two lengths and the measure of one opposite angle, or the measures of two angles and one opposite side.
To demonstrate how we can use the law of sines, letβs consider triangle with , , and , where we want to identify the measure of .
In this triangle, we have been given two sides and a nonincluded angle, so we begin by setting up the proportion of the opposite sides and the sines of their angles. That is,
Then, we solve for the unknown:
Therefore,
We can sketch the triangle as follows.
We can also use the law of sines to calculate an unknown length. For example, for a triangle , where , , and , we might want to find the length of to the nearest centimetre.
To do this, we can substitute our known values directly into the formula:
Therefore, to the nearest whole centimetre.
We can sketch this second triangle as follows.
Now, if we compare these two triangles, we notice that they have two corresponding sides that are equal and one corresponding angle that is equal. This is particularly interesting; had we tried to define the triangle by these two sides and this angle, we would have been in a position where we could have sketched either of these triangles. Therefore, the information would have been ambiguous.
This means that, in particular cases, the information that we are given to describe a triangle could be ambiguous, and we have to be careful when performing any subsequent calculations. Thankfully, these cases are well defined, so we can perform careful checks before performing any calculations to determine if more than one triangle could exist.
Definition: The Ambiguous Case of the Law of Sines
Using the law of sines to find an unknown length can give rise to an ambiguous answer due to the possibility of two solutions (namely, when you are given two side lengths and a nonincluded, acute angle). If angle is acute and , two possible triangles, and , exist.
The ambiguous case of the law of sines stems from the fact that two different angles can have the same sine value.
It is important to emphasize that this case may only occur when we are given two sides and a nonincluded angle, however, there are three possible outcomes that could occur from this case: no triangles exist, one triangle exists, or two triangles exist.
We have summarized the possible cases for two known side lengths, and , and a nonincluded angle, , in the table below.
Letβs explore some of these cases in our first few examples.
Example 1: Using the Law of Sines to Determine How Many Triangles Can Be Formed
For a triangle , , , and . How many triangles can be formed?
Answer
Here, we have been given two sides and a nonincluded angle, so we can use the law of sines, namely,
Substituting the values given in the question, we have that
Therefore,
However, this is not possible because the sine of an angle cannot be equal to 1.4; the largest value that the sine of an angle can have is 1.
Also, note that is acute, and if we calculate the height of this triangle, we see ; because this is larger than , we cannot draw this triangle. This is demonstrated in the following diagram.
No triangles can be formed.
Example 2: Using the Law of Sines to Determine How Many Measures an Angle May Take
For a triangle , , , and . Find all the possible measures of to the nearest degree.
Answer
We can actually clarify how many triangles exist from these measurements using the fact that is acute and the height of the triangle is . Then, as , this tells us that two possible triangles exist.
As we have been given two sides and a nonincluded angle, we may use the law of sines to solve this problem. That is,
Substituting the values given in the question, we have that
So, , to the nearest degree.
Now, because the sine of an obtuse angle also equals the sine of its supplement, there is another value for for which , so we must consider this. The supplementary angle for is .
We do need to check that this answer would make sense. We have that , so this is a possible value for angle .
Therefore, there are two possible measures for , and , so two triangles can be formed. These are as follows.
In a case where there is more than one value for the angle we are working out, a calculator will not give both possible values, so we must work out the alternative angle and then verify whether it is possible.
Example 3: Using the Law of Sines to Determine How Many Triangles Can Be Formed
For a triangle , , , and . How many triangles can be formed?
Answer
We firstly observe that is acute and that . Therefore, only one triangle exists for this problem.
We have not actually been asked to calculate the angle measures in this question, but we can confirm our previous result by doing so. We have been given two sides and a nonincluded angle, so we can use the law of sines, namely,
Substituting the values given in the question, we have that
Therefore,
Now, we check for the other angle that also has a sine equal to . That is, . However, we must check that this would actually make sense. If and , then these two angles alone add up to more than .
This confirms that there is only one possible triangle that can be formed.
These three examples have demonstrated the three outcomes that we can have when we deal with a case of the ambiguous law of sines. In summary, if we consider a triangle with height , we have the following:
- No triangles exist (such as when the sine value is out of range). This will occur when is acute and , or is obtuse and or .
- 1 triangle exists (the alternative angle is out of range). This will occur when is acute and or , or when is obtuse and .
- 2 triangles exist (the alternative angle is in range). This will occur when is acute and .
Example 4: Using the Law of Sines to Calculate All Possible Values of a Length in a Triangle
is a triangle, where , , and . Find all possible values of length giving the answer to three decimal places.
Answer
We can begin by drawing a quick sketch of what the triangle may look like to help visualize the problem.
We cannot directly use the law of sines to combine the ratios of the sine of angle and side with the sine of angle and side because we do not have a value for angle . However, if we first of all use then we can find angle :
Now, we need to check for the other angle that also has a sine equal to , that is, the supplement angle for :
This alternative angle for would also give a valid triangle because .
There are two possible triangles that we can draw.
The two possible values for angle will give us two possible values for angle . These are and
Therefore, we have found that the possible lengths of are and .
For our final example, letβs apply what we have learned about ambiguous triangles to, firstly, determine if a triangle is ambiguous and, secondly, calculate its possible perimeters.
Example 5: Finding the Perimeter for Each Solution of a Triangle Using the Law of Sines
is a triangle where , , and . Find all the possible values for the perimeter giving the answer to two decimal places.
Answer
First of all, note that is acute and that the βheightβ of this triangle (in this case, we consider it the line joining to ) is . Then, we have that , which tells us that there are two possible solutions to this problem.
Letβs draw a quick sketch of what this triangle may look like.
To find the perimeter, we would need the length of the side opposite angle . To find this length using the law of sines, we need the measure of angle first. To find this, we begin by finding the measure of angle :
Now, check for the other angle that also has a sine equal to , that is, the supplement angle for :
This gives us the two possible triangles.
The alternative triangle with is valid because .
This gives us two possible measures for angle : and
We can now use these values for angle to find the length opposite angle .
When ,
We can represent this using a diagram as follows.
When ,
We can represent our second triangle as follows.
In the first case, where and , the perimeter of the triangle is
In the second case, where and , the perimeter of the triangle is
Letβs finish by recalling some key points from this explainer.
Key Points
- For a triangle with height , there are 3
possible outcomes when using the law of sines with a side-side-angle (SSA) type of problem
that determine whether a triangle is ambiguous, is possible, or is unique:
- No triangles exist (such as when the sine value is out of range) when is acute and , or is obtuse and or .
- 1 triangle exists (the alternative angle is out of range) when is acute and or , or when is obtuse and .
- 2 triangles exist (the alternative angle is in range) when is acute and .
- Before starting any calculations to determine unknown measures in a triangle (when faced with an SSA problem), its advised to first check the number of possible triangles using the classifications outlined above. This can help avoid unnecessary calculations and provide insight into the calculations that need to be completed.
