5.4: The Ambiguous Case
Learning Objectives
 Find possible triangles given two sides and an angle (SSA).
 Use the Law of Cosines and Sines in various ambiguous cases.
In previous sections, we learned about the Law of Cosines and the Law of Sines. We learned that we can use the Law of Cosines when:
 we know all three sides of a triangle (SSS) and
 we know two sides and the included angle (SAS).
We learned that we can use the Law of Sines when:
 we know two angles and a nonincluded side (AAS) and
 we know two angles and the included side (ASA).
However, we have not explored how to approach a triangle when we know two sides and a nonincluded angle (SSA). In this section, we will look at why the SSA case is called the ambiguous case, the possible triangles formed by the SSA case, and how to apply the Law of Sines and the Law of Cosines when we encounter the SSA case.
Possible Triangles with SSA
In Geometry, you learned that two sides and a nonincluded angle do not necessarily define a unique triangle. Consider the following cases given \begin{align*}a, b\end{align*}, and \begin{align*}\angle{A}\end{align*}:
Case 1: No triangle exists \begin{align*}(a < b)\end{align*}
In this case \begin{align*}a < b\end{align*} and side \begin{align*}a\end{align*} is too short to reach the base of the triangle. Since no triangle exists, there is no solution.
Case 2: One triangle exists \begin{align*}(a < b)\end{align*}
In this case, \begin{align*}a < b\end{align*} and side \begin{align*}a\end{align*} is perpendicular to the base of the triangle. Since this situation yields exactly one triangle, there is exactly one solution.
Case 3: Two triangles exist \begin{align*}(a < b)\end{align*}
In this case, \begin{align*}a < b\end{align*} and side \begin{align*}a\end{align*} meets the base at exactly two points. Since two triangles exist, there are two solutions.
Case 4: One triangle exists \begin{align*}(a = b)\end{align*}
In this case \begin{align*}a = b\end{align*} and side \begin{align*}a\end{align*} meets the base at exactly one point. Since there is exactly one triangle, there is one solution.
Case 5: One triangle exists \begin{align*}(a > b)\end{align*}
In this case, \begin{align*}a > b\end{align*} and side \begin{align*}a\end{align*} meets the base at exactly one point. Since there is exactly one triangle, there is one solution.
Case 3 is referred to as the Ambiguous Case because there are two possible triangles and two possible solutions. One way to check to see how many possible solutions (if any) a triangle will have is to compare sides \begin{align*}a\end{align*} and \begin{align*}b\end{align*}. If you are faced with the first situation, where \begin{align*}a < b\end{align*}, we can still tell how many solutions there will be by using \begin{align*}a\end{align*} and \begin{align*}b \sin A\end{align*}.
If:  Then:  

a.  \begin{align*}a < b\end{align*}  No solution, one solution, two solutions 
i.  \begin{align*}a < b \sin A\end{align*}  No solution 
ii.  \begin{align*}a = b \sin A\end{align*}  One solution 
iii.  \begin{align*}a > b \sin A\end{align*}  Two solutions 
b.  \begin{align*}a = b\end{align*}  One solution 
c.  \begin{align*}a > b\end{align*}  One solution 
Example 1: Determine if the sides and angle given determine no, one or two triangles. All sets contain an angle, its opposite side and the side between them.
a. \begin{align*}a = 5, b = 8, A = 62.19^\circ\end{align*}
b. \begin{align*}c = 14, b = 10, B = 15.45^\circ\end{align*}
c. \begin{align*}d = 16, g = 11, D = 44.94^\circ\end{align*}
d. \begin{align*}a = 9, b = 7, B = 51.06^\circ\end{align*}
Solution: Even though \begin{align*}a, b\end{align*} and \begin{align*}\angle{A}\end{align*} are not used in every example, follow the same pattern from the table by multiplying the nonopposite side (of the angle) by the angle.
a. \begin{align*}5 < 8, 8 \sin 62.19^\circ = 7.076\end{align*}. So \begin{align*}5 < 7.076\end{align*}, which means there is no solution.
b. \begin{align*}10 < 14, 14 \sin 15.45^\circ = 3.73\end{align*}. So \begin{align*}10 > 3.73\end{align*}, which means there are two solutions.
c. \begin{align*}16 > 11\end{align*}, there is one solution.
d. \begin{align*}7 < 9, 9 \sin 51.06^\circ = 7.00\end{align*}. So \begin{align*}7 = 7\end{align*}, which means there is one solution.
In the next two sections we will look at how to use the Law of Cosines and the Law of Sines when faced with the various cases above.
Using the Law of Sines
In triangle ABC below, we know two sides and a nonincluded angle. Remember that the Law of Sines states: \begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b}\end{align*}. Since we know \begin{align*}a, b\end{align*}, and \begin{align*}\angle{A}\end{align*}, we can use the Law of Sines to find \begin{align*}\angle{B}\end{align*}. However, since this is the SSA case, we have to watch out for the Ambiguous case. Since \begin{align*}a < b\end{align*}, we could be faced with either Case 1, Case 2, or Case 3 above.
Example 2: Find \begin{align*}\angle{B}\end{align*}.
Solution: Use the Law of Sines to determine the angle.
\begin{align*}\frac{\sin 41}{12} & = \frac{\sin B}{23} \\ 23 \sin 41 & = 12 \sin B \\ \frac{23 \sin 41}{12} & = \sin B \\ 1.257446472 & = \sin B\end{align*}
Since no angle exists with a sine greater than 1, there is no solution to this problem.
We also could have compared \begin{align*}a\end{align*} and \begin{align*}b \sin A\end{align*} beforehand to see how many solutions there were to this triangle.
\begin{align*}a = 12, b \sin A = 15.1\end{align*}: since \begin{align*}12 < 15.1, a < b \sin A\end{align*} which tells us there are no solutions.
Example 3: In triangle \begin{align*}ABC, a = 15, b = 20\end{align*}, and \begin{align*}\angle{A} = 30^\circ\end{align*}. Find \begin{align*}\angle{B}\end{align*}.
Solution: Again in this case, \begin{align*}a < b\end{align*} and we know two sides and a nonincluded angle. By comparing \begin{align*}a\end{align*} and \begin{align*}b \sin A\end{align*}, we find that \begin{align*}a = 15, b \sin A = 10\end{align*}. Since \begin{align*}15 > 10\end{align*} we know that there will be two solutions to this problem.
\begin{align*}\frac{\sin 30}{15} & = \frac{\sin B}{20} \\ 20 \sin 30 & = 15 \sin B \\ \frac{20 \sin 30}{15} & = \sin B \\ 0.6666667 & = \sin B \\ \angle{B} & = 41.8^\circ\end{align*}
There are two angles less than \begin{align*}180^\circ\end{align*} with a sine of 0.6666667, however. We found the first one, \begin{align*}41.8^\circ\end{align*}, by using the inverse sine function. To find the second one, we will subtract \begin{align*}41.8^\circ\end{align*} from \begin{align*}180^\circ, \angle{B} = 180^\circ  41.8^\circ = 138.2^\circ\end{align*}.
To check to make sure \begin{align*}138.2^\circ\end{align*} is a solution, we will use the Triangle Sum Theorem to find the third angle. Remember that all three angles must add up to \begin{align*}180^\circ\end{align*}.
\begin{align*}180^\circ  (30^\circ + 41.8^\circ) = 108.2^\circ && \text{or} && 180^\circ  (30^\circ + 138.2^\circ) = 11.8^\circ\end{align*}
This problem yields two solutions. Either \begin{align*}\angle{B} = 41.8^\circ\end{align*} or \begin{align*}138.2^\circ\end{align*}.
Example 4: A boat leaves lighthouse \begin{align*}A\end{align*} and travels 63km. It is spotted from lighthouse \begin{align*}B\end{align*}, which is 82km away from lighthouse \begin{align*}A\end{align*}. The boat forms an angle of \begin{align*}65.1^\circ\end{align*} with both lighthouses. How far is the boat from lighthouse \begin{align*}B\end{align*}?
Solution: In this problem, we again have the SSA angle case. In order to find the distance from the boat to the lighthouse (a) we will first need to find the measure of angle \begin{align*}A\end{align*}. In order to find angle \begin{align*}A\end{align*}, we must first use the Law of Sines to find angle \begin{align*}B\end{align*}. Since \begin{align*}c > b\end{align*}, this situation will yield exactly one answer for the measure of angle \begin{align*}B\end{align*}.
\begin{align*}\frac{\sin 65.1^\circ}{82} & = \frac{\sin B}{63} \\ \frac{63 \sin 65.1^\circ}{82} & = \sin B \\ 0.6969 & \approx \sin B \\ \angle{B} & = 44.2^\circ\end{align*}
Now that we know the measure of angle \begin{align*}B\end{align*}, we can find the measure of angle \begin{align*}A, \angle{A} = 180^\circ  65.1^\circ  44.2^\circ = 70.7^\circ\end{align*}. Finally, we can use \begin{align*}\angle{A}\end{align*} to find side \begin{align*}a\end{align*}.
\begin{align*}\frac{\sin 65.1^\circ}{82} & = \frac{\sin 70.7^\circ}{a} \\ \frac{82 \sin 70.7^\circ}{\sin 65.1^\circ} & = a \\ a & = 85.3 \end{align*}
The boat is approximately 85.3 km away from lighthouse \begin{align*}B\end{align*}.
Using the Law of Cosines
Example 5: In a game of pool, a player must put the eight ball into the bottom left pocket of the table. Currently, the eight ball is 6.8 feet away from the bottom left pocket. However, due to the position of the cue ball, she must bank the shot off of the right side bumper. If the eight ball is 2.1 feet away from the spot on the bumper she needs to hit and forms a \begin{align*}168^\circ\end{align*} angle with the pocket and the spot on the bumper, at what angle does the ball need to leave the bumper?
Note: This is actually a trick shot performed by spinning the eight ball, and the eight ball will not actually travel in straightline trajectories. However, to simplify the problem, assume that it travels in straight lines.
Solution: In the scenario above, we have the SAS case, which means that we need to use the Law of Cosines to begin solving this problem. The Law of Cosines will allow us to find the distance from the spot on the bumper to the pocket \begin{align*}(y)\end{align*}. Once we know \begin{align*}y\end{align*}, we can use the Law of Sines to find the angle \begin{align*}(X)\end{align*}.
\begin{align*}y^2 & = 6.8^2 + 2.1^2  2(6.8)(2.1) \cos 168^\circ \\ y^2 & = 78.59 \\ y & = 8.86\ feet\end{align*}
The distance from the spot on the bumper to the pocket is 8.86 feet. We can now use this distance and the Law of Sines to find angle \begin{align*}X\end{align*}. Since we are finding an angle, we are faced with the SSA case, which means we could have no solution, one solution, or two solutions. However, since we know all three sides this problem will yield only one solution.
\begin{align*}\frac{\sin 168^\circ}{8.86} & = \frac{\sin X}{6.8} \\ \frac{6.8 \sin 168^\circ}{8.86} & = \sin X \\ 0.1596 & \approx \sin B \\ \angle{B} & = 8.77^\circ\end{align*}
In the previous example, we looked at how we can use the Law of Sines and the Law of Cosines together to solve a problem involving the SSA case. In this section, we will look at situations where we can use not only the Law of Sines and the Law of Cosines, but also the Pythagorean Theorem and trigonometric ratios. We will also look at another realworld application involving the SSA case.
Example 6: Three scientists are out setting up equipment to gather data on a local mountain. Person 1 is 131.5 yards away from Person 2, who is 67.8 yards away from Person 3. Person 1 is 72.6 yards away from the mountain. The mountains forms a \begin{align*}103^\circ\end{align*} angle with Person 1 and Person 3, while Person 2 forms a \begin{align*}92.7^\circ\end{align*} angle with Person 1 and Person 3. Find the angle formed by Person 3 with Person 1 and the mountain.
Solution: In the triangle formed by the three people, we know two sides and the included angle (SAS). We can use the Law of Cosines to find the remaining side of this triangle, which we will call \begin{align*}x\end{align*}. Once we know \begin{align*}x\end{align*}, we will two sides and the nonincluded angle (SSA) in the triangle formed by Person 1, Person 2, and the mountain. We will then be able to use the Law of Sines to calculate the angle formed by Person 3 with Person 1 and the mountain, which we will refer to as \begin{align*}Y\end{align*}.
To find \begin{align*}x\end{align*}:
\begin{align*}x^2 & = 131.5^2 + 67.8^2 2(131.5)(67.8) \cos 92.7 \\ x^2 & = 22729.06397 \\ x & = 150.8\ yds\end{align*}
Now that we know \begin{align*}x = 150.8\end{align*}, we can use the Law of Sines to find \begin{align*}Y\end{align*}. Since this is the SSA case, we need to check to see if we will have no solution, one solution, or two solutions. Since \begin{align*}150.8 > 72.6\end{align*}, we know that we will have only one solution to this problem.
\begin{align*}\frac{\sin 103}{150.8} & = \frac{\sin Y}{72.6} \\ \frac{72.6 \sin 103}{150.8} & = \sin Y \\ 0.4690932805 & = \sin Y \\ 28.0 & \approx \angle{Y}\end{align*}
Points to Consider
 Why is there only one possible solution to the SSA case if \begin{align*}a > b\end{align*}?
 Explain why \begin{align*}b > a > b \sin A\end{align*} yields two possible solutions to a triangle.
 If we have a SSA angle case with two possible solutions, how can we check both solutions to make sure they are correct?
Review Questions
 Using the table below, determine how many solutions there would be to each problem based on the given information and by calculating \begin{align*}b \sin A\end{align*} and comparing it with \begin{align*}a\end{align*}. Sketch an approximate diagram for each problem in the box labeled “diagram.”
Given  \begin{align*}a >, =\end{align*}, or \begin{align*}< b \sin A\end{align*}  Diagram  Number of solutions 

a. \begin{align*}A = 32.5^\circ, a = 26, b = 37\end{align*}  
b. \begin{align*}A = 42.3^\circ, a = 16, b = 26\end{align*}  
c. \begin{align*}A = 47.8^\circ, a = 13.48,b = 18.2\end{align*}  
d. \begin{align*}A = 51.5^\circ, a = 3.4,b = 4.2\end{align*} 
 Using the information in the table above, find all possible measures of angle \begin{align*}B\end{align*} if any exist.
 Prove using the Law of Sines: \begin{align*}\frac{ac}{c} = \frac{\sin A  \sin C}{\sin C}\end{align*}
 Give the measure of a nonincluded angle and the lengths of two sides so that two triangles exist. Explain why two triangles exist for the measures you came up with.
 If \begin{align*}a = 22\end{align*} and \begin{align*}b = 31\end{align*}, find the values of \begin{align*}A\end{align*} so that:
 There is no solution
 There is one solution
 There are two solutions
 In the figure below, \begin{align*}AB = 13.7, AD = 9.8\end{align*}, and \begin{align*}\angle{C} = 42.6^\circ\end{align*}. Find \begin{align*}\angle{A}, \angle{B}, BC\end{align*}, and \begin{align*}AC\end{align*}.
 In the figure below, \begin{align*}\angle{C} = 21.8^\circ, BE = 9.9, BD = 10.2, ED = 7.6\end{align*}, and \begin{align*}\angle{ABC} = 109.6^\circ\end{align*}. Find the following:
 \begin{align*}\angle{EBD}\end{align*}
 \begin{align*}\angle{BDE}\end{align*}
 \begin{align*}\angle{DEB}\end{align*}
 \begin{align*}\angle{BDC}\end{align*}
 \begin{align*}\angle{BEA}\end{align*}
 \begin{align*}\angle{DBC}\end{align*}
 \begin{align*}\angle{ABE}\end{align*}
 \begin{align*}\angle{BAE}\end{align*}
 BC
 AB
 AE
 DC
 AC
 Radio detection sensors for tracking animals have been placed at three different points in a wildlife preserve. The distance between Sensor 1 and Sensor 2 is 4500ft. The distance between Sensor 1 and Sensor 3 is 4000ft. The angle formed by Sensor 3 with Sensors 1 and 2 is \begin{align*}56^\circ\end{align*}. If the range of Sensor 3 is 6000ft, will it be able to detect all movement from its location to the location of Sensor 2?
 In problem 8 above, a fourth sensor is placed in the wildlife preserve. Sensor 2 forms a 36 angle with Sensors 3 and 4, and Sensor 3 forms a 49 angle with Sensors 2 and 4. How far away is Sensor 4 from Sensors 2 and 3?
Review Answers
Given  \begin{align*}a >, =\end{align*}, or \begin{align*}< b \sin A\end{align*}  Diagram  Number of solutions 

a. \begin{align*}A = 32.5^\circ, a = 26, b = 37\end{align*}  \begin{align*}26 > 19.9\end{align*}  2  
b. \begin{align*}A = 42.3^\circ, a = 16, b = 26\end{align*}  \begin{align*}16 < 17.5\end{align*}  0  
c. \begin{align*}A = 47.8^\circ, a = 13.48,b = 18.2\end{align*}  \begin{align*}13.48 = 13.48\end{align*}  1  
d. \begin{align*}A = 51.5^\circ, a = 3.4,b = 4.2\end{align*}  \begin{align*}3.4 > 3.3\end{align*}  2 

 \begin{align*}\frac{\sin 32.5^\circ}{26} = \frac{\sin B}{37} \rightarrow B = 49.9^\circ\end{align*} or \begin{align*}180^\circ  49.9^\circ = 130.1^\circ\end{align*}
 no solution
 \begin{align*}\frac{\sin 47.8^\circ}{13.48} = \frac{\sin B}{18.2} \rightarrow B = 90^\circ\end{align*}
 \begin{align*}\frac{\sin 51.5^\circ}{3.4} = \frac{\sin B}{4.2} \rightarrow B = 75.2^\circ\end{align*} or \begin{align*}180^\circ  75.2^\circ = 104.8^\circ\end{align*}
 \begin{align*}\frac{\sin A}{a} & = \frac{\sin C}{c} \\ c \sin A & = a \sin C \\ c \sin A  c \sin C & = a \sin C  c \sin C \\ c (\sin A  \sin C) & = \sin C (ac) \\ \frac{\sin A  \sin C}{\sin C} & = \frac{ac}{c}\end{align*}
 Student answers will vary. Student should mention using \begin{align*}a > b \sin A\end{align*} in their explanation.
 The answers can be determined as follows:
 \begin{align*}a < b \sin A \rightarrow \frac{a}{b} < \sin A \rightarrow \frac{22}{31} < \sin A \rightarrow A > 45.2^\circ\end{align*}
 \begin{align*}a = b \sin A \rightarrow \frac{a}{b} = \sin A \rightarrow \frac{22}{31} = \sin A \rightarrow A = 45.2^\circ\end{align*}
 \begin{align*}a > b \sin A \rightarrow \frac{a}{b} > \sin A \rightarrow \frac{22}{31} > \sin A \rightarrow A < 45.2^\circ\end{align*}
 This problem can be done entirely with right triangle trig, but there are several different ways to solve this particular problem. \begin{align*}BD & = \sqrt{13.7^2  9.8^2} = 9.6 && \tan 42.6^\circ = \frac{9.8}{DC} \rightarrow DC = 10.7 \\
\sin 42.6^\circ & = \frac{9.8}{AC} \rightarrow AC = 14.5 && BC = 9.6 + 10.7 = 20.3 \\
\sin B & = \frac{9.8}{13.7} \rightarrow \angle{B} = 45.7^\circ && \angle{A} = 180^\circ  45.7^\circ  42.6^\circ = 91.7^\circ\end{align*}
 \begin{align*}\angle{EBD} \Rightarrow 7.6^2 = 9.9^2 + 10.2^2  2 \cdot 9.9 \cdot 10.2 \cos EBD \Rightarrow 44.4^\circ\end{align*}
 \begin{align*}\angle{BDE} \Rightarrow \frac{\sin BDE}{9.9} = \frac{\sin 44.4^\circ}{7.6} \Rightarrow 65.7^\circ\end{align*}
 \begin{align*}\angle{DEB} \Rightarrow 180^\circ  65.7^\circ  44.4^\circ \Rightarrow 69.9^\circ\end{align*}
 \begin{align*}\angle{BDC} \Rightarrow 180^\circ  65.7^\circ \Rightarrow 114.3^\circ\end{align*}
 \begin{align*}\angle{BEA} \Rightarrow 180^\circ  69.9^\circ \Rightarrow 110.1^\circ\end{align*}
 \begin{align*}\angle{DBC} \Rightarrow 180^\circ  114.3^\circ  21.8^\circ \Rightarrow 43.9^\circ\end{align*}
 \begin{align*}\angle{ABE} \Rightarrow 109.6^\circ  43.9^\circ  44.4^\circ \Rightarrow 21.3^\circ\end{align*}
 \begin{align*}\angle{BAE} \Rightarrow 180^\circ  21.3^\circ  110.1^\circ \Rightarrow 48.6^\circ\end{align*}
 \begin{align*}BC \Rightarrow \frac{\sin 114.3^\circ}{BC} = \frac{\sin 21.8^\circ}{10.2} \Rightarrow 25.0\end{align*}
 \begin{align*}AB \Rightarrow \frac{\sin 110.1^\circ}{AB} = \frac{\sin 48.6^\circ}{9.9} \Rightarrow 12.4\end{align*}
 \begin{align*}AE \Rightarrow \frac{\sin 21.3^\circ}{AE} = \frac{\sin 48.6^\circ}{9.9} \Rightarrow 4.8\end{align*}
 \begin{align*}DC \Rightarrow \frac{\sin 43.9^\circ}{DC} = \frac{\sin 21.8^\circ}{9.9} \Rightarrow 19.0\end{align*}
 \begin{align*}AC = 19 + 4.8 + 7.6 = 31.4\end{align*}
 We need to find the distance between sensors 2 and 3. If it is less than 6000 ft, then the sensor will be able to detect all motion between the two. First, \begin{align*}4500 > 4000\end{align*}, so there is going to be one solution. \begin{align*}\frac{\sin S 2}{4000} = \frac{\sin 56^\circ}{4500} \rightarrow S2 = 47.47^\circ && S1 + 47.47^\circ + 56^\circ = 180^\circ \rightarrow S1 = 76.53^\circ && \frac{\sin 76.53^\circ}{x} = \frac{\sin 56^\circ}{4500} \rightarrow x = 5279\end{align*} Since \begin{align*}x < 6000\end{align*} ft., Sensor 3 will be able to pick up all movement from its location to the location of Sensor 2.
 The length from \begin{align*}S3\end{align*} to \begin{align*}S4\end{align*} is \begin{align*}x\end{align*} and from \begin{align*}S4\end{align*} to \begin{align*}S2\end{align*} is \begin{align*}y\end{align*}. \begin{align*}180^\circ  36^\circ  49^\circ = 95^\circ\end{align*}, which is the angle at \begin{align*}S4\end{align*}. \begin{align*}\frac{\sin 95^\circ}{5279} = \frac{\sin 36^\circ}{x} = \frac{\sin 49^\circ}{y} && x = 3114.8, y = 3999.3\end{align*}
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