Why We Need the Laws of Sines and Cosines
Basic right-triangle trigonometry (SOH CAH TOA) only works when one angle is exactly 90°. But most triangles in the real world — in surveying, navigation, engineering, and physics — are oblique, meaning they have no right angle. To solve these, we need two more powerful tools: the Law of Sines and the Law of Cosines.
Together, they let you find unknown sides or angles in any triangle as long as you know enough information to determine it.
The five classic cases
| Case | What you know | Which law |
|---|---|---|
| ASA | two angles and the included side | Law of Sines |
| AAS | two angles and a non-included side | Law of Sines |
| SSA | two sides and a non-included angle — the ambiguous case | Law of Sines |
| SAS | two sides and the included angle | Law of Cosines |
| SSS | three sides | Law of Cosines |
A single problem often requires both — use one to find a first unknown, then the other to find the rest.
a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. The three angles always sum to 180°.The Law of Sines
sin(A)/a = sin(B)/b = sin(C)/c. Equivalently: a/sin(A) = b/sin(B) = c/sin(C).Use this whenever you have a complete angle-side pair (one angle and its opposite side) plus one more piece of information.
In triangle ABC, A = 35°, B = 70°, a = 10. Find b and c.
First find C = 180° − 35° − 70° = 75°.
Now apply the Law of Sines: b = a · sin(B)/sin(A) = 10 · sin(70°)/sin(35°) ≈ 10 · 0.9397/0.5736 ≈ 16.38.
c = a · sin(C)/sin(A) = 10 · sin(75°)/sin(35°) ≈ 10 · 0.9659/0.5736 ≈ 16.84.
b ≈ 16.38, c ≈ 16.84A = 40°, B = 60°, c = 20.
C = 180° − 40° − 60° = 80°.
a = c · sin(A)/sin(C) = 20 · sin(40°)/sin(80°) ≈ 20 · 0.6428/0.9848 ≈ 13.06.
b = c · sin(B)/sin(C) = 20 · sin(60°)/sin(80°) ≈ 20 · 0.8660/0.9848 ≈ 17.59.
a ≈ 13.06, b ≈ 17.59The Law of Sines is efficient for ASA and AAS, but SSA is trickier — see the next section.
The Ambiguous Case (SSA)
When you know two sides and a non-included angle (SSA), the Law of Sines can give zero, one, or two valid triangles. This is the ambiguous case, and it trips up many students.
Suppose you know a, b, and A, and want to find B. Apply sin(B) = b · sin(A)/a.
Three scenarios
| Case | Condition | Result | Why |
|---|---|---|---|
| 1 | sin(B) > 1 | No triangle | The given sides can't form a triangle with that angle. |
| 2 | sin(B) = 1, or a ≥ b | One triangle | Exactly one triangle (right triangle at B). Also one triangle if the side opposite the given angle is at least as long as the other given side. |
| 3 | sin(B) < 1 and a < b and A is acute | Two triangles | There are two possible values of B (one acute, one obtuse, since sin(180° − B) = sin(B)). |
a = 7, b = 10, A = 30°.
sin(B) = 10 · sin(30°)/7 = 10 · 0.5/7 ≈ 0.7143.
So B ≈ 45.58° OR B ≈ 180° − 45.58° = 134.42°.
Check both: for B₁ = 45.58°: C = 180° − 30° − 45.58° = 104.42°, valid.
For B₂ = 134.42°: C = 180° − 30° − 134.42° = 15.58°, also valid.
Both triangles existThis is why SSA is called 'the ambiguous case' — you must check whether both solutions are geometrically valid.
The Law of Cosines
a, b, c and opposite angles A, B, C: a² = b² + c² − 2bc · cos(A), b² = a² + c² − 2ac · cos(B), c² = a² + b² − 2ab · cos(C).Notice that if A = 90°, cos(A) = 0 and the formula collapses to the Pythagorean theorem.
Use the Law of Cosines for SAS (given two sides and the included angle) to find the third side, or for SSS (given three sides) to find any angle.
b = 8, c = 11, A = 47°. Find a.
a² = 8² + 11² − 2 · 8 · 11 · cos(47°) = 64 + 121 − 176 · 0.6820 ≈ 185 − 120.03 ≈ 64.97.
a ≈ 8.06.
Then use Law of Sines for the remaining angles: sin(B) = b · sin(A)/a = 8 · sin(47°)/8.06 ≈ 8 · 0.7314/8.06 ≈ 0.7259.
B ≈ 46.57°.
C = 180° − 47° − 46.57° ≈ 86.43°.
a ≈ 8.06, B ≈ 46.57°, C ≈ 86.43°a = 6, b = 8, c = 10. Find all angles.
Solve for A: cos(A) = (b² + c² − a²)/(2bc) = (64 + 100 − 36)/160 = 128/160 = 0.8. A = cos⁻¹(0.8) ≈ 36.87°.
Solve for B: cos(B) = (a² + c² − b²)/(2ac) = (36 + 100 − 64)/120 = 72/120 = 0.6. B = cos⁻¹(0.6) ≈ 53.13°.
C = 180° − 36.87° − 53.13° = 90°. (This is a 6-8-10 right triangle, a scaled 3-4-5.)
A ≈ 36.87°, B ≈ 53.13°, C = 90°The Law of Cosines never has an ambiguous case — it always gives a unique answer.
When to Use Which Law
Choosing between the two laws is based on what you're given:
Quick decision table
| Given | Start with |
|---|---|
AAS | Law of Sines |
ASA | Find third angle, then Law of Sines |
SSA | Law of Sines (check ambiguous case) |
SAS | Law of Cosines (find third side) |
SSS | Law of Cosines (find any angle) |
Notice that AAA (only angles given) has no unique triangle — you need at least one side length to determine size.
After using one law to find a first unknown, you usually have a complete angle-side pair, and can switch to Law of Sines for the rest (which is less computation than another Law of Cosines application).
When solving for angles with Law of Cosines, always solve for the largest angle first (opposite the longest side) — this tells you immediately whether the triangle is acute or obtuse, and removes SSA-style ambiguity later.
Real-World Applications and Worked Example
The Laws of Sines and Cosines appear anywhere triangles do:
| Field | How the laws are used |
|---|---|
| Surveying | Measuring distances across rivers or ravines where you can't reach both endpoints directly. |
| Navigation | Calculating course corrections when winds or currents push you off the direct path. |
| Astronomy | Parallax calculations rely on the Law of Sines to compute stellar distances from Earth's orbit geometry. |
| Engineering | Force-vector decomposition, truss analysis, and robotics all use oblique triangle math. |
A surveyor wants to measure the distance across a lake from point A to point B, but can't walk across. She picks point C on dry land, measures AC = 450 m, BC = 520 m, and angle ACB = 78°. Find AB.
This is SAS. Using Law of Cosines with C = 78° and opposite side c = AB: c² = 450² + 520² − 2 · 450 · 520 · cos(78°).
c² = 202,500 + 270,400 − 468,000 · 0.2079 ≈ 472,900 − 97,296 ≈ 375,604.
c ≈ 612.87 m.
The lake is about 613 meters across.A plane flies 200 km from A to B on a heading, then turns through an angle of 40° and flies 150 km to C. How far is C from A?
Side AB = 200, side BC = 150, the interior angle at B is 180° − 40° = 140°.
Law of Cosines: AC² = 200² + 150² − 2 · 200 · 150 · cos(140°) = 40,000 + 22,500 − 60,000 · (−0.7660) ≈ 62,500 + 45,963 ≈ 108,463.
AC ≈ 329.34 km.
For the heading-back angle, use Law of Sines.
AC ≈ 329.34 kmIf you want to verify any triangle computation, scan the problem with Solver AI — it identifies the case, picks the correct law, and shows all the computation steps.