The Laws of Sines and Cosines - Precalculus

SECTION 1

Historical Context — Triangles Beyond the Right Angle

For centuries, trigonometry was limited to right triangles. Builders and astronomers could use sine and cosine ratios only when a 90° angle was present, which is a significant constraint when you are trying to survey a field, navigate across an ocean, or calculate the distance to a planet. The drive to solve oblique triangles — triangles with no right angle at all — motivated mathematicians across cultures to discover two elegant relationships that extend trigonometry far beyond its right-angle origins.

c. 100 CE

The Greek astronomer Menelaus of Alexandria establishes early results about spherical triangles, hinting at relationships between sides and angles that go beyond Euclid's right-triangle framework.

c. 1000 CE

Persian mathematician Abū al-Wafāʾ and later al-Bīrūnī formalize the Law of Sines for both plane and spherical triangles, enabling more accurate astronomical calculations across the Islamic Golden Age.

c. 1400 CE

al-Kāshī, working in Samarkand, provides one of the earliest explicit statements of the Law of Cosines in a form recognizable today, applying it to geodesy and architecture.

1596

The French mathematician François Viète publishes systematic trigonometric identities, making both laws accessible to European navigation and cartography during the Age of Exploration.

Modern Era

Today, the Laws of Sines and Cosines form the backbone of surveying, navigation, physics, and engineering — any discipline that must solve triangles that aren't conveniently right-angled.

The underlying question these mathematicians pursued was deceptively simple: Given some combination of sides and angles in a triangle, how do you find the rest? The Law of Sines and the Law of Cosines are the two tools that answer this question for every possible triangle.

SECTION 2

Core Principles and Definitions

Before we prove anything, let's set up the language. In any triangle △ABC, we use the convention that side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. This "opposite" pairing is central to both laws. You should also recall that the interior angles of any triangle sum to 180°, a fact we will lean on repeatedly.

Law of Sines

The ratio of any side to the sine of its opposite angle is the same for all three side–angle pairs in a triangle.

Law of Cosines

Relates one side of a triangle to the other two sides and the cosine of the included angle — a generalization of the Pythagorean Theorem.

Oblique Triangles

Any triangle that does not contain a right angle. These laws let us solve oblique triangles where SOHCAHTOA alone cannot help.

Solving a Triangle

Finding all unknown sides and angles given at least three pieces of information (with at least one side), using the appropriate law.

✦

Key Takeaway

Think of the Law of Sines as a "proportional recipe" — if you know one side–angle pair and part of another, you can scale to find the missing ingredient. The Law of Cosines, by contrast, works like an upgraded Pythagorean Theorem — it handles the extra term that appears when the angle between two sides isn't 90°.

SECTION 3

Visual Proof — Law of Sines

The proof of the Law of Sines relies on a beautifully simple idea: drop an altitude from one vertex to the opposite side, creating two right triangles. In each right triangle, you can express the altitude using a basic sine ratio, then set those two expressions equal to each other.

Figure 1 — Altitude h from vertex C creates two right triangles, yielding two expressions for h that prove the Law of Sines.

Consider triangle △ABC with altitude h dropped from vertex C to point D on side c. In right triangle △ACD, the sine of angle A gives us sin A = h / b, so h = b sin A. In right triangle △BCD, the sine of angle B gives us sin B = h / a, so h = a sin B. Since both expressions equal h, we can write b sin A = a sin B, which rearranges to:

LAW OF SINES — DERIVED

a / sin A = b / sin B

By dropping an altitude from vertex A instead, the same argument gives b / sin B = c / sin C, completing the full law.

The elegance of this proof lies in its simplicity: no advanced tools are required, just the definition of sine in a right triangle and the willingness to construct a helpful altitude. Note that this proof works whether triangle ABC is acute or obtuse — in the obtuse case the altitude foot D falls outside segment AB, but the algebra still holds because sin(180° − θ) = sin θ.

SECTION 4

Mathematical Framework — Proving and Stating the Laws

The Law of Sines (Complete Statement)

LAW OF SINES

a / sin A = b / sin B = c / sin C

where a, b, c are side lengths and A, B, C are their respective opposite angles.

This law tells you that the ratio of each side to the sine of its opposite angle is a constant for any given triangle. In fact, that constant equals the diameter of the triangle's circumscribed circle (the circle passing through all three vertices), a beautiful geometric connection you may encounter later. You should use the Law of Sines when you know an angle–opposite-side pair along with one additional angle or side — cases commonly labeled AAS (angle–angle–side), ASA (angle–side–angle), or the ambiguous case SSA (side–side–angle).

Proof of the Law of Cosines

The Law of Cosines requires a coordinate-based approach. Place triangle △ABC with vertex A at the origin and side c along the positive x-axis, so B sits at the point (c, 0). Vertex C is at some point we can express using angle A:

C = (b cos A, b sin A)

Now compute the distance from B to C using the distance formula. Since B = (c, 0) and C = (b cos A, b sin A):

DISTANCE FORMULA — SIDE a

a² = (b cos A − c)² + (b sin A)²

Expanding the right side: a² = b² cos² A − 2bc cos A + c² + b² sin² A. Notice that b² cos² A + b² sin² A = b²(cos² A + sin² A) = b², thanks to the Pythagorean identity. This simplifies everything to:

LAW OF COSINES

a² = b² + c² − 2bc cos A

Equivalently: b² = a² + c² − 2ac cos B, and c² = a² + b² − 2ab cos C.

Notice what happens when A = 90°: since cos 90° = 0, the equation reduces to a² = b² + c², which is exactly the Pythagorean Theorem. The term −2bc cos A is the "correction factor" that accounts for the angle not being a right angle. When A is acute, cos A is positive and the term subtracts from the sum of squares (making a shorter than it would be in the right-angle case). When A is obtuse, cos A is negative and the term adds to the sum, making a longer.

Use the Law of Cosines when you know SAS (two sides and the included angle) or SSS (all three sides). In either case, the Law of Sines alone would leave you stuck, but the Law of Cosines gives you an entry point.

✦

Key Takeaway

The Law of Cosines is the Pythagorean Theorem's grown-up sibling. When the included angle is exactly 90°, the cosine term vanishes and you're back to a² + b² = c². For any other angle, that extra −2bc cos A term adjusts the result — like adding or subtracting a correction for how "off" from 90° your angle is.

SECTION 5

When to Use Which Law — A Decision Flowchart

One of the most practical skills in this topic is recognizing which law to apply. The choice depends entirely on what information you start with. The flowchart below walks you through the decision process, and the table that follows summarizes the cases.

Figure 2 — Decision flowchart for choosing between the Law of Sines and the Law of Cosines based on given information.

Given Info	Case Name	Which Law?	Notes
Two angles + one side	AAS / ASA	Law of Sines	Find the third angle first (angles sum to 180°), then use proportions.
Two sides + non-included angle	SSA	Law of Sines	Ambiguous case — may yield 0, 1, or 2 solutions. Always check.
Two sides + included angle	SAS	Law of Cosines	Find the third side, then switch to Law of Sines for remaining angles.
All three sides	SSS	Law of Cosines	Solve for the largest angle first (avoids ambiguous-supplement issues).

SECTION 6

Worked Example — Solving a Triangle

In triangle △ABC, you are given: b = 14, c = 9, and A = 62°. Find all remaining sides and angles. Since we know two sides and the included angle (SAS), we start with the Law of Cosines.

Solving a Triangle (SAS)

1
Step 1 — Find side a using the Law of Cosines

Apply the formula a² = b² + c² − 2bc cos A: a² = 14² + 9² − 2(14)(9) cos 62° a² = 196 + 81 − 252 × 0.46947 a² = 277 − 118.31 = 158.69

Taking the square root: a ≈ 12.60
2
Step 2 — Find angle B using the Law of Sines

Now that we have a side–angle pair (a and A), we can use the Law of Sines to find B: sin B / b = sin A / a sin B = b × sin A / a = 14 × sin 62° / 12.60 sin B = 14 × 0.88295 / 12.60 = 0.9811 So B = sin⁻¹(0.9811)

B ≈ 78.85° (Since we used the Law of Cosines first and found a < b, we know B > A, and B must be acute here because A + B would exceed 180° if B were obtuse. No ambiguity.)
3
Step 3 — Find angle C

Use the angle sum property: C = 180° − A − B = 180° − 62° − 78.85°

C ≈ 39.15°
4
Step 4 — Verify and Interpret

Let's verify with the Law of Sines: a / sin A = 12.60 / sin 62° = 12.60 / 0.8830 ≈ 14.27. Also c / sin C = 9 / sin 39.15° = 9 / 0.6312 ≈ 14.26. These ratios match (the tiny difference is rounding), confirming our solution.

The complete solution is: a ≈ 12.60, B ≈ 78.85°, C ≈ 39.15°

SECTION 7

Strengths, Limitations, and Comparisons

Each law has distinct advantages and pitfalls. Understanding these will save you time and help you avoid errors, especially the notorious ambiguous case.

Feature	Law of Sines	Law of Cosines
Simplicity	Simple proportions — easy to set up and solve	More algebraic — requires squaring and square roots
Best for	AAS, ASA; also finding remaining angles after one is known	SAS and SSS — the cases Law of Sines can't start
Ambiguity risk	High in SSA — inverse sine gives two possible angles	None — inverse cosine always gives a unique angle in (0°, 180°)
Connection to Pythagorean Thm	No direct connection	Reduces to Pythagorean Theorem when included angle = 90°
Finding area	Can be combined with area = ½ab sin C	Can find angle first, then use ½ab sin C

✦

Key Takeaway

Think of the Law of Cosines as the "safe" opener — it never introduces ambiguity because cos⁻¹ always returns exactly one angle between 0° and 180°. Once you've used it to crack the triangle open, the Law of Sines becomes a quick, clean tool for finding everything else. A good strategy: start with the Law of Cosines whenever you can (SAS or SSS), then switch to the Law of Sines for the remaining pieces.

SECTION 8

Connections to Advanced Theory

The Laws of Sines and Cosines are not dead ends — they are springboards to deeper mathematics. Here are several directions in which these ideas extend.

The Law of Sines reveals that a / sin A = 2R, where R is the circumradius — the radius of the circle passing through all three vertices. This connection between a triangle and its circumscribed circle is fundamental in geometry and appears again in topics like inscribed angle theorems and the study of cyclic quadrilaterals.

The Law of Cosines connects to vector mathematics. If you place two sides of a triangle as vectors u and v, the third side is their difference, and its length squared is |u − v|² = |u|² + |v|² − 2|u||v| cos θ. This is exactly the Law of Cosines restated in vector language, and it naturally leads to the concept of the dot product in linear algebra and physics.

This Lesson	Advanced Extension
a / sin A = b / sin B = c / sin C	= 2R (circumradius relationship), extended sine rule
a² = b² + c² − 2bc cos A	Dot product: u · v = \|u\|\|v\| cos θ (vectors in physics)
Plane triangle solutions	Spherical trigonometry (navigation, astronomy)
Area = ½ab sin C	Heron's formula: Area = √[s(s−a)(s−b)(s−c)]

In spherical trigonometry — used for navigation on the curved surface of the Earth — analogous laws of sines and cosines exist, but with sines and cosines applied to the sides as well as the angles. If you continue into calculus-based physics, you will find the Law of Cosines hiding inside every dot product calculation and every force decomposition that involves non-perpendicular components.

SECTION 9

Practice Problems

Question 1: Conceptual

Explain why the Law of Cosines reduces to the Pythagorean Theorem when the included angle is 90°. What specific mathematical fact makes the extra term disappear?

Show answer

The Law of Cosines states a² = b² + c² − 2bc cos A. When A = 90°, we have cos 90° = 0, so the term −2bc cos A becomes −2bc(0) = 0. The equation simplifies to a² = b² + c², which is the Pythagorean Theorem. The key mathematical fact is that the cosine of a right angle is exactly zero, which eliminates the correction term entirely. This confirms that the Pythagorean Theorem is a special case of the Law of Cosines, not a separate rule.

Question 2: Basic Calculation

In triangle △ABC, angle A = 40°, angle B = 75°, and side a = 10. Find side b using the Law of Sines. Round to two decimal places.

Show answer

By the Law of Sines: a / sin A = b / sin B, so b = a × sin B / sin A. b = 10 × sin 75° / sin 40° b = 10 × 0.9659 / 0.6428 b = 10 × 1.5027 b ≈ 15.03 This makes sense because angle B (75°) is larger than angle A (40°), and in any triangle, a larger angle is always opposite a longer side. Since B > A, we expect b > a, and indeed 15.03 > 10.

Question 3: Intermediate

In triangle △PQR, p = 8, q = 11, and r = 15. Find the measure of angle R to the nearest degree.

Show answer

Since we know all three sides (SSS), use the Law of Cosines solved for angle R: r² = p² + q² − 2pq cos R. Rearranging: cos R = (p² + q² − r²) / (2pq) cos R = (64 + 121 − 225) / (2 × 8 × 11) cos R = (185 − 225) / 176 cos R = −40 / 176 = −0.2273 R = cos⁻¹(−0.2273) R ≈ 103° The negative cosine tells us angle R is obtuse, which makes sense since side r = 15 is the longest side and is therefore opposite the largest angle. In any triangle where the longest side squared exceeds the sum of the squares of the other two sides (225 > 185), the angle opposite the longest side must be obtuse.

Question 4: Applied (Multi-Step)

Two fire lookout towers are 18 km apart. Tower A spots a fire at a bearing of N 52° E, and Tower B (due east of A) spots the same fire at a bearing of N 38° W. How far is the fire from each tower? (Round to two decimal places.)

Show answer

Set up a triangle with vertices at Tower A, Tower B, and the Fire (F). Side AB = 18 km. Tower B is due east of A, so the line AB runs east–west. At Tower A, the bearing N 52° E means the angle from the north direction, measured clockwise, to the fire is 52°. The interior angle of the triangle at A (measured from AB) is: angle A = 90° − 52° = 38°. At Tower B, the bearing N 38° W means the angle from north measured counter-clockwise (westward) is 38°. The interior angle at B is: angle B = 90° − 38° = 52°. The third angle: angle F = 180° − 38° − 52° = 90°. (Interesting — it's a right triangle at the fire!) Now use the Law of Sines: AB / sin F = AF / sin B = BF / sin A 18 / sin 90° = AF / sin 52° = BF / sin 38° 18 / 1 = 18 AF = 18 × sin 52° = 18 × 0.7880 ≈ 14.18 km BF = 18 × sin 38° = 18 × 0.6157 ≈ 11.08 km The fire is approximately 14.18 km from Tower A and 11.08 km from Tower B. Note that the Law of Sines was ideal here because the problem naturally gives us an AAS configuration.

Question 5: Challenge (Synthesis)

Prove that the area of any triangle can be written as Area = ½ab sin C by combining what you know about altitudes and the Law of Sines proof technique. Then use this formula, together with the Law of Cosines, to derive Heron's formula: Area = √[s(s − a)(s − b)(s − c)], where s = (a + b + c) / 2. (Hint: start by writing sin²C = 1 − cos²C, and substitute the Law of Cosines expression for cos C.)

Show answer

Part 1 — Proving Area = ½ab sin C: Drop an altitude h from vertex C to side c. By definition of sine in the right triangle formed, h = a sin B (or equivalently h = b sin A). But we can also express h using angle C by dropping an altitude from A to side a: h' = b sin C. Using the standard base–height area formula with base a and height h' (the altitude from B to side b), we get Area = ½ × a × (b sin C) = ½ab sin C. ✓ Part 2 — Deriving Heron's Formula: Start with Area² = (½ab sin C)² = ¼a²b² sin²C. Since sin²C = 1 − cos²C, substitute: Area² = ¼a²b²(1 − cos²C). From the Law of Cosines: cos C = (a² + b² − c²) / (2ab). So cos²C = (a² + b² − c²)² / (4a²b²). Then: Area² = ¼a²b² × [1 − (a² + b² − c²)² / (4a²b²)] = ¼a²b² − (a² + b² − c²)² / 16 = [4a²b² − (a² + b² − c²)²] / 16 Factor the numerator as a difference of squares: (2ab + a² + b² − c²)(2ab − a² − b² + c²). The first factor = (a + b)² − c² = (a + b + c)(a + b − c). The second factor = c² − (a − b)² = (c + a − b)(c − a + b). Let s = (a + b + c)/2. Then: a + b + c = 2s, a + b − c = 2(s − c), c + a − b = 2(s − b), c − a + b = 2(s − a). So: Area² = [2s × 2(s−c) × 2(s−b) × 2(s−a)] / 16 = s(s−a)(s−b)(s−c). Therefore: Area = √[s(s−a)(s−b)(s−c)]. This is Heron's formula. ✓ This derivation shows how the Law of Cosines, the area formula, and the Pythagorean identity work together to produce a result that depends only on the three side lengths — a remarkable synthesis.

SUMMARY

Lesson Summary

Summary

The Law of Sines — a/sin A = b/sin B = c/sin C — states that in any triangle, the ratio of a side to the sine of its opposite angle is constant. We proved it by dropping an altitude to create two right triangles and equating two expressions for the altitude's length. It is the tool of choice for AAS and ASA cases, and it also applies (with care) to the ambiguous SSA case, where zero, one, or two triangles may satisfy the given conditions.

The Law of Cosines — a² = b² + c² − 2bc cos A — generalizes the Pythagorean Theorem to non-right triangles. We proved it using coordinate placement and the distance formula, invoking the identity sin²θ + cos²θ = 1 to simplify. It is ideal for SAS and SSS cases and produces no ambiguity since cos⁻¹ returns exactly one angle in (0°, 180°). Together, these two laws form a complete toolkit: given any three independent measurements of a triangle (with at least one side), you can always find all remaining sides and angles.