The Law of Sines
A surveyor stands at one end of a pond, sights a tree across the water, walks a measured distance along the bank, and sights the tree again. Two angles and one distance are known, but no right angle is in sight. The Law of Sines turns those measurements into the missing distances.
By the end of this page you can:
- State the Law of Sines and identify which parts of a triangle it relates.
- Solve a triangle given two angles and a side (the AAS and ASA cases).
- Recognize the ambiguous SSA case and check for a possible second triangle.
The floor. In any triangle, each side divided by the sine of its opposite angle gives the same ratio. Match a known angle to its opposite side, set up a proportion, and solve.
Quick Reference
| Field | Value |
|---|---|
| Textbook | Stewart, Redlin, Watson 7e |
| Chapter | 6 |
| Section | 6.5 The Law of Sines |
| Open alternate | OpenStax Precalculus 2e, 8.1 Non-right Triangles: Law of Sines |
| Difficulty | Core |
| Time | 30 minutes |
Before You Start
The Idea
Right triangle trig only works when a triangle has a right angle. Most triangles do not. The Law of Sines fills that gap for any triangle by pairing each side with the angle directly across from it. Label the angles A, B, C and the side opposite each one a, b, c. The claim is that the three ratios (side over the sine of its opposite angle) are all equal.
This single relationship solves a triangle whenever you know an angle and the side across from it, plus one more piece. That covers two angles and a side (AAS and ASA). It also handles two sides and an angle opposite one of them (SSA), which is the case that can hide a surprise.
Key Formulas
For a triangle with angles A, B, C and opposite sides a, b, c:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \]
The same statement, written with the sides on top, is equivalent:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Source: Stewart 7e, Section 6.5 The Law of Sines.
In the SSA case, where you know two sides and an angle opposite one of them, solving for the second angle can produce zero, one, or two valid triangles. After you take the inverse sine, check whether the supplement of that angle (180 degrees minus it) still leaves the angle sum below 180 degrees. If it does, a second triangle exists.
Worked Example
Solve the triangle with A = 40 degrees, C = 75 degrees, and side a = 12.
- Find the third angle. The angles sum to 180 degrees, so B = 180 - 40 - 75 = 65 degrees.
- Pair side a with angle A, since both are known. Use that ratio to find side c, which is opposite the known angle C.
- Set up the proportion: c / sin C = a / sin A, so c = a (sin C / sin A) = 12 (sin 75 / sin 40).
Show the answer
B = 65 degrees. Using sin 75 = 0.9659 and sin 40 = 0.6428:
c = 12 (0.9659 / 0.6428) = 18.03 (rounded to two decimal places).
Side b follows the same way: b = 12 (sin 65 / sin 40) = 16.92.
This was an ASA setup (two angles with the included side), so it gives exactly one triangle. No ambiguity check is needed here.
Connections
This builds on right triangle trigonometry, which set up sine ratios for triangles that contain a right angle. It also leans on inverse trigonometry to recover an angle from a sine value, and on reference angles to handle an obtuse angle in the SSA case.
It leads next to the Law of Cosines, the tool for the two cases the Law of Sines cannot start: two sides with the included angle (SAS) and all three sides (SSS). Together, the two laws solve any triangle.
Resources
- Primary: Stewart, Redlin, Watson 7e, Section 6.5 The Law of Sines.
- Alternate: OpenStax Precalculus 2e, Section 8.1 Non-right Triangles: Law of Sines.
Last updated: 2026-06-24