Law of Sines

SOH CAH TOA only works for right triangles. But most real-world triangles — a plot of land, a roof truss, a surveying problem — are not right triangles. The law of sines lets you solve any triangle when you have a matched side-angle pair.

The Formula

For any triangle with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

This can also be written with sines in the numerator:

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

The law says: the ratio of any side to the sine of its opposite angle is the same for all three sides of the triangle.

When to Use the Law of Sines

You need the law of sines when you have a matched pair — a side and the angle directly opposite it — plus one more piece of information:

Memory aid: If you know a side and the angle across from it, start with the law of sines. If you don’t have a matched pair (e.g., SAS or SSS), use the law of cosines instead.

Worked Examples

Example 1: Solving a Triangle (AAS)

In triangle $ABC$: $A = 40°$, $B = 75°$, and $a = 10$. Find side $b$.

Step 1 — Find the third angle:

$C = 180° - 40° - 75° = 65°$

Step 2 — Set up the law of sines:

$\frac{a}{\sin A} = \frac{b}{\sin B}$

$\frac{10}{\sin 40°} = \frac{b}{\sin 75°}$

Step 3 — Solve for $b$:

$b = \frac{10 \times \sin 75°}{\sin 40°} = \frac{10 \times 0.9659}{0.6428} = \frac{9.659}{0.6428} = 15.03$

Answer: $b \approx 15.03$

Example 2: Finding All Remaining Parts (ASA)

In triangle $ABC$: $A = 50°$, $C = 70°$, and $b = 12$ (the side between angles A and C). Find all remaining parts.

Step 1 — Find angle $B$:

$B = 180° - 50° - 70° = 60°$

Step 2 — Find side $a$:

$\frac{a}{\sin 50°} = \frac{12}{\sin 60°}$

$a = \frac{12 \times \sin 50°}{\sin 60°} = \frac{12 \times 0.7660}{0.8660} = 10.61$

Step 3 — Find side $c$:

$c = \frac{12 \times \sin 70°}{\sin 60°} = \frac{12 \times 0.9397}{0.8660} = 13.02$

Answer: $B = 60°$, $a \approx 10.61$, $c \approx 13.02$

The Ambiguous Case (SSA)

When you know two sides and an angle opposite one of them (SSA), the law of sines can produce zero, one, or two valid triangles. This is called the ambiguous case, and it is the trickiest part of the law of sines.

Why It Happens

The equation $\sin B = \frac{b \sin A}{a}$ can have:

• No solution — if the result is greater than 1 (impossible for sine)
• One solution — if $B = 90°$, or if the second possible angle makes the angle sum exceed 180°
• Two solutions — if both $B$ and $180° - B$ produce valid triangles

Example 3: SSA with Two Solutions

In triangle $ABC$: $a = 8$, $b = 12$, and $A = 30°$. Find angle $B$.

$\frac{\sin A}{a} = \frac{\sin B}{b}$

$\sin B = \frac{b \sin A}{a} = \frac{12 \times \sin 30°}{8} = \frac{12 \times 0.5}{8} = 0.75$

$B = \sin^{-1}(0.75) = 48.59°$

But sine is also positive in Quadrant II, so $B$ could also be:

$B_2 = 180° - 48.59° = 131.41°$

Check both solutions:

• Solution 1: $A + B = 30° + 48.59° = 78.59°$ → $C = 101.41°$ ✓ (valid)
• Solution 2: $A + B_2 = 30° + 131.41° = 161.41°$ → $C = 18.59°$ ✓ (also valid)

Answer: This triangle has two valid solutions — $B \approx 48.59°$ or $B \approx 131.41°$.

Quick Check for the Ambiguous Case

After finding $\sin B$:

1. If $\sin B > 1$ → no triangle exists
2. If $\sin B = 1$ → one right triangle ($B = 90°$)
3. If $\sin B < 1$ → calculate $B$ and $180° - B$, then check if each makes the angle sum valid

Real-World Application: Surveying — Distance Across a River

A surveyor needs to find the distance to a point across a river without crossing it.

Setup: The surveyor stands at point $A$ and spots a landmark at point $B$ across the river. They walk 100 meters along the riverbank to point $C$. They measure angle $A = 85°$ (between the riverbank and the line of sight to $B$) and angle $C = 62°$.

Step 1 — Find angle $B$:

$B = 180° - 85° - 62° = 33°$

Step 2 — Apply the law of sines. Side $AC = 100$ m is opposite angle $B$.

$\frac{AC}{\sin B} = \frac{AB}{\sin C}$

$AB = \frac{100 \times \sin 62°}{\sin 33°} = \frac{100 \times 0.8829}{0.5446} = 162.1 \text{ m}$

Answer: The landmark is approximately 162 meters from the surveyor’s starting point — measured without ever crossing the river.

Common Mistakes

1. Forgetting the ambiguous case. When using SSA, always check for two possible solutions.
2. Using the law of sines without a matched pair. You need to know a side AND its opposite angle. If you only have SAS or SSS, use the law of cosines.
3. Not finding the third angle first. In AAS/ASA problems, always compute the missing angle before applying the law of sines.
4. Dividing instead of cross-multiplying. Set up $\frac{a}{\sin A} = \frac{b}{\sin B}$, then solve: $b = \frac{a \sin B}{\sin A}$.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The law of sines states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
• Use it when you have a matched side-angle pair (AAS, ASA, or SSA)
• The SSA (ambiguous) case can produce 0, 1, or 2 valid triangles — always check
• If you don’t have a matched pair (SAS or SSS), use the law of cosines instead
• The law of sines is essential for surveying, navigation, and any indirect measurement problem

Return to Trigonometry for more topics in this section.