Worked example

A = 30°, a = 6, b = 10 — solve the triangle

How to decide, before computing

Two sides and an angle that is not between them: the sine rule returns a sine value, and a sine value belongs to two angles. Decide first whether the second angle is allowed at all — the angle sum of a triangle is the referee — and only then compute the sides.

Steps

1. h = b·sin A = 10 · 0.5 = 5
   The height from C onto AB — the number that decides how many triangles exist.
2. h < a < b, so 5 < 6 < 10: two triangles
   The side opposite the given angle reaches past the height but not past the other side.
3. sin B = b·sin A / a = 5/6
   Law of sines, solved for sin B. · Law of sines
4. B ≈ 56.44° or B ≈ 123.56°
   A sine value has two angles in (0°, 180°); both keep A + B < 180°, so both survive.
5. C = 180° − A − B ≈ 93.56° or 26.44°
   The angle sum closes each triangle. · Angles of a triangle add to 180°
6. c = a·sin C / sin A ≈ 11.98 or 5.34
   Law of sines once more — and both answers check out in the cosine rule. · Law of sines

Numbers computed with the site's own engine on real triangles: in both solutions a/sin A and b/sin B give 12.000000000000, and √(b² + c² − 2bc·cos A) returns a = 6.

Identities used

• Law of sines — You have one complete pair — a side together with the angle opposite it — and you need an angle or a side from a second pair. It is also the formula that produces the ambiguous case: with two sides and a non-included angle, sin B can come from two different angles.
• Angles of a triangle add to 180° — Two angles are known and the third must be found before any side work starts — this is the step that turns 'two angles' into a usable pair for the sine rule. It also tells you when a proposed triangle cannot exist.

Check it yourself

Open the verifier with the main identity pre-filled — substituting your own numbers into a step is the fastest way to find out which line you did not follow.

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