Law of Sines & Law of Cosines: Solving Any Triangle
Solve any triangle — not just right triangles — with the Law of Sines and the Law of Cosines, including how to choose between them, with worked ACT problems.
The Short Version
- SOH-CAH-TOA only handles right triangles; these two laws handle any triangle.
- Law of Sines: a/sin A = b/sin B = c/sin C — use with a known angle-side pair.
- Law of Cosines: c² = a² + b² − 2ab·cos C — use for SAS or SSS.
- Choose by what you're given: an angle opposite a known side → Sines; two sides + included angle (or all three sides) → Cosines. An ACT / Pre-Calc topic.
SOH-CAH-TOA is powerful, but it has a hard limit: it only works on right triangles. Plenty of triangles — on the ACT and in the real world — have no right angle at all. The Law of Sines and the Law of Cosines remove that limitation, letting you find missing sides and angles in any triangle. The two laws cover different situations, so the real skill is recognizing which one a problem calls for.
This guide explains both laws, how to label a triangle, and how to choose between them, with worked and practice problems matched to real ACT difficulty at Northside Tutoring.
Why These Laws Matter
The ACT includes trigonometry questions involving non-right ("oblique") triangles, where SOH-CAH-TOA simply doesn't apply. The Law of Sines and Law of Cosines are the tools for those problems, and they're core Pre-Calculus content. (The SAT generally doesn't test them, and they're well beyond the SSAT.)
Labeling Any Triangle
The standard convention: label the angles A, B, C and the side opposite each angle with the matching lowercase letter — side a is across from angle A, and so on.
Each lowercase side sits opposite the capital angle of the same letter — the key to applying both laws correctly.
The Law of Sines
The Law of Sines relates each side to the sine of its opposite angle:
Use it when you know an angle and its opposite side, plus one more piece — for example, two angles and a side (AAS or ASA), or two sides and a non-included angle. Set two of the ratios equal and solve.
The Law of Cosines
The Law of Cosines is a generalization of the Pythagorean theorem that works for any triangle:
(The same pattern works for any side: just put the side you want on the left and its opposite angle in the cosine.) Notice that when C = 90°, cos C = 0 and it reduces to a² + b² = c² — the Pythagorean theorem. Use it for two sides and the included angle (SAS), or all three sides (SSS).
Which Law to Use
| What you're given | Use |
|---|---|
| Two angles + a side (AAS, ASA) | Law of Sines |
| A side, its opposite angle, + one more | Law of Sines |
| Two sides + the included angle (SAS) | Law of Cosines |
| All three sides (SSS) | Law of Cosines |
The quick decision
If you have an angle paired with the side across from it, reach for the Law of Sines. If you're missing that pairing — you have two sides and the angle between them, or all three sides — use the Law of Cosines.
A Note on the Tricky Case
The Law of Sines has one well-known complication: the "ambiguous case" (given two sides and a non-included angle, SSA), where there can be two possible triangles. On the ACT this is rare, but be aware that solving for an angle with the Law of Sines can sometimes have two valid answers (an angle and its supplement). When in doubt, sketch the triangle to see which fits.
Where You'll See This — Test by Test
No reference sheet provides these laws — you must know them. The ACT tests trigonometry of non-right triangles; the SAT generally doesn't, and they're beyond the SSAT. They're core Pre-Calculus content.
ACT
The ACT tests oblique-triangle trig — the Law of Sines and Law of Cosines — that SOH-CAH-TOA can't handle.
Explore ACT Tutoring → College AdmissionsSAT
The SAT generally doesn't test these laws; this is more of an ACT and Pre-Calculus topic.
Explore SAT Tutoring → K-12 CurriculumPre-Calculus
Solving triangles is a core Pre-Calculus unit, extending right-triangle trigonometry.
Explore Math Tutoring → K-12 CurriculumAlgebra II
Builds on the trig from Algebra II and the unit circle.
Explore Algebra Tutoring →Watch the Lesson
Sometimes a diagram needs a voice. In the short video below, one of our Northside tutors walks through the core idea and works through test-style problems in real time.
Law of Sines & Cosines — In Plain English
A live walkthrough from our tutoring team.
— Featuring a Northside Tutoring instructor
Worked Example Problems
These problems are calibrated to the difficulty you'll actually see on test day. Try each one before opening the solution.
In a triangle, A = 40°, B = 60°, and side a = 10. Find side b. (sin 40° ≈ 0.643, sin 60° ≈ 0.866)
Show solution
Law of Sines: a/sin A = b/sin B, so b = a·sin B / sin A = 10(0.866)/0.643 ≈ 13.5.
A triangle has a = 5, b = 7, and included angle C = 60°. Find side c.
Show solution
Law of Cosines: c² = 5² + 7² − 2(5)(7)cos 60° = 25 + 49 − 70(0.5) = 74 − 35 = 39. So c = √39 ≈ 6.24.
You're given all three sides of a triangle and asked for an angle. Which law applies?
Show solution
The Law of Cosines — it handles SSS. Rearrange it to solve for the cosine of the angle.
A triangle has sides a = 6, b = 8, c = 10. Find angle C using the Law of Cosines.
Show solution
cos C = (a² + b² − c²)/(2ab) = (36 + 64 − 100)/(2·6·8) = 0/96 = 0, so C = 90°. (It's a right triangle — the law confirms it.)
Given A = 35°, B = 80°, and side c = 12, which law finds side a, and what's the first step?
Show solution
Law of Sines. First find C = 180 − 35 − 80 = 65°, then a/sin A = c/sin C gives a = 12·sin 35°/sin 65°.
Common Mistakes to Avoid
Three traps that catch students every year
- Using SOH-CAH-TOA on a non-right triangle. Those ratios only work with a right angle — switch to the Law of Sines or Cosines.
- Mislabeling sides and angles. Each lowercase side is opposite its matching capital angle; pairing them wrong breaks both laws.
- Picking the wrong law. Angle-opposite-side pairing → Sines; SAS or SSS → Cosines.
Practice Problems — You Try
Three problems below. Work each before checking the solution.
A = 45°, a = 12, B = 60°. Find b. (sin 45° ≈ 0.707, sin 60° ≈ 0.866)
Show solution
b = a·sin B/sin A = 12(0.866)/0.707 ≈ 14.7.
Which law do you use given two sides and the angle between them (SAS)?
Show solution
The Law of Cosines.
A triangle has a = 7, b = 9, and included angle C = 40°. Find side c. (cos 40° ≈ 0.766)
Show solution
c² = 7² + 9² − 2(7)(9)cos 40° = 49 + 81 − 126(0.766) = 130 − 96.5 = 33.5. So c = √33.5 ≈ 5.79.
The Northside Method — How We Teach This 1-on-1
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