Law of Sines
Solve any triangle when you know an angle and its opposite side
If you have ever looked at a triangle that does not have a right angle and thought, “How am I supposed to work with this?” you are not alone. The Pythagorean Theorem and basic trigonometry ratios are wonderful tools, but they only work when you have that convenient $90°$ angle to anchor everything. Most triangles in the real world do not cooperate like that. The good news? There is a beautiful formula called the Law of Sines that works for any triangle, right-angled or not. And here is the best part: if you already understand ratios and basic sine calculations, you already have everything you need to master this.
Surveyors use the Law of Sines to measure distances across rivers they cannot cross. Astronomers use it to calculate distances to stars billions of miles away. Navigation systems use it to pinpoint your location from cell tower signals. This one formula unlocks a whole world of triangle problems that seemed unsolvable before.
Core Concepts
Oblique Triangles: Life Beyond Right Angles
An oblique triangle is simply any triangle that does not contain a right angle. It might have all three angles less than $90°$ (an acute triangle), or it might have one angle greater than $90°$ (an obtuse triangle). Either way, the standard right-triangle tools (SOHCAHTOA, the Pythagorean Theorem) do not apply directly.
But do not let that discourage you. The Law of Sines gives us a powerful way to work with these triangles, as long as we know the right combination of parts.
The Law of Sines: One Formula to Rule Them All
In any triangle with angles $A$, $B$, and $C$, and opposite sides $a$, $b$, and $c$ respectively, the following relationship holds:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
This can also be written as:
$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$
Both forms say the same thing: in any triangle, the ratio of a side to the sine of its opposite angle is the same for all three side-angle pairs.
Think about what this means. If you know one side and its opposite angle, you have established the “magic ratio” for the entire triangle. Every other side-angle pair must follow the same ratio. That is incredibly powerful.
Important convention: In a triangle, we always label the angle with a capital letter ($A$, $B$, $C$) and the side opposite that angle with the corresponding lowercase letter ($a$, $b$, $c$). So side $a$ is always opposite angle $A$, side $b$ is opposite angle $B$, and side $c$ is opposite angle $C$.
When Does the Law of Sines Apply?
The Law of Sines works when you know certain combinations of parts. We classify these by which pieces of information you start with:
AAS (Angle-Angle-Side): You know two angles and a side that is NOT between them.
- Example: You know angles $A = 40°$, $B = 60°$, and side $a = 10$.
ASA (Angle-Side-Angle): You know two angles and the side that IS between them.
- Example: You know angles $A = 42°$, $C = 75°$, and side $b = 12$ (side $b$ is between angles $A$ and $C$).
SSA (Side-Side-Angle): You know two sides and an angle opposite one of them (not the angle between them).
- Example: You know sides $a = 10$, $b = 15$, and angle $A = 30°$.
For AAS and ASA cases, the Law of Sines always gives you exactly one solution. The SSA case is trickier, which is why it has earned the nickname “the ambiguous case.” More on that soon.
When does the Law of Sines NOT apply? When you have:
- SSS (three sides, no angles): Use the Law of Cosines instead
- SAS (two sides and the included angle): Use the Law of Cosines instead
Solving AAS Triangles
When you have two angles and a non-included side (AAS), solving the triangle is straightforward:
Step 1: Find the third angle using the fact that angles sum to $180°$. $$C = 180° - A - B$$
Step 2: Use the Law of Sines to find the unknown sides.
Since you know at least one complete side-angle pair (a side and its opposite angle), you can set up proportions to find everything else.
Solving ASA Triangles
ASA triangles work almost the same way. The only difference is which side you start with.
Step 1: Find the third angle. $$\text{Third angle} = 180° - \text{(sum of known angles)}$$
Step 2: Use the Law of Sines. You will need to identify which angle is opposite your known side, then build out from there.
The key insight: once you know all three angles and at least one side, you can find any other side using the Law of Sines.
The Ambiguous Case (SSA): When One Triangle Is Not Enough
Here is where things get interesting. When you know two sides and an angle opposite one of them (SSA), something strange can happen: the given information might describe zero triangles, exactly one triangle, or even two different triangles.
This is not a flaw in the mathematics. It is a genuine geometric reality. Think of it this way: if someone tells you two sides and an angle, there might be multiple ways to assemble those pieces into a valid triangle, or no way at all.
Let us say you know side $a$, side $b$, and angle $A$ (where $A$ is opposite side $a$). Here is how to figure out what you are dealing with:
Case 1: No triangle exists This happens when the given side opposite the known angle ($a$) is too short to reach the other side ($b$) and form a triangle.
If $A$ is acute and $a < b \sin A$, no triangle is possible. The side $a$ is simply too short.
If $A$ is obtuse and $a \leq b$, no triangle is possible either.
Case 2: Exactly one triangle exists
- If $A$ is acute and $a = b \sin A$, you get exactly one right triangle.
- If $A$ is acute and $a \geq b$, you get exactly one triangle.
- If $A$ is obtuse and $a > b$, you get exactly one triangle.
Case 3: Two triangles exist If $A$ is acute and $b \sin A < a < b$, two different triangles are possible.
How do two triangles arise? When you use the Law of Sines to find angle $B$, you get $\sin B = \text{some value}$. But here is the thing: if $\sin B = 0.5$, for example, then $B$ could be $30°$ or $150°$ (since sine is positive in both the first and second quadrants). Both values might lead to valid triangles.
Practical approach to SSA:
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Use the Law of Sines to find $\sin B$: $$\sin B = \frac{b \sin A}{a}$$
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Check if $\sin B > 1$. If so, no triangle exists (sine cannot exceed 1).
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If $\sin B \leq 1$, find angle $B$ using inverse sine: $B = \sin^{-1}(\sin B)$.
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Check if a second triangle is possible: Calculate $B’ = 180° - B$. Then check if $A + B’ < 180°$. If yes, the second triangle is valid.
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For each valid value of $B$, find $C = 180° - A - B$, then use the Law of Sines to find side $c$.
Area of a Triangle Using Sine
Here is a bonus that comes naturally from the Law of Sines. You can find the area of any triangle if you know two sides and the included angle:
$$\text{Area} = \frac{1}{2}ab \sin C$$
where $a$ and $b$ are two sides and $C$ is the angle between them.
This formula works because in any triangle, the height can be expressed using sine. If you drop a perpendicular from one vertex to the opposite side, the height is $h = b \sin C$ (or $a \sin B$, depending on how you set it up). Since area equals $\frac{1}{2} \times \text{base} \times \text{height}$, you get $\frac{1}{2} \times a \times b \sin C$.
This is extremely useful when you do not know the height directly but you do know the sides and angles.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| AAS | Two angles and non-included side | Know $A$, $B$, and side $a$ |
| ASA | Two angles and included side | Know $A$, $C$, and side $b$ (between $A$ and $C$) |
| SSA | Two sides and non-included angle | Know $a$, $b$, and angle $A$; the ambiguous case |
| Ambiguous case | SSA can give 0, 1, or 2 triangles | Must check all possibilities |
| Oblique triangle | Triangle with no right angle | Any acute or obtuse triangle |
| Opposite side | The side across from a given angle | Side $a$ is opposite angle $A$ |
| Included side | The side between two angles | Side $b$ is included between angles $A$ and $C$ |
| Included angle | The angle between two sides | Angle $C$ is included between sides $a$ and $b$ |
Examples
In triangle $ABC$, $A = 40°$, $B = 60°$, and $a = 10$. Find side $b$.
Solution:
We have two angles and the side opposite one of them. This is an AAS situation.
First, note that we already have a complete pair: angle $A = 40°$ and its opposite side $a = 10$. We want to find side $b$, which is opposite angle $B = 60°$.
Set up the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$
Substitute the known values: $$\frac{10}{\sin 40°} = \frac{b}{\sin 60°}$$
Solve for $b$: $$b = \frac{10 \cdot \sin 60°}{\sin 40°}$$
Calculate: $$b = \frac{10 \cdot 0.8660}{0.6428}$$ $$b = \frac{8.660}{0.6428} \approx 13.47$$
Side $b \approx 13.47$.
Find angle $B$ if $a = 8$, $A = 30°$, and $b = 12$.
Solution:
We know two sides and an angle opposite one of them. We want to find the angle opposite the other known side.
Use the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$
Substitute: $$\frac{8}{\sin 30°} = \frac{12}{\sin B}$$
Since $\sin 30° = 0.5$: $$\frac{8}{0.5} = \frac{12}{\sin B}$$ $$16 = \frac{12}{\sin B}$$
Solve for $\sin B$: $$\sin B = \frac{12}{16} = 0.75$$
Find angle $B$: $$B = \sin^{-1}(0.75) \approx 48.59°$$
Angle $B \approx 48.59°$.
(Note: Since this is an SSA situation, we should check for a second possible triangle. The other possible value is $B’ = 180° - 48.59° = 131.41°$. Check: $A + B’ = 30° + 131.41° = 161.41° < 180°$, so yes, a second triangle exists with $B \approx 131.41°$. For this problem, we found the acute solution.)
Solve the triangle with $A = 42°$, $B = 63°$, and $c = 18$.
Solution:
We have two angles and a side. Since side $c$ is opposite angle $C$ (the unknown angle), and the two known angles are $A$ and $B$, this is an ASA case.
Step 1: Find angle $C$. $$C = 180° - A - B = 180° - 42° - 63° = 75°$$
Step 2: Now we have a complete pair: $C = 75°$ and $c = 18$. Use the Law of Sines to find sides $a$ and $b$.
Find side $a$: $$\frac{a}{\sin A} = \frac{c}{\sin C}$$ $$\frac{a}{\sin 42°} = \frac{18}{\sin 75°}$$ $$a = \frac{18 \cdot \sin 42°}{\sin 75°} = \frac{18 \cdot 0.6691}{0.9659} \approx \frac{12.04}{0.9659} \approx 12.47$$
Find side $b$: $$\frac{b}{\sin B} = \frac{c}{\sin C}$$ $$\frac{b}{\sin 63°} = \frac{18}{\sin 75°}$$ $$b = \frac{18 \cdot \sin 63°}{\sin 75°} = \frac{18 \cdot 0.8910}{0.9659} \approx \frac{16.04}{0.9659} \approx 16.60$$
Complete solution:
- $A = 42°$, $B = 63°$, $C = 75°$
- $a \approx 12.47$, $b \approx 16.60$, $c = 18$
Determine how many triangles are possible with $a = 10$, $b = 15$, and $A = 30°$.
Solution:
This is an SSA case (two sides and a non-included angle), so we need to check for the ambiguous case.
Given: $a = 10$, $b = 15$, $A = 30°$
Step 1: Use the Law of Sines to find $\sin B$. $$\frac{a}{\sin A} = \frac{b}{\sin B}$$ $$\sin B = \frac{b \sin A}{a} = \frac{15 \cdot \sin 30°}{10} = \frac{15 \cdot 0.5}{10} = \frac{7.5}{10} = 0.75$$
Since $0 < 0.75 < 1$, a valid angle $B$ exists.
Step 2: Find the primary value of $B$. $$B = \sin^{-1}(0.75) \approx 48.59°$$
Step 3: Check for a second triangle. The supplementary angle is: $$B’ = 180° - 48.59° = 131.41°$$
Check if $A + B’ < 180°$: $$30° + 131.41° = 161.41° < 180°$$
Yes! Both values of $B$ are valid.
Conclusion: Two triangles are possible.
- Triangle 1: $B \approx 48.59°$, $C \approx 101.41°$
- Triangle 2: $B \approx 131.41°$, $C \approx 18.59°$
Given $a = 10$, $b = 15$, and $A = 30°$, solve both possible triangles completely.
Solution:
From Example 4, we found that $\sin B = 0.75$, giving us two possible values for $B$.
Triangle 1: $B_1 = 48.59°$
Find $C_1$: $$C_1 = 180° - 30° - 48.59° = 101.41°$$
Find side $c_1$ using the Law of Sines: $$\frac{c_1}{\sin C_1} = \frac{a}{\sin A}$$ $$c_1 = \frac{a \cdot \sin C_1}{\sin A} = \frac{10 \cdot \sin 101.41°}{\sin 30°}$$ $$c_1 = \frac{10 \cdot 0.9806}{0.5} = \frac{9.806}{0.5} = 19.61$$
Triangle 1 solution:
- $A = 30°$, $B \approx 48.59°$, $C \approx 101.41°$
- $a = 10$, $b = 15$, $c \approx 19.61$
Triangle 2: $B_2 = 131.41°$
Find $C_2$: $$C_2 = 180° - 30° - 131.41° = 18.59°$$
Find side $c_2$: $$c_2 = \frac{a \cdot \sin C_2}{\sin A} = \frac{10 \cdot \sin 18.59°}{\sin 30°}$$ $$c_2 = \frac{10 \cdot 0.3186}{0.5} = \frac{3.186}{0.5} = 6.37$$
Triangle 2 solution:
- $A = 30°$, $B \approx 131.41°$, $C \approx 18.59°$
- $a = 10$, $b = 15$, $c \approx 6.37$
Notice how different these two triangles are. Triangle 1 is fairly “spread out” with a large angle $C$, while Triangle 2 is more “squished” with angle $B$ being obtuse.
Find the area of a triangle with $b = 12$, $c = 15$, and $A = 52°$.
Solution:
We have two sides and the included angle (the angle between them). This is perfect for the area formula:
$$\text{Area} = \frac{1}{2}bc \sin A$$
Substitute the values: $$\text{Area} = \frac{1}{2}(12)(15) \sin 52°$$
Calculate: $$\text{Area} = \frac{1}{2}(180)(0.7880)$$ $$\text{Area} = 90 \cdot 0.7880$$ $$\text{Area} \approx 70.92 \text{ square units}$$
The area of the triangle is approximately $70.92$ square units.
Why this works: Imagine dropping a perpendicular from vertex $B$ to side $c$. The height $h$ of this perpendicular is $h = b \sin A = 12 \sin 52° \approx 9.46$. Then the area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(15)(9.46) \approx 70.92$, which matches our formula.
Key Properties and Rules
The Law of Sines
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
Or equivalently: $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$
When to Use the Law of Sines
| Given Information | Classification | Number of Solutions |
|---|---|---|
| Two angles and any side | AAS or ASA | Exactly 1 |
| Two sides and opposite angle | SSA | 0, 1, or 2 |
The Ambiguous Case (SSA) Decision Tree
Given sides $a$, $b$ and angle $A$ (opposite side $a$):
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Calculate $\sin B = \frac{b \sin A}{a}$
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If $\sin B > 1$: No triangle exists
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If $\sin B = 1$: One right triangle ($B = 90°$)
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If $\sin B < 1$:
- Find $B = \sin^{-1}(\sin B)$
- Find $B’ = 180° - B$
- If $A + B’ \geq 180°$: One triangle (only $B$ is valid)
- If $A + B’ < 180°$: Two triangles (both $B$ and $B’$ are valid)
Area Formula
$$\text{Area} = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B$$
Use any two sides and their included angle.
Key Relationships to Remember
- The largest side is always opposite the largest angle
- The smallest side is always opposite the smallest angle
- The sum of all angles in a triangle is exactly $180°$
- Sine values are always between $-1$ and $1$
Real-World Applications
Surveying Land
Surveyors frequently need to measure distances that cannot be measured directly, such as across a river, a canyon, or someone else’s property. Using the Law of Sines, they can set up two observation points on their side, measure the distance between them (the baseline), and measure the angles to a target point across the obstacle.
For example, a surveyor wants to find the distance across a river. They mark two points $A$ and $B$ on their side, 100 meters apart. From point $A$, the angle to a tree $C$ on the other side is $72°$. From point $B$, the angle to the same tree is $53°$. Using the Law of Sines, they can calculate the distance from either observation point to the tree without ever crossing the water.
Navigation and Maritime Distance
Ships at sea use the Law of Sines to calculate distances to landmarks, other ships, or ports. By taking bearing measurements from two different positions along their course, navigators can triangulate their position or the position of another object.
If a lighthouse is spotted at a bearing of $32°$ from a ship, and then again at a bearing of $58°$ from a point 5 miles further along the ship’s course, the Law of Sines can determine exactly how far the ship is from the lighthouse. This was crucial for maritime navigation before GPS existed, and it remains a valuable backup technique.
Astronomy and Stellar Parallax
When astronomers measure the distance to nearby stars, they use stellar parallax, which relies on the same principle as the Law of Sines. Earth’s orbit creates a baseline of about 186 million miles (the diameter of Earth’s orbit). By measuring the tiny angle shift of a star as seen from opposite sides of Earth’s orbit (six months apart), astronomers can calculate the star’s distance.
The angles involved are incredibly small (measured in fractions of an arcsecond), but the mathematics is the same Law of Sines you are learning now. It is remarkable that the same formula helps surveyors measure a river and astronomers measure the universe.
Cell Tower Triangulation
Your smartphone can determine your location by measuring signal strength from multiple cell towers. If the angles from two towers to your phone can be estimated, and the distance between towers is known, the Law of Sines helps calculate your position. This is one of several techniques used for location services, especially as a backup to GPS or in urban areas where GPS signals may be blocked by buildings.
Construction and Architecture
Architects and engineers use the Law of Sines when designing structures with non-right angles, such as roof trusses, bridges, and modern architectural features. When a roof has multiple slopes meeting at unusual angles, the Law of Sines helps calculate the exact lengths of beams needed.
Self-Test Problems
Problem 1: In triangle $ABC$, $A = 35°$, $C = 62°$, and $a = 7$. Find side $c$.
Show Answer
First, find angle $B$: $B = 180° - 35° - 62° = 83°$
Use the Law of Sines: $$\frac{a}{\sin A} = \frac{c}{\sin C}$$ $$\frac{7}{\sin 35°} = \frac{c}{\sin 62°}$$ $$c = \frac{7 \cdot \sin 62°}{\sin 35°} = \frac{7 \cdot 0.8829}{0.5736} \approx \frac{6.18}{0.5736} \approx 10.78$$
Side $c \approx 10.78$
Problem 2: Find angle $A$ if $a = 14$, $b = 10$, and $B = 42°$.
Show Answer
Use the Law of Sines: $$\frac{\sin A}{a} = \frac{\sin B}{b}$$ $$\sin A = \frac{a \cdot \sin B}{b} = \frac{14 \cdot \sin 42°}{10} = \frac{14 \cdot 0.6691}{10} = 0.9368$$
$$A = \sin^{-1}(0.9368) \approx 69.52°$$
(Note: The supplement $180° - 69.52° = 110.48°$ would give $A + B = 110.48° + 42° = 152.48°$, leaving $C = 27.52°$, which is valid. So there could be a second triangle with $A \approx 110.48°$.)
Primary answer: $A \approx 69.52°$
Problem 3: Determine how many triangles exist with $a = 6$, $b = 10$, and $A = 150°$.
Show Answer
Since $A = 150°$ is obtuse, we check: is $a > b$?
Here, $a = 6$ and $b = 10$, so $a < b$.
When the angle is obtuse and the side opposite it ($a$) is less than or equal to the other given side ($b$), no triangle is possible.
Alternatively, calculate $\sin B$: $$\sin B = \frac{b \sin A}{a} = \frac{10 \cdot \sin 150°}{6} = \frac{10 \cdot 0.5}{6} = \frac{5}{6} \approx 0.833$$
This gives $B \approx 56.44°$. But then $A + B = 150° + 56.44° = 206.44° > 180°$.
No triangle exists.
Problem 4: Solve the triangle: $A = 55°$, $B = 45°$, $c = 20$.
Show Answer
Step 1: Find angle $C$: $$C = 180° - 55° - 45° = 80°$$
Step 2: Find side $a$: $$\frac{a}{\sin A} = \frac{c}{\sin C}$$ $$a = \frac{c \cdot \sin A}{\sin C} = \frac{20 \cdot \sin 55°}{\sin 80°} = \frac{20 \cdot 0.8192}{0.9848} \approx 16.64$$
Step 3: Find side $b$: $$b = \frac{c \cdot \sin B}{\sin C} = \frac{20 \cdot \sin 45°}{\sin 80°} = \frac{20 \cdot 0.7071}{0.9848} \approx 14.36$$
Solution:
- $A = 55°$, $B = 45°$, $C = 80°$
- $a \approx 16.64$, $b \approx 14.36$, $c = 20$
Problem 5: Find the area of a triangle with sides $a = 8$, $b = 11$, and included angle $C = 67°$.
Show Answer
Use the area formula: $$\text{Area} = \frac{1}{2}ab \sin C$$ $$\text{Area} = \frac{1}{2}(8)(11) \sin 67°$$ $$\text{Area} = \frac{1}{2}(88)(0.9205)$$ $$\text{Area} = 44 \cdot 0.9205 \approx 40.50$$
The area is approximately $40.50$ square units.
Problem 6: Two angles of a triangle are $38°$ and $72°$. The side opposite the $38°$ angle is 15 units. Find the perimeter of the triangle.
Show Answer
Let $A = 38°$, $B = 72°$, and $a = 15$.
Step 1: Find angle $C$: $$C = 180° - 38° - 72° = 70°$$
Step 2: Find side $b$: $$b = \frac{a \cdot \sin B}{\sin A} = \frac{15 \cdot \sin 72°}{\sin 38°} = \frac{15 \cdot 0.9511}{0.6157} \approx 23.18$$
Step 3: Find side $c$: $$c = \frac{a \cdot \sin C}{\sin A} = \frac{15 \cdot \sin 70°}{\sin 38°} = \frac{15 \cdot 0.9397}{0.6157} \approx 22.90$$
Step 4: Calculate perimeter: $$P = a + b + c = 15 + 23.18 + 22.90 \approx 61.08$$
The perimeter is approximately $61.08$ units.
Problem 7: Given $a = 12$, $b = 8$, and $A = 40°$. Does this produce 0, 1, or 2 triangles? Explain your reasoning.
Show Answer
Calculate $\sin B$: $$\sin B = \frac{b \sin A}{a} = \frac{8 \cdot \sin 40°}{12} = \frac{8 \cdot 0.6428}{12} = \frac{5.14}{12} \approx 0.4285$$
Since $0 < 0.4285 < 1$, angle $B$ exists.
$$B = \sin^{-1}(0.4285) \approx 25.38°$$
Check the supplement: $B’ = 180° - 25.38° = 154.62°$
Check if second triangle is valid: $A + B’ = 40° + 154.62° = 194.62° > 180°$
Since $A + B’ > 180°$, the second triangle is not valid.
Exactly one triangle exists with $B \approx 25.38°$.
Another way to see this: since $a = 12 > b = 8$ (the side opposite the known angle is longer than the other given side), and $A$ is acute, there can only be one triangle.
Summary
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The Law of Sines states that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ for any triangle. This ratio is constant throughout the triangle.
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Use the Law of Sines when you have AAS (two angles and a non-included side) or ASA (two angles and the included side). These cases always produce exactly one triangle.
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The SSA case (two sides and a non-included angle) is called the ambiguous case because it can produce zero, one, or two valid triangles. Always check by computing $\sin B$ and testing whether the supplementary angle also produces a valid triangle.
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An oblique triangle is any triangle without a right angle. The Law of Sines handles these cases elegantly.
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The area formula $\text{Area} = \frac{1}{2}ab \sin C$ lets you find the area of any triangle when you know two sides and their included angle.
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In any triangle, the largest angle is always opposite the largest side, and the smallest angle is opposite the smallest side.
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Real-world applications include surveying (measuring inaccessible distances), navigation (determining positions at sea), astronomy (measuring stellar distances), and cell tower triangulation (locating mobile devices).
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Remember the labeling convention: angle $A$ is opposite side $a$, angle $B$ is opposite side $b$, and angle $C$ is opposite side $c$. This consistency makes setting up Law of Sines proportions much easier.
The Law of Sines is your first major tool for working with oblique triangles. It handles many situations beautifully, though for SSS and SAS cases, you will need its companion, the Law of Cosines. Together, these two laws let you solve any triangle problem you will encounter.