The two possible sine-rule angles

For an angle B in a triangle, 0° < B < 180°. If sin B = k with 0 < k < 1, the candidates are the acute angle arcsin k and its supplement 180° − arcsin k. Both have the same sine. When two sides and an angle opposite one of them are supplied, either candidate may be geometrically possible.

Subtract each candidate and the known angle from 180° to find the remaining angle. It must be positive. Use the sine rule again to obtain the remaining side for each valid triangle. If k > 1, no such triangle exists; if k = 1, the angle is 90° and the two branches coincide. A single calculator inverse-sine output does not settle the number of triangles.

Worked example

In triangle ABC, A = 30°, side a = 5 and side b = 8. Find the possible angles B and C to one decimal place.

1. The sine rule gives sin B = b sin A/a = 8(1/2)/5 = 0.8.
2. The candidates are B ≈ 53.1301° and B ≈ 126.8699°.
3. Subtract A and each candidate from 180°: C ≈ 96.8699° or C ≈ 23.1301°. Both are positive.

Answer: Two triangles: (B,C) ≈ (53.1°,96.9°) or (126.9°,23.1°).

Revise first: Sine rule, cosine rule and triangle area, Trigonometric ratios in each quadrant.