Solve any triangle using the Law of Sines. Enter two angles and one side or two sides and one angle to find all missing values, with ambiguous-case detection.
The Law of Sines states that the ratio of the length of a side to the sine of its opposite angle is the same for all three sides of a triangle: a/sin A = b/sin B = c/sin C. This powerful relationship allows you to solve any triangle when you know at least one side and two other measurements (angles or sides). Our Law of Sines Triangle Solver handles three common input configurations: two angles and one side (AAS or ASA), and two sides and one angle (SSA). The SSA case is particularly interesting because it can produce zero, one, or two valid triangles—known as the ambiguous case. This calculator automatically detects the ambiguous case and presents both possible solutions when they exist. Beyond finding the missing sides and angles, it also computes the triangle's area, perimeter, circumradius, inradius, and all three altitudes. Whether you're studying trigonometry, surveying land, or solving engineering problems, this tool gives you a complete picture of any triangle from minimal input. Eight built-in presets let you explore classic configurations instantly, and a reference table summarises the key formulas for quick review.
The Law of Sines is straightforward in clean textbook cases, but the SSA configuration can become messy fast because you must check whether no triangle, one triangle, or two different triangles satisfy the inputs. This calculator is useful because it handles that branch logic automatically and then reports not just the missing sides and angles, but also the downstream properties that matter in applications such as area, radii, and altitudes.
Law of Sines: a / sin A = b / sin B = c / sin C. Area = ½ · a · b · sin C. Circumradius R = a / (2 sin A). Inradius r = Area / s, where s = (a + b + c) / 2.
Result: A = 40°, B = 60°, C = 80°, b ≈ 13.47, c ≈ 15.32, area ≈ 66.31
Given angle A = 40°, angle B = 60°, and side a = 10: angle C = 180° − 40° − 60° = 80°. By the Law of Sines, b = 10 · sin 60° / sin 40° ≈ 13.47, and c = 10 · sin 80° / sin 40° ≈ 15.32. Area ≈ ½ · 10 · 13.47 · sin 80° ≈ 66.31.
The Law of Sines is most useful when you know at least one side-angle opposite pair. That makes it ideal for AAS, ASA, and many SSA problems. In contrast, SSS and SAS problems usually point to the Law of Cosines first. Recognizing that distinction saves time and helps prevent circular work where you try to force the wrong formula onto the wrong kind of triangle.
SSA inputs are special because the sine function does not distinguish between an acute angle and its supplementary obtuse partner. If side a, side b, and angle A are compatible, angle B may come out as either B or 180° − B. Sometimes both options create valid triangles, sometimes only one works, and sometimes neither works because the geometry cannot close. That is why the calculator may return two solutions, and why a manual setup must always check the angle sum after using inverse sine.
The computed perimeter, area, circumradius, inradius, and altitudes are not just bonus values. They show how one triangle solution can imply very different geometric behavior from another in the ambiguous case. Two SSA solutions can share the same given data yet produce different heights, different third sides, and different enclosed areas. Looking at those outputs side by side gives a better geometric intuition than solving only for the missing angle and stopping there.
The ambiguous case arises when you know two sides and a non-included angle (SSA). Depending on the measurements, there may be zero, one, or two valid triangles that satisfy the given information.
Use the Law of Sines when you know at least one side-angle pair (a side and its opposite angle). Use the Law of Cosines when you have SAS or SSS configurations.
Yes, but basic trigonometric ratios (SOH-CAH-TOA) are usually simpler for right triangles. The Law of Sines is most useful for oblique (non-right) triangles.
The circumradius R is the radius of the circle that passes through all three vertices. It equals a / (2 sin A) for any side a and its opposite angle A.
Once you know two sides and their included angle, use Area = ½ · a · b · sin C. You can derive this even if the included angle was not given directly.
It means no triangle exists with those measurements. The given side opposite the known angle is too short to form a triangle.