Law of Sines and Law of Cosines
Solve any triangle, not just right triangles
If you have spent time with right triangles, you know how powerful SOH-CAH-TOA can be. Opposite over hypotenuse, adjacent over hypotenuse, opposite over adjacent; these ratios let you find missing sides and angles with ease. But here is the catch: most triangles you encounter in the real world are not right triangles. Architects designing roof pitches, surveyors mapping property lines, pilots calculating flight paths; they all work with triangles that have no 90-degree angle in sight.
So what do you do when there is no right angle to anchor your calculations? This is exactly where the Law of Sines and the Law of Cosines come to the rescue. These two tools extend your triangle-solving abilities to every triangle imaginable. By the end of this lesson, you will be able to look at any triangle, identify what information you have, and choose the right method to find what you need.
Core Concepts
The Setup: Labeling a Triangle
Before diving into the laws themselves, let us establish a naming convention. In any triangle, we label the vertices (corners) with capital letters A, B, and C. The side opposite each vertex gets the corresponding lowercase letter:
- Side a is opposite angle A
- Side b is opposite angle B
- Side c is opposite angle C
This labeling system is not arbitrary; it is essential for the Law of Sines and Law of Cosines because these laws depend on the relationship between each angle and its opposite side.
The Law of Sines
The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
You can also write this with the sines in the numerator:
$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$
Both forms are equivalent; use whichever is more convenient for your problem. The first form is often easier when solving for a side, while the second form is often easier when solving for an angle.
Why does this work? Imagine the triangle inscribed in a circle. The Law of Sines actually relates to the diameter of this circumscribed circle. While the proof involves some geometry, the intuition is that larger angles “pull” their opposite sides into longer arcs.
When to Use the Law of Sines
The Law of Sines works best when you know an angle and its opposite side, plus one other piece of information. Specifically, use it for:
- AAS (Angle-Angle-Side): You know two angles and a side not between them
- ASA (Angle-Side-Angle): You know two angles and the side between them
- SSA (Side-Side-Angle): You know two sides and an angle opposite one of them (but watch out for the ambiguous case!)
The key is having a “matched pair”; an angle and its opposite side. Once you have that, you can set up a proportion to find the missing piece.
The Ambiguous Case (SSA)
Here is where things get interesting. When you have SSA (two sides and an angle opposite one of them), something unusual can happen: you might get zero solutions, one solution, or two different triangles that both satisfy your given information.
Let us say you know angle A, side a (opposite to A), and side b. Depending on the values, here is what can happen:
Case 1: No triangle exists If angle A is acute and $a < b \sin A$, then side a is too short to even reach the base. No triangle can be formed.
Case 2: Exactly one right triangle If angle A is acute and $a = b \sin A$, then side a exactly reaches the base at a right angle. One triangle exists.
Case 3: Two triangles exist If angle A is acute and $b \sin A < a < b$, then side a can swing to either side of the perpendicular, creating two different valid triangles.
Case 4: Exactly one triangle If angle A is acute and $a \geq b$, there is only one way to form the triangle. Similarly, if angle A is obtuse and $a > b$, exactly one triangle exists.
Case 5: No triangle (obtuse angle) If angle A is obtuse and $a \leq b$, no triangle exists.
This might seem like a lot to remember, but there is a practical approach: just try to solve the problem. If you get a sine value greater than 1, no triangle exists. If you get a valid answer, check whether a second triangle is possible by seeing if the supplementary angle also works.
The Law of Cosines
The Law of Cosines is a generalization of the Pythagorean theorem. Remember that $c^2 = a^2 + b^2$ only works for right triangles. The Law of Cosines adds a correction term that accounts for angles that are not 90 degrees:
$$c^2 = a^2 + b^2 - 2ab\cos C$$
Similarly for the other sides:
$$a^2 = b^2 + c^2 - 2bc\cos A$$ $$b^2 = a^2 + c^2 - 2ac\cos B$$
Notice: When $C = 90°$, we have $\cos 90° = 0$, so the formula becomes $c^2 = a^2 + b^2$; exactly the Pythagorean theorem! The Law of Cosines truly is a generalization.
Why the minus sign? The term $-2ab\cos C$ adjusts for how the angle differs from 90 degrees. If C is acute, the cosine is positive, making $c^2$ smaller than $a^2 + b^2$. If C is obtuse, the cosine is negative, making $c^2$ larger. This matches our geometric intuition: triangles with obtuse angles have their longest side opposite that obtuse angle.
When to Use the Law of Cosines
Use the Law of Cosines when you have:
- SAS (Side-Angle-Side): You know two sides and the angle between them
- SSS (Side-Side-Side): You know all three sides and need to find an angle
Unlike the Law of Sines, the Law of Cosines never produces an ambiguous case. When you use it, you get exactly one answer.
Area Formulas
While we are working with general triangles, let us cover two important area formulas.
Formula 1: Using Two Sides and the Included Angle
$$\text{Area} = \frac{1}{2}ab\sin C$$
This says: take half the product of two sides times the sine of the angle between them. You can use any two sides with their included angle:
$$\text{Area} = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C$$
Why does this work? Recall that $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. If you drop a perpendicular from one vertex to the opposite side, you create a right triangle where the height equals (adjacent side) $\times$ sine of the angle.
Formula 2: Heron’s Formula
What if you only know the three sides? Heron’s formula has you covered.
First, calculate the semi-perimeter (half the perimeter):
$$s = \frac{a + b + c}{2}$$
Then the area is:
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$
This remarkable formula, discovered over 2000 years ago, lets you find a triangle’s area using only its side lengths.
Notation and Terminology
| Term | Symbol | Meaning |
|---|---|---|
| Vertices | A, B, C | The corners of the triangle |
| Sides | a, b, c | The side lengths; lowercase letter opposite the corresponding uppercase angle |
| Opposite | Side a is opposite angle A, etc. | |
| Included angle | The angle formed by two given sides | |
| Semi-perimeter | s | Half the perimeter: $s = \frac{a+b+c}{2}$ |
| Ambiguous case | SSA configuration that might produce 0, 1, or 2 triangles | |
| AAS | Angle-Angle-Side: two angles and a non-included side known | |
| ASA | Angle-Side-Angle: two angles and the included side known | |
| SAS | Side-Angle-Side: two sides and the included angle known | |
| SSA | Side-Side-Angle: two sides and a non-included angle known | |
| SSS | Side-Side-Side: all three sides known |
Examples
In triangle ABC, angle A = 42 degrees, angle B = 73 degrees, and side a = 12. Find side b.
Step 1: First, let us identify what we have. We know two angles and a side opposite one of them. This is AAS, perfect for the Law of Sines.
Step 2: Set up the proportion using the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$
Step 3: Substitute the known values: $$\frac{12}{\sin 42°} = \frac{b}{\sin 73°}$$
Step 4: Solve for b: $$b = \frac{12 \times \sin 73°}{\sin 42°}$$
Step 5: Calculate: $$b = \frac{12 \times 0.9563}{0.6691} \approx \frac{11.476}{0.6691} \approx 17.15$$
Answer: Side b is approximately 17.15 units.
In triangle ABC, side a = 8, side b = 11, and angle C = 37 degrees. Find side c.
Step 1: We have two sides and the angle between them (SAS). This calls for the Law of Cosines.
Step 2: Write the Law of Cosines for side c: $$c^2 = a^2 + b^2 - 2ab\cos C$$
Step 3: Substitute the values: $$c^2 = 8^2 + 11^2 - 2(8)(11)\cos 37°$$
Step 4: Calculate: $$c^2 = 64 + 121 - 176(0.7986)$$ $$c^2 = 185 - 140.55$$ $$c^2 = 44.45$$
Step 5: Take the square root: $$c = \sqrt{44.45} \approx 6.67$$
Answer: Side c is approximately 6.67 units.
A triangle has sides a = 7, b = 10, and c = 12. Find angle C.
Step 1: We know all three sides (SSS). Use the Law of Cosines, rearranged to solve for the angle.
Step 2: Start with: $$c^2 = a^2 + b^2 - 2ab\cos C$$
Step 3: Rearrange to solve for $\cos C$: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$
Step 4: Substitute the values: $$\cos C = \frac{7^2 + 10^2 - 12^2}{2(7)(10)}$$ $$\cos C = \frac{49 + 100 - 144}{140}$$ $$\cos C = \frac{5}{140} = \frac{1}{28} \approx 0.0357$$
Step 5: Find the angle: $$C = \cos^{-1}(0.0357) \approx 87.95°$$
Answer: Angle C is approximately 88 degrees.
Check: Since $\cos C$ is small and positive, the angle should be close to 90 degrees but slightly less. This matches our answer.
In triangle ABC, angle A = 35 degrees, side a = 7, and side b = 10. Find angle B.
Step 1: We have SSA. Let us check for the ambiguous case. First, calculate $b \sin A$: $$b \sin A = 10 \times \sin 35° = 10 \times 0.5736 = 5.736$$
Step 2: Compare: Since $5.736 < 7 < 10$ (that is, $b \sin A < a < b$), we are in the ambiguous case. Two triangles are possible.
Step 3: Use the Law of Sines: $$\frac{\sin B}{b} = \frac{\sin A}{a}$$ $$\sin B = \frac{b \sin A}{a} = \frac{10 \times \sin 35°}{7} = \frac{5.736}{7} \approx 0.8194$$
Step 4: Find angle B. Since $\sin B = 0.8194$: $$B_1 = \sin^{-1}(0.8194) \approx 55.0°$$
Step 5: Check for a second solution. The sine function also equals 0.8194 at the supplementary angle: $$B_2 = 180° - 55.0° = 125.0°$$
Step 6: Verify both solutions work:
- For $B_1 = 55°$: Angle $C = 180° - 35° - 55° = 90°$ (valid triangle)
- For $B_2 = 125°$: Angle $C = 180° - 35° - 125° = 20°$ (valid triangle)
Answer: There are two possible triangles. In one, B is approximately 55 degrees; in the other, B is approximately 125 degrees.
Find the area of a triangle with sides a = 9 and b = 14, and included angle C = 62 degrees.
Step 1: We have two sides and the angle between them. Use the area formula: $$\text{Area} = \frac{1}{2}ab\sin C$$
Step 2: Substitute: $$\text{Area} = \frac{1}{2}(9)(14)\sin 62°$$
Step 3: Calculate: $$\text{Area} = \frac{1}{2}(126)(0.8829)$$ $$\text{Area} = 63 \times 0.8829 \approx 55.62$$
Answer: The area is approximately 55.62 square units.
A triangular garden has sides of length 13 meters, 14 meters, and 15 meters. Find its area.
Step 1: Calculate the semi-perimeter: $$s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$$
Step 2: Apply Heron’s formula: $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$ $$\text{Area} = \sqrt{21(21-13)(21-14)(21-15)}$$ $$\text{Area} = \sqrt{21 \times 8 \times 7 \times 6}$$
Step 3: Calculate the product under the radical: $$21 \times 8 = 168$$ $$168 \times 7 = 1176$$ $$1176 \times 6 = 7056$$
Step 4: Take the square root: $$\text{Area} = \sqrt{7056} = 84$$
Answer: The area of the garden is exactly 84 square meters.
Note: This turned out to be a perfect square! The triangle with sides 13-14-15 is a special triangle with a nice integer area.
In triangle ABC, angle B = 28 degrees, side b = 9, and side c = 15. Solve the triangle completely (find all missing sides and angles).
Step 1: Identify the configuration. We have angle B, side b (opposite to B), and side c. This is SSA; watch for the ambiguous case.
Step 2: Check for ambiguity. Calculate $c \sin B$: $$c \sin B = 15 \times \sin 28° = 15 \times 0.4695 = 7.04$$
Since $b = 9 > c \sin B = 7.04$, a triangle exists. Since $b = 9 < c = 15$, we might have two triangles.
Step 3: Find angle C using the Law of Sines: $$\frac{\sin C}{c} = \frac{\sin B}{b}$$ $$\sin C = \frac{c \sin B}{b} = \frac{15 \times 0.4695}{9} = \frac{7.04}{9} \approx 0.7822$$
$$C_1 = \sin^{-1}(0.7822) \approx 51.5°$$ $$C_2 = 180° - 51.5° = 128.5°$$
Step 4: Check both solutions:
- For $C_1 = 51.5°$: $A = 180° - 28° - 51.5° = 100.5°$ (valid)
- For $C_2 = 128.5°$: $A = 180° - 28° - 128.5° = 23.5°$ (valid)
Both triangles exist.
Step 5: Find side a for each triangle using the Law of Sines.
Triangle 1 (A = 100.5 degrees, C = 51.5 degrees): $$\frac{a}{\sin A} = \frac{b}{\sin B}$$ $$a = \frac{b \sin A}{\sin B} = \frac{9 \times \sin 100.5°}{\sin 28°} = \frac{9 \times 0.9833}{0.4695} \approx 18.85$$
Triangle 2 (A = 23.5 degrees, C = 128.5 degrees): $$a = \frac{9 \times \sin 23.5°}{\sin 28°} = \frac{9 \times 0.3987}{0.4695} \approx 7.64$$
Answer:
- Triangle 1: A = 100.5 degrees, B = 28 degrees, C = 51.5 degrees, a = 18.85, b = 9, c = 15
- Triangle 2: A = 23.5 degrees, B = 28 degrees, C = 128.5 degrees, a = 7.64, b = 9, c = 15
Key Properties
Choosing the Right Method
Here is a quick reference for deciding which law to use:
| Given Information | Method to Use | Notes |
|---|---|---|
| AAS or ASA | Law of Sines | First find the third angle (angles sum to 180 degrees) |
| SSA | Law of Sines | Watch for the ambiguous case |
| SAS | Law of Cosines | Finds the side opposite the known angle |
| SSS | Law of Cosines | Rearrange to find any angle |
Important Relationships
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The largest side is always opposite the largest angle, and the smallest side is opposite the smallest angle.
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The sum of any two sides must be greater than the third side (Triangle Inequality). If this fails, no triangle exists.
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The sum of angles in any triangle is exactly 180 degrees. Use this to find a third angle when you know two.
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For the ambiguous case (SSA): If the given angle is obtuse and the opposite side is not the longest, no triangle exists.
Area Formula Summary
| Known Information | Formula |
|---|---|
| Base and height | $A = \frac{1}{2}bh$ |
| Two sides and included angle | $A = \frac{1}{2}ab\sin C$ |
| All three sides (SSS) | $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$ |
Real-World Applications
Surveying and Land Measurement
Surveyors frequently use the Law of Cosines and Law of Sines to calculate distances that cannot be measured directly. Imagine needing to find the width of a river. You cannot stretch a tape measure across it, but you can:
- Mark two points A and B on your side of the river
- Identify a landmark C on the opposite bank
- Measure the distance AB
- Measure angles CAB and CBA
With this information (ASA), you can calculate the distance across the river using the Law of Sines.
Navigation
Pilots and sailors use these laws constantly. If an aircraft travels 200 miles in one direction, turns 40 degrees, and travels another 150 miles, how far is it from its starting point? This is a classic SAS problem solved with the Law of Cosines.
The area formula is also crucial in navigation; finding the area of visibility or the region covered by a search pattern.
Construction and Architecture
When building a roof truss or designing a bridge, engineers work with non-right triangles constantly. Knowing the lengths of rafters and the angles at which they meet allows calculation of spans, heights, and the amount of material needed.
Astronomy
Astronomers use triangulation to calculate distances to nearby stars. By measuring angles to a star from opposite sides of Earth’s orbit (six months apart), they form a triangle with a base of about 300 million kilometers. The Law of Sines helps convert tiny angle measurements into vast cosmic distances.
Sports
In golf, if you know the distance to the hole from two different positions and the angle between those sightlines, you can calculate how far your playing partners are from the hole. In billiards, calculating bank shot angles involves understanding triangle geometry.
Self-Test Problems
Problem 1: In triangle ABC, angle A = 47 degrees, angle C = 55 degrees, and side a = 15. Find side c.
Show Answer
First, this is AAS, so we use the Law of Sines.
Set up the proportion: $$\frac{a}{\sin A} = \frac{c}{\sin C}$$
Solve for c: $$c = \frac{a \sin C}{\sin A} = \frac{15 \times \sin 55°}{\sin 47°} = \frac{15 \times 0.8192}{0.7314} \approx \frac{12.29}{0.7314} \approx 16.80$$
Answer: Side c is approximately 16.80 units.
Problem 2: In triangle ABC, a = 6, b = 8, and c = 11. Find angle A.
Show Answer
This is SSS, so we use the Law of Cosines rearranged to solve for the angle: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$
Substitute: $$\cos A = \frac{8^2 + 11^2 - 6^2}{2(8)(11)} = \frac{64 + 121 - 36}{176} = \frac{149}{176} \approx 0.8466$$
$$A = \cos^{-1}(0.8466) \approx 32.2°$$
Answer: Angle A is approximately 32.2 degrees.
Problem 3: In triangle ABC, angle A = 40 degrees, side a = 5, and side b = 7. How many triangles are possible? Find all possible values of angle B.
Show Answer
This is SSA. First check: $b \sin A = 7 \times \sin 40° = 7 \times 0.6428 = 4.50$
Since $4.50 < 5 < 7$ (that is, $b \sin A < a < b$), two triangles are possible.
Using the Law of Sines: $$\sin B = \frac{b \sin A}{a} = \frac{7 \times 0.6428}{5} = 0.8999$$
$$B_1 = \sin^{-1}(0.8999) \approx 64.1°$$ $$B_2 = 180° - 64.1° = 115.9°$$
Check both:
- $A + B_1 = 40° + 64.1° = 104.1° < 180°$ (valid)
- $A + B_2 = 40° + 115.9° = 155.9° < 180°$ (valid)
Answer: Two triangles are possible. Angle B is either approximately 64.1 degrees or approximately 115.9 degrees.
Problem 4: Two sides of a triangle are 12 and 18, with an included angle of 48 degrees. Find the area of the triangle.
Show Answer
Use the area formula with two sides and included angle: $$\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(12)(18)\sin 48°$$
$$\text{Area} = \frac{1}{2}(216)(0.7431) = 108 \times 0.7431 \approx 80.25$$
Answer: The area is approximately 80.25 square units.
Problem 5: A triangle has sides of length 5, 12, and 13. Find its area using Heron’s formula. Then verify by recognizing what type of triangle this is.
Show Answer
Using Heron’s formula:
Semi-perimeter: $s = \frac{5 + 12 + 13}{2} = 15$
$$\text{Area} = \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15 \times 10 \times 3 \times 2} = \sqrt{900} = 30$$
Verification: Notice that $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. This is a right triangle (a 5-12-13 Pythagorean triple)!
For a right triangle: $\text{Area} = \frac{1}{2} \times 5 \times 12 = 30$
Answer: The area is exactly 30 square units.
Problem 6: A pilot flies 120 miles from city A to city B on a bearing of N35 degrees E. She then flies 180 miles from city B to city C on a bearing of S55 degrees E. How far is city C from city A?
Show Answer
First, we need to find the angle at B. The bearing N35 degrees E means 35 degrees east of north. The bearing S55 degrees E means 55 degrees east of south.
The angle between these directions (at point B) is: $180° - 35° - 55° = 90°$
Wait, let us think more carefully. N35 degrees E puts the first leg 35 degrees from north. S55 degrees E puts the second leg 55 degrees from south. The angle ABC between the paths going into B and out of B is $35° + 55° = 90°$. But the interior angle of the triangle at B is $180° - 90° = 90°$.
Actually, for the interior angle at B: the pilot arrives heading roughly northeast, then leaves heading roughly southeast. The change in direction creates an interior angle of $35° + 55° = 90°$.
With SAS (AB = 120, angle B = 90 degrees, BC = 180), we can use the Law of Cosines. But since the angle is 90 degrees, this is actually a right triangle:
$$AC^2 = AB^2 + BC^2 = 120^2 + 180^2 = 14400 + 32400 = 46800$$
$$AC = \sqrt{46800} \approx 216.3$$
Answer: City C is approximately 216.3 miles from city A.
Summary
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Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Use for AAS, ASA, or SSA problems.
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The ambiguous case (SSA) can produce 0, 1, or 2 triangles. When using the Law of Sines with SSA, always check whether a second triangle exists by testing the supplementary angle.
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Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$. Use for SAS or SSS problems. It generalizes the Pythagorean theorem to all triangles.
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To find an angle with Law of Cosines, rearrange: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
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Area formulas:
- With two sides and included angle: $\text{Area} = \frac{1}{2}ab\sin C$
- With three sides (Heron’s formula): $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$
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Strategy for solving triangles:
- Identify what you know (which combination of sides and angles)
- Choose Law of Sines (need matched pair) or Law of Cosines (need SAS or SSS)
- For SSA, always check for the ambiguous case
- Use the angle sum property (A + B + C = 180 degrees) to find remaining angles
- Verify your answer makes sense (largest side opposite largest angle, etc.)
With the Law of Sines and Law of Cosines in your toolkit, you can now solve any triangle you encounter. These are not just classroom exercises; they are the same tools used by surveyors, navigators, engineers, and scientists to solve real problems where right angles are the exception rather than the rule.