The formula
In any triangle, each side divided by the sine of its opposite angle gives the same number:
a / sin A = b / sin B = c / sin C
Here lowercase letters are side lengths and capital letters are the angles opposite those sides. That pairing is the whole trick: side a faces angle A, side b faces angle B, side c faces angle C. Bigger angles always face bigger sides, and the Law of Sines makes that relationship exact.
Where the Formula Actually Comes From
The Law of Sines isn't a rule to memorize blindly — it falls straight out of one simple trick: dropping an altitude and writing its length two different ways.
Drop an altitude from C to side c (AB)
Call the foot of that altitude D, and its length h. This splits triangle ABC into two right triangles, ADC and BDC, sharing the same height h.
Write h two different ways
In right triangle ADC, h is opposite angle A with hypotenuse b, so h = b sin A. In right triangle BDC, h is opposite angle B with hypotenuse a, so h = a sin B.
Set the two expressions equal
b sin A = a sin B, which rearranges to a/sin A = b/sin B. Drop a second altitude — say from A to side a — and the same argument gives b/sin B = c/sin C, completing the full three-way ratio.
The same setup explains the "Good to Know" fact above: since Area = ½ab sin C = ½bc sin A = ½ac sin B, and each of those area expressions can be rewritten in terms of the circumradius R, the shared ratio a/sin A always equals 2R. It's the same underlying geometry showing up twice.
When do you need it?
SOH-CAH-TOA only works in right triangles. The Law of Sines works in every triangle. You can use it whenever you know one complete angle-side pair (an angle and its opposite side), plus one more angle or side:
- ASA — two angles and the side between them → Law of Sines solves the rest.
- AAS — two angles and a side not between them → same idea, still solvable.
- SSA — two sides and a non-included angle → Law of Sines, but watch for the ambiguous case below.
- Right triangle? SOH-CAH-TOA is faster, but the Law of Sines still holds true.
Worked example
Problem: A triangle has angle A = 50°, angle B = 60°, and side a = 12 cm. Find side b.
Write the ratio with the known pair
a / sin A = 12 / sin 50°.
Set it equal to the target's ratio
12 / sin 50° = b / sin 60°.
Solve for b
b = 12 × sin 60° / sin 50° = 12 × 0.866 / 0.766 ≈ 13.57 cm.
Sanity check: B (60°) is bigger than A (50°), so side b should be longer than side a — and 13.57 > 12. It checks out.
Finding the third angle first
Because a triangle's angles always sum to 180°, two known angles instantly give the third: with 50° and 60°, angle C = 180 − 110 = 70°. Then side c follows from the same ratio: c = 12 × sin 70° / sin 50° ≈ 14.72 cm. One known pair unlocks the entire triangle.
The shared ratio in a/sin A = b/sin B = c/sin C actually equals the diameter of the triangle's circumscribed circle (2R). It's a neat identity worth remembering for harder problems.
The ambiguous case (SSA) — the classic exam trap
When you know two sides and a non-included angle, the sine ratio can produce two valid triangles. Because sin θ = sin (180° − θ), an answer of, say, B = 48° also allows B = 132°. Always check: does the second angle still fit inside 180° with the given angle? If yes, there are two solutions; state both unless the context rules one out.
Calculators only return the acute angle from an inverse sine. In SSA problems, always test whether 180° minus that angle also fits — skipping this step is the single most common Law of Sines exam error.
Worked Example: A Genuine Two-Triangle Case
Problem: A triangle has side a = 20, side b = 26, and angle A = 40° (opposite side a). Find all possible values of angle B.
Check whether the ambiguous case applies
Compare a to b sin A: b sin A = 26 × sin 40° ≈ 26 × 0.643 ≈ 16.7. Since 16.7 < a (20) < b (26), two distinct triangles are possible.
Solve for the first (acute) angle B
sin B = b sin A / a = 16.7 / 20 ≈ 0.836, so B₁ ≈ 56.7°. Then C₁ = 180° − 40° − 56.7° ≈ 83.3°, and c₁ = a sin C₁ / sin A ≈ 30.9.
Solve for the second (obtuse) angle B
B₂ = 180° − 56.7° ≈ 123.3°. This still leaves room for a positive third angle: C₂ = 180° − 40° − 123.3° ≈ 16.7°, and c₂ = a sin C₂ / sin A ≈ 8.9.
State both triangles
Triangle 1: B ≈ 56.7°, C ≈ 83.3°, c ≈ 30.9. Triangle 2: B ≈ 123.3°, C ≈ 16.7°, c ≈ 8.9. Both satisfy every original condition (a = 20, b = 26, A = 40°) — that's exactly what makes SSA "ambiguous."
Given sides a, b and angle A opposite a: if a < b sin A, no triangle exists (the side is too short to reach the base). If a = b sin A, exactly one right triangle exists. If b sin A < a < b, two triangles exist (the case above). If a ≥ b, only one triangle exists, because the second solution's angle would push the angle sum past 180°.
Worked Example: Finding a Height You Can't Directly Measure
Problem: A surveyor stands at point P and measures the angle of elevation to the top of a tower as 30°. Walking 50 m directly toward the tower to point Q, the angle of elevation increases to 45°. How tall is the tower?
This is the exact kind of problem the Law of Sines was built for: you have two angles and one side, but no right triangle to drop SOH-CAH-TOA into right away — because you don't yet know the distance from either point to the tower's base.
Build a triangle out of what you know
Let T be the top of the tower. In triangle PQT, angle P = 30° (the elevation from P). Since P, Q, and the tower's base are in a straight line, the interior angle at Q inside triangle PQT is the supplement of its elevation angle: angle Q = 180° − 45° = 135°.
Find the third angle and apply the Law of Sines
Angle T = 180° − 30° − 135° = 15°. Now PQ (=50) is opposite angle T, and PT (the slant distance to the top) is opposite angle Q, so: PT / sin(135°) = 50 / sin(15°).
Solve for PT
PT = 50 × sin(135°) / sin(15°) ≈ 50 × 0.7071 / 0.2588 ≈ 136.6 m.
Drop into a right triangle for the final height
Now that PT is known, the height is a simple right-triangle step: h = PT × sin(30°) ≈ 136.6 × 0.5 ≈ 68.3 m.
This "two angles of elevation, one baseline" setup is one of the most common real-world uses of the Law of Sines — surveying, navigation, and astronomy all lean on the same idea: turn an unreachable distance into an angle-side pair you can measure, then let the ratio do the rest. The full method, with more practice problems, is covered in the Heights & Distances lesson.
Law of Sines vs Law of Cosines
Use this decision rule:
• Know an angle-side pair? → Law of Sines (ASA, AAS, SSA).
• Know two sides + the included angle (SAS), or all three sides (SSS)? → Law of Cosines: c² = a² + b² − 2ab·cos C.
• Right triangle? Plain SOH-CAH-TOA is fastest (though both laws still hold).
| Tool | What you must already know | Core formula | Watch out for |
|---|---|---|---|
| Law of Sines | One full angle-side pair (ASA, AAS, or SSA) | a/sin A = b/sin B = c/sin C | SSA can have 0, 1, or 2 solutions |
| Law of Cosines | Two sides + the included angle (SAS), or all three sides (SSS) | c² = a² + b² − 2ab·cos C | Always gives one unambiguous answer — no second-solution check needed |
| SOH-CAH-TOA | A right angle, plus one more side or angle | sin/cos/tan of the known angle | Only valid when a 90° angle is actually present |
How to Solve Any Oblique Triangle: A 4-Step Checklist
Whatever the problem looks like, the same short checklist gets you to the right tool every time.
List what you're given
Write down every known side and angle, and label them clearly (a, b, c and A, B, C). Half of all triangle-solving mistakes come from mislabeling which side is opposite which angle before any calculation even starts.
Check for a complete angle-side pair
Do you have one angle and its opposite side, both known? If yes, you're in Law of Sines territory (ASA, AAS, or SSA). If no — meaning you only have SAS or SSS — go to the Law of Cosines instead.
Set up and solve the ratio
Write the known ratio, set it equal to the ratio containing your unknown, and cross-multiply. If you're solving for an angle, remember the inverse sine only returns one of two possible answers.
If it's SSA, run the ambiguous-case check
Compare your given side to b sin A (see the info box above) before you commit to a single answer. Then verify: do your three final angles sum to exactly 180°? If not, a rounding or setup error slipped in somewhere.
Mistakes That Aren't the Ambiguous Case
The SSA trap gets most of the attention, but it isn't the only place students lose marks on Law of Sines problems.
- Degrees vs. radians. A calculator left in the wrong angle mode will silently return a nonsense value with no error message — always confirm the mode before you start, not after you get a strange answer.
- Mismatched side-angle pairs. The ratio only works when a side is paired with the angle directly opposite it. Pairing side a with angle B by mistake is one of the most common setup errors.
- Rounding too early. Carrying a rounded sine value into two or three further steps compounds the error. Keep full calculator precision until the final answer, then round once.
- Forgetting the angle-sum check. After solving a triangle, add all three angles. If they don't total 180°, work back through your steps — this single check catches most arithmetic slips.
- Assuming SSA always has two solutions. It doesn't — use the comparison in the info box above rather than always defaulting to "there must be two answers."
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What to Remember
The Essential Points
- a/sin A = b/sin B = c/sin C holds true for every triangle, not just right triangles
- You need one complete angle-side pair before the Law of Sines can solve the rest
- In SSA problems, always check for the ambiguous second (obtuse) solution
- SAS and SSS need the Law of Cosines instead — no angle-side pair means Sines can't start
- The shared ratio equals 2R, the diameter of the triangle's circumscribed circle
- The formula comes from writing one altitude's length two different ways in two right triangles — it isn't arbitrary
- For SSA, compare your given side to b sin A first: that single comparison tells you whether to expect 0, 1, or 2 triangles
Quick FAQ
Q: Does the Law of Sines apply to all triangles? Yes — acute, obtuse and right triangles alike. You just need one angle-side pair to start.
Q: Why does my calculator give the wrong angle in SSA problems? It returns the acute solution only; check whether the obtuse supplement also forms a valid triangle.
Q: What does the shared ratio actually equal? The diameter of the triangle's circumscribed circle (a/sin A = 2R) — a neat fact worth remembering.
Q: How do I find a triangle's area using the Law of Sines setup? If you know two sides and the included angle, Area = ½ab sin C (or the equivalent with any other side pair and their included angle) — no altitude needs to be drawn separately.
Q: How many solutions can an SSA (ambiguous case) problem have? Zero, one, or two, depending on how the given side compares to b sin A. Zero if it's too short to reach the base, one if it exactly matches (a right angle), and two if it falls between b sin A and b itself.
Q: Does the Law of Sines still work if I use radians instead of degrees? Yes — the formula itself doesn't care which angle unit you use. Just make sure your calculator's mode matches whichever unit you're working in, since mixing the two is a common source of wrong answers.
Further Practice on BitWithBite
Law of Sines — full lesson, part of the free BitWithBite Trigonometry course.
Law of Sines worksheet — printable PDF with a complete answer key.
Heights & Distances — a real-world application that uses the Law of Sines directly.
This article is for educational purposes. Free lessons, worksheets, and answer keys are available on BitWithBite for every topic covered here.