Algebra 2 / Precalculus · Step-by-step guide

Law of Sines

Use the Law of Sines for non-right triangles, including the ambiguous SSA case.

By Tyler Blovat··9 min read
The short answer

Start with the central idea

For triangle sides opposite angles A, B, and C, a/sin⁡A=b/sin⁡B=c/sin⁡C\displaystyle a/\sin A=b/\sin B=c/\sin C. Use it when you know an opposite side-angle pair and another side or angle.

asin⁡A=bsin⁡B=csin⁡C\displaystyle \frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}
01

How it works

Build the method from meaning before memorizing the moves.

These are the decisions that make the procedure reliable. Read them once, then look for each idea in the worked examples below.

1

Pair opposites

Each side must be matched with the sine of the angle directly across from it.

2

Choose known pairs

Build a proportion with one complete side-angle pair and the unknown’s pair.

3

Check the ambiguous case

SSA data can produce zero, one, or two triangles because sine has the same value in Quadrants I and II.

02

See the structure

A picture makes the relationships easier to remember.

baseheight
The labeled lengths and right-angle marker connect the diagram to the triangle formulas in the guide.
03

Worked examples

Follow the reason for each line, then try to reproduce it without looking.

Example 1Find a side

A=30∘,a=8,B=75∘\displaystyle A=30^\circ,a=8,B=75^\circ. Find b.

  1. Write 8/sin⁡30=b/sin⁡75\displaystyle 8/\sin30=b/\sin75.
  2. Solve b=8sin⁡75/sin⁡30\displaystyle b=8\sin75/\sin30.
  3. Round only at the end.
Answerb≈15.45\displaystyle b\approx15.45
Example 2Find an angle

a=10,b=7,A=50∘\displaystyle a=10,b=7,A=50^\circ. Find B.

  1. sin⁡B/7=sin⁡50/10\displaystyle \sin B/7=\sin50/10.
  2. Compute sin⁡B=0.7sin⁡50\displaystyle \sin B=0.7\sin50.
  3. Use inverse sine and check possible supplementary angle.
AnswerB≈32.4∘\displaystyle B\approx32.4^\circ
Example 3Impossible data

A=30∘,a=4,b=10\displaystyle A=30^\circ,a=4,b=10.

  1. sin⁡B=10sin⁡30/4=1.25\displaystyle \sin B=10\sin30/4=1.25.
  2. Sine cannot exceed 1.
  3. No triangle fits the data.
AnswerNo triangle
04

Common mistakes—and how to avoid them

Accuracy improves fastest when you know what to check.

  • Pairing a side with an adjacent angle.

  • Ignoring the second possible SSA angle.

  • Rounding intermediate trig values.

05

Turn the explanation into a skill

Reading creates recognition. Independent practice creates recall.

Free printable worksheet · Answer key included

Law of Sines and Cosines

Practice choosing the right non-right-triangle formula.

Preview the worksheet

A strong practice loop: solve one example with the guide open, solve a similar problem without it, explain the method aloud, and return the next day for a short mixed review.

06

Practice problems with worked solutions

Solve each problem on paper, use the hint only if needed, then compare every step.

1

Given A = 30°, a = 8, and B = 45°, find b.

A = 30°B = 45°a = 8b = ?c
Show hint

Use a/sin⁡A=b/sin⁡B\displaystyle a/\sin A=b/\sin B.

Show worked solution
  1. 8/sin⁡30°=b/sin⁡45°\displaystyle 8/\sin30°=b/\sin45°.
  2. b=8(2/2)/(1/2)\displaystyle b=8(\sqrt2/2)/(1/2).

Answer: 82\displaystyle 8\sqrt2

2

Given a = 10, A = 40°, and b = 7, find B.

A = 40°B = ?a = 10b = 7
Show hint

Solve for sin⁡B\displaystyle \sin B.

Show worked solution
  1. sin⁡B=7sin⁡40°/10\displaystyle \sin B=7\sin40°/10.
  2. sin⁡B≈0.450\displaystyle \sin B\approx0.450.
  3. B≈26.7°\displaystyle B\approx26.7°.

Answer: Approximately 26.7°.

3

Given A = 52°, B = 71°, and a = 12, find c.

A = 52°B = 71°a = 12C = ?c = ?
Show hint

Find C first.

Show worked solution
  1. C=180°−52°−71°=57°\displaystyle C=180°-52°-71°=57°.
  2. c/sin⁡57°=12/sin⁡52°\displaystyle c/\sin57°=12/\sin52°.
  3. c≈12.77\displaystyle c\approx12.77.

Answer: Approximately 12.8.

4

Can A = 30°, a = 4, b = 10 form a triangle?

A = 30°b = 10a = 4 reachno intersection
Show hint

Check the sine value for B.

Show worked solution
  1. sin⁡B=bsin⁡A/a=10(0.5)/4=1.25\displaystyle \sin B=b\sin A/a=10(0.5)/4=1.25.
  2. A sine value cannot exceed 1.

Answer: No triangle.

5

Given A = 35°, a = 12, b = 15, explain why two triangles may occur.

A = 35°b = 15a = 12a = 12B₁ ≈ 45.8°B₂ ≈ 134.2°
Show hint

This is the SSA ambiguous case.

Show worked solution
  1. sin⁡B=15sin⁡35°/12≈0.717\displaystyle \sin B=15\sin35°/12\approx0.717.
  2. Both B≈45.8°\displaystyle B\approx45.8° and 180°−45.8°=134.2°\displaystyle 180°-45.8°=134.2° can pair with A because each leaves a positive third angle.

Answer: Two possible triangles.

07

Frequently asked questions

Quick answers before you move on.

What should I remember first?

For triangle sides opposite angles A, B, and C, a/sin⁡A=b/sin⁡B=c/sin⁡C\displaystyle a/\sin A=b/\sin B=c/\sin C. Use it when you know an opposite side-angle pair and another side or angle.

How do I know whether I understand this topic?

You should be able to name the method, explain why each step is valid, complete a new example without copying, and check whether your answer is reasonable.

What should I do if I keep making the same mistake?

Write the error as a specific checkpoint—for example, “I will identify the hypotenuse before substituting.” Then use that checkpoint on three short problems rather than repeating a full page without feedback.

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