Recipe

Choose the right law and solve an oblique triangle

Use when: you're given three pieces of a triangle (some mix of angles and sides), no right angle is guaranteed, and you need to find the rest

Goal: every angle and side of the triangle, using the correct law from the start instead of guessing and backtracking

Taught in Law of Cosines: SAS and SSS triangles

Quick recipe

Count what's given: angles vs. sides
AAS/ASA: find the third angle, then Law of Sines twice
SSA: run the swing-arm test before solving anything
SAS/SSS: reach for the Law of Cosines first

1
Sort the three known pieces into angles and sides. That count alone decides which law applies, before any arithmetic starts.
- Two angles and any one side (AAS or ASA) — go to Step 2.
- Two sides and a non-included angle (SSA) — go to Step 3.
- Two sides and the included angle (SAS), or three bare sides (SSS) — go to Step 4.
Why: the Law of Sines needs a complete angle-side pair to build its proportion; the Law of Cosines needs sides squeezed around a known (or solvable) angle instead. Counting angles versus sides tells you immediately which ingredient you have.
2
Two angles pin the third down instantly: $180°$ minus their sum, with no ambiguity possible. Once all three angles are known, every side has a matching angle across from it, so apply the Law of Sines twice to find the two missing sides.

Example

For $A=40°$ , $B=75°$ , $a=15$ : $C=180°-40°-75°=65°$ . Then $b=\dfrac{15\sin75°}{\sin40°}\approx22.54$ and $c=\dfrac{15\sin65°}{\sin40°}\approx21.15$ .

Why: two known angles always determine the third with total certainty, so this case never branches and never produces more than one triangle.
3
Compute the test height $h=b\sin A$ first. Compare side $a$ to $h$ and to $b$ : $a<h$ means no triangle; $a=h$ means exactly one (a right triangle at the tangent point); $h<a<b$ means two triangles; $a\ge b$ means exactly one.
- zero or one triangle — solve that single triangle with the ordinary Law of Sines.
- two triangles — solve both candidate triangles separately (using both $\sin^{-1}$ solutions), and keep both sets of answers unless the problem's own context rules one out.
Why: SSA is the one setup where jumping straight to $\sin^{-1}$ without checking the triangle count first can silently throw away a second valid triangle, or claim a triangle exists where none does.
4
With two sides and the included angle, or three sides alone, there's no complete angle-side pair to build a Law of Sines proportion from. For SAS, use the Law of Cosines to find the missing side, then finish with a second Law of Cosines application and the angle sum. For SSS, find the angle facing the longest side first, since that's the only angle capable of being obtuse.

⚡ Shortcut

Once one angle is found by the Law of Cosines rearranged to isolate cosine, it's tempting to finish the triangle with the Law of Sines instead of a second Law of Cosines pass. That's fine for the last angle (found from the $180°$ sum, no inverse trig needed), but risky for a second inverse-trig angle, since $\sin^{-1}$ can't tell an angle from its supplement. Stick with $\cos^{-1}$ for any angle you still need to solve for directly.

Why: with fewer than two known angle-side pairs, the Law of Sines has nothing to divide by, while the Law of Cosines only ever needs sides plus one angle, known or solvable.

Common mistakes

Reaching for the Law of Cosines before checking whether two angles are already given; the Law of Sines is faster whenever AAS or ASA applies.
Solving an SSA triangle with the Law of Sines without running the $h$ -versus- $a$ -versus- $b$ test first, and missing a second valid triangle.
Using $\sin^{-1}$ to find a second angle after a Law of Cosines solution, instead of the safer $\cos^{-1}$ , which never needs a supplement check.
Forgetting that AAS and ASA run the identical procedure regardless of where the given side sits: there's no separate "included side" algorithm to look up.