Law of Sines Ambiguous Case Lesson

In the last lesson, we learned how to solve oblique triangles involving SAA (side-angle-angle) or ASA (angle-side-angle) using the law of sines. For that situation, we are given two angles and one side.

Law of Sines

$$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}=\frac{c}{\text{sin}\hspace{.1em}C}$$ In some cases, we may need an alternative form: $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}B}{b}=\frac{\text{sin}\hspace{.1em}C}{c}$$

Applying the Law of Sines

• For any angle β of a triangle:
  • 0 < sin β ≤ 1
  • sin β = 1, β = 90°
  • sin β = sin(180° - β)
    • Supplementary angles have the same sine value
• Assuming the triangle has sides that are all of different lengths:
  • The smallest angle is opposite the shortest side
  • The largest angle is opposite the longest side
  • The middle-valued angle is opposite the intermediate side

Law of Sines - Ambiguous Case

When we are given the lengths of two sides and the angle opposite one of them SSA (side-side-angle), then zero, one, or two such triangles may exist. This creates a situation where we need to dig into the details of the problem.
Let's start with a triangle ABC where A is an acute angle. If we know the measure of the acute angle A, the length of side a (side opposite of angle A), and the length of side b, there will be 4 possible outcomes.

Case 1: No Solution

First, let's think about the situation where there are 0 possible triangles. This happens when the law of sines leads to: $$\text{sin}\hspace{.1em}B > 1, a < h < b$$ Note: $$h=b \cdot \text{sin}\hspace{.1em}A$$ Let's look at an example.
Example #1: Solve each triangle. Round your answer to the nearest tenth. $$A=61°, b=29\hspace{.1em}\text{cm}, a=20\hspace{.1em}\text{cm}$$ First, we can check to see if a is less than h. $$h=29 \hspace{.1em}\text{cm}\cdot \text{sin}\hspace{.1em}61°$$ $$h ≈ 25.4 \hspace{.1em}\text{cm}$$ We see that a is clearly less than h, therefore, no such triangle can exist. We can also show this using the law of sines to try and find B: $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}B}{b}$$ $$\frac{\text{sin}\hspace{.1em}61°}{20 \hspace{.1em}\text{cm}}=\frac{\text{sin}\hspace{.1em}B}{29 \hspace{.1em}\text{cm}}$$ $$\text{sin}\hspace{.1em}B=\frac{29 \hspace{.1em}\text{cm}\cdot \text{sin}\hspace{.1em}61°}{20 \hspace{.1em}\text{cm}}≈ 1.3$$ The sine of B can never be greater than 1, therefore, no such triangle will exist.

Case 2: Right Triangle

Second, we will think about the case, where we have a right triangle. This happens when the law of sines leads to: $$\text{sin}\hspace{.1em}B=1, a=h, h < b$$ Note: $$h=b \cdot \text{sin}\hspace{.1em}A$$ Let's look at an example.
Example #2: Solve each triangle. Round your answer to the nearest tenth. $$A=30°, c=2 \sqrt{5}\hspace{.1em}\text{in}, a=\sqrt{5}\hspace{.1em}\text{in}$$ Note: c will take the place of b in our formula above.
Again, we can use our formula to find h. $$h=2 \sqrt{5}\hspace{.1em}\text{in}\cdot \text{sin}\hspace{.1em}30°$$ $$h=2 \sqrt{5}\hspace{.1em}\text{in}\cdot \frac{1}{2}$$ $$h=2 \sqrt{5}\hspace{.1em}\text{in}\cdot \frac{1}{2}$$ $$h=\sqrt{5}\hspace{.1em}\text{in}$$ We can see that a and h are equal in value. Also, a is less than c. This tells us we have a right triangle. We can also show this using the law of sines. $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}C}{c}$$ $$\frac{\text{sin}\hspace{.1em}30°}{\sqrt{5}\hspace{.1em}\text{in}}=\frac{\text{sin}\hspace{.1em}C}{2\sqrt{5}\hspace{.1em}\text{in}}$$ $$\text{sin}\hspace{.1em}C=2\sqrt{5}\hspace{.1em}\text{in}\cdot \frac{\text{sin}\hspace{.1em}30°}{\sqrt{5}\hspace{.1em}\text{in}}=1$$ This tells us that C is a right angle or 90° angle. $$\text{sin}^{-1}(1)=90°$$ For angle B, we can use our triangle sum property: $$B=180° - 30° - 90°=60°$$ For side b, let's again turn to our law of sines: $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}$$ $$\frac{\sqrt{5}\hspace{.1em}\text{in}}{\text{sin}\hspace{.1em}30°}=\frac{b}{\text{sin}\hspace{.1em}60°}$$ $$b=\text{sin}\hspace{.1em}60° \cdot \frac{\sqrt{5}\hspace{.1em}\text{in}}{\text{sin}\hspace{.1em}30°}=\sqrt{15}\hspace{.1em}\text{in}$$

Case 3: Exactly One Non-Right Triangle

Third, we will think about the case, where we have exactly one solution, which is not a right triangle. This happens when the law of sines leads to: $$0 < \text{sin}\hspace{.1em}B < 1, a ≥ b$$ Let's look at an example.
Example #3: Solve each triangle. Round your answer to the nearest tenth. $$C=39°, b=28 \hspace{.1em}\text{in}, c=30 \hspace{.1em}\text{in}$$ Note: C will take the place of A and c will take the place of a in our formula above.
Here, c is larger than b. This tells us we will have one solution.
Let's use the law of sines to find angle B. $$\frac{\text{sin}\hspace{.1em}C}{c}=\frac{\text{sin}\hspace{.1em}B}{b}$$ $$\frac{\text{sin}\hspace{.1em}39°}{30 \hspace{.1em}\text{in}}=\frac{\text{sin}\hspace{.1em}B}{28 \hspace{.1em}\text{in}}$$ $$\text{sin}\hspace{.1em}B=28 \hspace{.1em}\text{in}\cdot \frac{\text{sin}\hspace{.1em}39°}{30 \hspace{.1em}\text{in}}$$ $$\text{sin}\hspace{.1em}B ≈ .58737$$ Let's use our inverse sine function. $$B ≈ \text{sin}^{-1}(.58737) ≈ 36°$$ We know from the concept of reference angles that there is another angle with a sine value of 0.58737. $$180° - 36°=144°$$ We know this isn't possible, if we add this to the measure given for angle C: $$39° + 144°=183°$$ The sum of all angles of a triangle is always 180°. So there isn't a second triangle that is possible. We could also note that in the triangle c > b, which means the measure of angle C must be larger than the measure of angle B. To find A and a, we can again turn to our law of sines. $$A ≈ 180° - 39° - 36°=105°$$ $$\frac{c}{\text{sin}\hspace{.1em}C}=\frac{a}{\text{sin}\hspace{.1em}A}$$ $$\frac{30 \hspace{.1em}\text{in}}{\text{sin}\hspace{.1em}39°}=\frac{a}{\text{sin}\hspace{.1em}105°}$$ $$a=\frac{30 \hspace{.1em}\text{in}\cdot \text{sin}\hspace{.1em}105°}{\text{sin}\hspace{.1em}39°}$$ $$a ≈ 46\hspace{.1em}\text{in}$$

Case 4: Exactly Two Triangles

Fourth, we will think about the case, where we have exactly two solutions or two triangles. This happens when the law of sines leads to: $$0 < \text{sin}\hspace{.1em}B_{1}< 1$$$$A + B_{2}< 180°$$$$h < a < b$$ $$h=b \cdot \text{sin}\hspace{.1em}A$$ Alternatively, we see another possible triangle of: Putting the two triangles together gives us: In the image above, c₁ is the length of the line segment AB₁, whereas, c₂ is the length of the line segment AB₂.
Let's look at an example.
Example #4: Solve each triangle. Round your answer to the nearest tenth. $$C=25°, b=14 \hspace{.1em}\text{km}, c=8\hspace{.1em}\text{km}$$ Note: C will take the place of A and c will take the place of a in our formula above.
$$h=14 \hspace{.1em}\text{km}\cdot \text{sin}\hspace{.1em}25°$$ $$h ≈ 5.9 \hspace{.1em}\text{km}$$ $$5.9 \hspace{.1em}\text{km}< 8\hspace{.1em}\text{km}< 14 \hspace{.1em}\text{km}$$ Let's use the law of sines to find angle B. $$\frac{\text{sin}\hspace{.1em}C}{c}=\frac{\text{sin}\hspace{.1em}B}{b}$$ $$\frac{\text{sin}\hspace{.1em}25°}{8 \hspace{.1em}\text{km}}=\frac{\text{sin}\hspace{.1em}B}{14\hspace{.1em}\text{km}}$$ $$\text{sin}\hspace{.1em}B=\frac{\text{sin}\hspace{.1em}25° \cdot 14\hspace{.1em}\text{km}}{8 \hspace{.1em}\text{km}}$$ $$\text{sin}\hspace{.1em}B ≈ 0.7396$$ $$B ≈ \text{sin}^{-1}(0.7369)$$ $$B ≈ 47.7°$$ Let's think about the other possible value for B here: $$180° - 47.7°=132.3°$$ Since 132.3° + 25° = 157.3° is less than 180°, it is a valid angle measure. This means we will need to provide two different solutions or triangles.
Triangle #1: $$A=180° - 25° - 47.7°=107.3°$$ $$C=25°, B=47.7°, A=107.3°$$ To find a, let's use the law of sines: $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{c}{\text{sin}\hspace{.1em}C}$$ $$\frac{a}{\text{sin}\hspace{.1em}107.3°}=\frac{8\hspace{.1em}\text{km}}{\text{sin}\hspace{.1em}25°}$$ $$a=\frac{8\hspace{.1em}\text{km}\cdot \text{sin}\hspace{.1em}107.3°}{\text{sin}\hspace{.1em}25°}$$ $$a ≈ 18.1 \hspace{.1em}\text{km}$$ $$c=8\hspace{.1em}\text{km}, b=14 \hspace{.1em}\text{km}, a ≈ 18.1 \hspace{.1em}\text{km}$$ Triangle #2: $$A=180° - 25° - 132.3°=22.7°$$ $$C=25°, B=132.3°, A=22.7°$$ To find a, let's use the law of sines: $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{c}{\text{sin}\hspace{.1em}C}$$ $$\frac{a}{\text{sin}\hspace{.1em}22.7°}=\frac{8\hspace{.1em}\text{km}}{\text{sin}\hspace{.1em}25°}$$ $$a=\frac{8\hspace{.1em}\text{km}\cdot \text{sin}\hspace{.1em}22.7°}{\text{sin}\hspace{.1em}25°}$$ $$a ≈ 7.3 \hspace{.1em}\text{km}$$ $$c=8\hspace{.1em}\text{km}, b=14 \hspace{.1em}\text{km}, a ≈ 7.3 \hspace{.1em}\text{km}$$

Law of Sines - Ambiguous Case (Obtuse Angle)

Let's now change things up and think about the case where A is now an obtuse angle. This will lead to two cases, one where there is no solution and another where there is exactly one solution.

Case 5: No Solution with a given obtuse angle

Fifth, let's think about the situation where the given angle is obtuse and there are 0 possible triangles. This happens when: $$a ≤ b$$ Let's look at an example.
Example #5: Solve each triangle. Round your answer to the nearest tenth. $$A=125°, c=18\hspace{.1em}\text{mi}, a=14 \hspace{.1em}\text{mi}$$ Since A is an obtuse angle, it is the largest angle, therefore, the longest side of the triangle must be a. Here, we see that c is greater than a, which is not possible. Therefore no such triangle ABC exists.
In this case, we can also show this with the law of sines. $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}C}{c}$$ $$\frac{\text{sin}\hspace{.1em}125°}{14 \hspace{.1em}\text{mi}}=\frac{\text{sin}\hspace{.1em}C}{18\hspace{.1em}\text{mi}}$$ $$\text{sin}\hspace{.1em}C=\frac{\text{sin}\hspace{.1em}125° \cdot 18\hspace{.1em}\text{mi}}{14 \hspace{.1em}\text{mi}}$$ $$\text{sin}\hspace{.1em}C ≈ 1.1$$ Since the sine of an angle can't be greater than 1, we know this triangle does not exist.

Case 6: One solution with a given obtuse angle

Sixth, we will think about the case, where we are given an obtuse angle and there is exactly one solution. This happens when the law of sines leads to: $$0 < \text{sin}\hspace{.1em}B < 1, a > b$$ Let's look at an example.
Example #6: Solve each triangle. Round your answer to the nearest tenth. $$A=114°, c=8\hspace{.1em}\text{ft}, a=35 \hspace{.1em}\text{ft}$$ Let's use the law of sines to find angle C. $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}C}{c}$$ $$\frac{\text{sin}\hspace{.1em}114°}{35 \hspace{.1em}\text{ft}}=\frac{\text{sin}\hspace{.1em}C}{8\hspace{.1em}\text{ft}}$$ $$\text{sin}\hspace{.1em}C=\frac{\text{sin}\hspace{.1em}114° \cdot 8\hspace{.1em}\text{ft}}{35 \hspace{.1em}\text{ft}}$$ $$\text{sin}\hspace{.1em}C ≈ 0.2088$$ $$C ≈ \text{sin}^{-1}(0.2088)$$ $$C ≈ 12.1°$$ $$B=180° - 114° - 12.1°=53.9°$$ Let's use the law of sines to find b. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}$$ $$\frac{35 \hspace{.1em}\text{ft}}{\text{sin}\hspace{.1em}114°}=\frac{b}{\text{sin}\hspace{.1em}53.9°}$$ $$b=\frac{35 \hspace{.1em}\text{ft}\cdot \text{sin}\hspace{.1em}53.9°}{\text{sin}\hspace{.1em}114°}$$ $$b ≈ 31 \hspace{.1em}\text{ft}$$

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