Solve for missing sides or angles in a triangle that isn't right-angled, using the two laws that extend SOHCAHTOA beyond right triangles. Every question comes with a written walkthrough of which law applied and why.
Without a matching angle-side pair given directly, the Law of Sines has no ratio to set equal to another — SAS and SSS situations need the Law of Cosines instead.
An SSA situation with the Law of Sines can yield two possible angles, one acute and one obtuse, that both satisfy the sine ratio — checking whether both keep the triangle's angles summing to less than 180° determines how many triangles are actually valid.
The angle in c² = a² + b² − 2ab·cos(C) has to be the one included between sides a and b — plugging in a different angle solves for the wrong side entirely.
Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.
In triangle ABC, a=7, b=9, and angle C=60°. Find side c. (Round to two decimal places.)
A — 8.19. Apply the Law of Cosines: c² = 7² + 9² − 2(7)(9)cos(60°) = 49+81−126(0.5) = 130−63 = 67, so c = √67 ≈ 8.19. 11.40 (B) comes from applying the Pythagorean theorem directly — treating the triangle as right-angled and skipping the −2ab·cos(C) correction term: √(49+81) ≈ 11.40. 13.89 (C) comes from adding the correction term instead of subtracting it: √(49+81+63) ≈ 13.89.
In triangle ABC, angle A=40°, angle B=75°, and side a=12. Find side b. (Round to two decimal places.)
A — 18.03. Apply the Law of Sines: a/sin(A) = b/sin(B), so 12/sin(40°) = b/sin(75°). Solving for b: b = 12·sin(75°)/sin(40°) ≈ 12(0.9659)/0.6428 ≈ 18.03. 7.99 (B) comes from inverting the ratio — computing b = 12·sin(40°)/sin(75°) instead, swapping which angle's sine belongs in the numerator.
In triangle ABC, a=10, b=14, and angle A=35°. How many distinct triangles can be formed with these measurements?
A — 2. This is an SSA setup, which requires checking the ambiguous case. Using the Law of Sines: sin(B) = b·sin(A)/a = 14·sin(35°)/10 ≈ 0.803, giving B ≈ 53.4° or B ≈ 180°−53.4° = 126.6°. Checking both: A+B = 35°+53.4° = 88.4° < 180°, and A+B = 35°+126.6° = 161.6° < 180° — both remain valid triangles, so there are 2 distinct triangles possible. 1 (C) comes from only checking the acute solution for B and assuming the obtuse supplement must automatically be invalid, without actually checking whether A+B stays under 180°.
Law of Sines and Law of Cosines problems are almost entirely about recognizing which law fits the given information — the trig calculations themselves are usually the easy part once the setup is right.
Before touching a formula, name what you have — SAS, SSS, AAS, ASA, or SSA — and match it to the right law. Untimed practice is where that judgment call gets fast.
Move to timed sessions once law selection is instant. These take longer than right-triangle trig since there's often a calculator step and a check to run.
Pair this with SOHCAHTOA and standard angles in a mock — both draw on the same trig ratio foundation.
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