Geometry • Trigonometric Triangle Solver

Law of Sines Calculator

Solve any oblique triangle for missing sides, angles, and area using the Law of Sines. Supports AAS, ASA, and the full SSA ambiguous case with dynamic 2D geometric triangle plots.

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Last Updated: September 2026
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Verified Mathematical Solution
Looking for SSS or SAS Triangles?

If you have three side lengths (SSS) or two sides and the included angle (SAS), use our companion Law of Cosines solver.

Open Law of Cosines Solver
Switch to Law of Cosines (SSS / SAS)
AAS / ASA Given Parameters
Preset Examples:
Triangle Solution Unique Triangle (1 Solution)
Solved Missing Parts
C = 80.00° | b = 15.80 | c = 17.17
Sine Ratio: a / sin(A) = 17.43
Angle C 80.00°
Side b 15.80
Side c 17.17
Area 77.80

Geometry Plot & Oblique Triangle

a/sin(A) = b/sin(B) = c/sin(C)
A (Vertex)
B (Vertex)
C (Vertex)

Step-by-Step Law of Sines Derivation Formula: a / sin(A) = b / sin(B) = c / sin(C)

Direct Answer & Overview
Verified Educational Guide

How to Solve a Triangle Using the Law of Sines

To solve an oblique triangle using the Law of Sines, set up the proportion a/sin(A) = b/sin(B) = c/sin(C). For AAS or ASA cases, first find the third angle using C = 180° - A - B, then cross-multiply to find missing sides: b = a·sin(B)/sin(A) and c = a·sin(C)/sin(A). For SSA cases, test the altitude h = b·sin(A) to check if 0, 1, or 2 triangles exist.

Primary Mathematical Formula Standard Mathematical Model
Standard Equation
ƒ(x)
Q.E.D.
a / sin(A) = b / sin(B) = c / sin(C) | Area = (1/2) ab sin(C) | h = b · sin(A)
Evaluated with exact mathematical formulation • Rigorously verified
Exact Formula
Input Parameters
Required
1
AAS / ASA Mode: Two known angles (A, B) and one side length (a)
2
SSA Ambiguous Mode: Two sides (a, b) and one non-included angle (A)
Expected Outputs
Calculated
Missing Angles: Calculated in degrees with exact angle sums (A + B + C = 180°)
Missing Sides: Solved side lengths b and c
Ambiguous Case Status: Identification of 0, 1, or 2 valid geometric triangles
Triangle Area: Computed via the sine area formula (1/2) ab sin(C)
Worked Numerical Example
Instant Verification
Solve triangle with A = 35°, B = 65°, a = 10
→ C = 180° - 35° - 65° = 80°. b = 10 · [sin(65°)/sin(35°)] = 15.80. c = 10 · [sin(80°)/sin(35°)] = 17.17
C = 80.00° | Side b = 15.80 | Side c = 17.17 | Area = 77.80 sq units

The Law of Sines Formula & Mathematical Ratio

The Law of Sines is a fundamental theorem of trigonometry that relates the lengths of the sides of any triangle (acute, obtuse, or right) to the sines of its angles:

a / sin(A) = b / sin(B) = c / sin(C) = 2R

Where $R$ is the circumradius of the triangle's circumscribed circle.

Solving AAS & ASA Triangle Cases

Whenever two angles are known, the third angle is immediately found via the Euclidean triangle angle-sum identity:

C = 180° − A − B

Once all three angles are established, missing sides are calculated using direct sine ratios without any ambiguity.

The Ambiguous Case (SSA): 0, 1, or 2 Triangles

When given two sides and a non-included acute angle $A$ (SSA), the altitude is $h = b \sin(A)$. The geometric possibilities depend on the length of side $a$:

0 Triangles (a < h)

Side $a$ is too short to reach the base line. No triangle exists.

1 Right Triangle (a = h)

Side $a$ meets the base line perpendicularly ($\sin B = 1, B = 90^\circ$).

2 Triangles (h < a < b)

Side $a$ can swing inward (obtuse $B_2 = 180^\circ - B_1$) or outward (acute $B_1$).

1 Unique Triangle (a ≥ b)

Side $a$ can only swing outward; the inward swing fails to form a valid triangle.

Comparing Law of Sines vs Law of Cosines

Given Information Recommended Method Primary Formula
AAS or ASA Law of Sines a / sin(A) = b / sin(B)
SSA (Ambiguous) Law of Sines sin(B) = (b · sin(A)) / a
SAS Law of Cosines c² = a² + b² - 2ab cos(C)
SSS Law of Cosines cos(A) = (b² + c² - a²) / 2bc

Step-by-Step Worked Examples (AAS & SSA Cases)

Two-Triangle SSA Solution Level: Advanced

Solve triangle with A = 30°, a = 7, b = 10.

1. Altitude: h = 10 × sin(30°) = 10 × 0.5 = 5.0.

2. Test: 5.0 < a (7) < b (10) → Two triangles exist!

3. sin(B) = (10 × 0.5) / 7 = 5/7 ≈ 0.7143.

4. Triangle 1: B₁ = sin⁻¹(0.7143) = 45.58°, C₁ = 180° − 30° − 45.58° = 104.42°, c₁ = 7 × [sin(104.42°)/sin(30°)] = 13.56.

5. Triangle 2: B₂ = 180° − 45.58° = 134.42°, C₂ = 180° − 30° − 134.42° = 15.58°, c₂ = 7 × [sin(15.58°)/sin(30°)] = 3.76.

Common Pitfalls in Sine Ratio Calculations

Pitfall 1: Forgetting the Second SSA Triangle

Standard handheld calculators only return the principal acute angle from $\arcsin(\theta)$. In SSA cases, you must always test whether $B_2 = 180^\circ - B_1$ forms a valid second triangle.

Pitfall 2: Degree vs Radian Mode

Ensure your calculator is in Degree mode when working with degree inputs like $35^\circ$ or $65^\circ$.

Fact-Checked & Verified • Computational Accuracy Standards
Updated July 2026 • Editorial Policy
Authored By
Sanjay Samanta

Lead Developer & Founder of Basic Math Tools. Specializes in browser-native computational algorithms and applied mathematics.

Reviewed & Verified By
Academic Review Board

Mathematics & curriculum specialists. Audited against standard algebraic and arithmetic principles.

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Frequently Asked Questions

What is the Law of Sines?
The Law of Sines states that the ratio of the sine of an angle to the length of its opposite side is constant for all three angles in any triangle: a / sin(A) = b / sin(B) = c / sin(C).
When should I use the Law of Sines versus the Law of Cosines?
Use the Law of Sines when you know: (1) Two angles and any side (AAS or ASA), or (2) Two sides and a non-included angle (SSA). Use the Law of Cosines when you know three sides (SSS) or two sides and the included angle between them (SAS).
What is the ambiguous case (SSA) in the Law of Sines?
When given two sides and a non-included acute angle (SSA), there may be: 0 triangles (if the opposite side a is shorter than the altitude h = b·sin(A)), exactly 1 right triangle (if a = h), exactly 1 triangle (if a ≥ b), or 2 distinct triangles (if h < a < b).
How do you find the second triangle in the SSA ambiguous case?
Calculate the first acute angle B₁ = arcsin((b·sin(A))/a). The second possible obtuse angle is B₂ = 180° - B₁. If A + B₂ < 180°, then a valid second triangle exists with remaining angle C₂ = 180° - A - B₂.
Does the Law of Sines work for right triangles?
Yes! For a right triangle where C = 90°, sin(90°) = 1, so a / sin(A) = c / 1 ⟹ sin(A) = a/c (Opposite / Hypotenuse), which matches standard SOH-CAH-TOA trigonometry.