Triangle calculator

Fill in any three of the six boxes and leave the other three empty. At least one has to be a side. The solver works out which case you've given it, solves the triangle and shows each step with the rule it used.

Side a always faces angle A, b faces B and c faces C. So if you know two sides and the angle where they meet, put the angle in the box whose letter isn't one of those sides. With sides a and b, the angle between them is C.

Sides can be whole numbers, decimals, fractions like 3/4 or roots like 5√2 (type sqrt(2) if √ is hard to reach). Angles go in degrees, or switch to radians and type 0.6 or π/3. Whatever you type stays in your browser.

Exact answers appear beside the decimals when the inputs allow it. Whole-number or decimal sides, and angles that are whole multiples of 15°, give results like 12√5 or 10√3 − 10. Anything else gets decimals only.

Take a = 7, b = 8 and c = 9. The law of cosines turns three sides into an angle: cos A = (b² + c² − a²) / 2bc = (64 + 81 − 49) / 144 = 2/3, so A = arccos(2/3) = 48.1897°.

The same rule gives B = 58.4119°. The angles of a triangle add up to 180°, so C = 180° − 48.1897° − 58.4119° = 73.3985°.

For the area, Heron's formula needs only the sides. The semiperimeter is s = (7 + 8 + 9)/2 = 12, and the area is √(12 × 5 × 4 × 3) = √720 = 12√5, about 26.8328.

The calculator works out the largest angle first. A cosine can tell an obtuse angle from an acute one and a sine can't, and only the largest angle can be obtuse.

With A = 45°, B = 60° and c = 10, the third angle comes first: C = 180° − 45° − 60° = 75°. Every angle is now known, and the law of sines scales the one side you have into the other two.

a = c sin A / sin C = 10 × sin 45° / sin 75° = 7.3205, and b = 10 × sin 60° / sin 75° = 8.9658. Since 75° is a multiple of 15°, its sine is exact, (√6 + √2)/4, and a works out to exactly 10√3 − 10.

Two sides and an angle that isn't between them can describe two different triangles. Try a = 6, b = 8 and A = 35°. The law of sines gives sin B = 8 sin 35° / 6 = 0.7648.

Two angles below 180° have that sine, 49.8864° and 130.1136°. Both leave room for a third angle, so both triangles exist. In one, C = 95.1136° and c = 10.419. In the other, C = 14.8864° and c = 2.6874.

Picture side a as a stick hinged at the top of side b, swinging down to the base line. If it's long enough to reach, it can land in two places. Shorten it to a = 3 and sin B would be 1.5295, which no angle has, so the stick misses the line and no triangle fits. Make it longer than b, as with a = 9, and only one landing point is left on the right side of the hinge.

The calculator shows both triangles when there are two, each with its own drawing and working.

The area is worked out two ways, as a check. Heron's formula uses the three sides, area = √(s(s − a)(s − b)(s − c)) with s half the perimeter. The other way is ½ab sin C, two sides and the angle between them.

Each height (altitude) runs from a corner at right angles to the opposite side, and h = 2 × area ÷ that side. In an obtuse triangle two of the heights land outside the triangle, on the opposite side extended, and the drawing shows the base extended as a dotted line when that happens to h_c.

The inradius r is the radius of the largest circle that fits inside, area ÷ s. The circumradius R is the radius of the circle through all three corners, a ÷ (2 sin A), and the law of sines says every side over the sine of its angle gives the same 2R.

Eight questions on one page, each with a sketch to scale. Two right-triangle questions for SOH CAH TOA come first, then one each for SSS, SAS, ASA and AAS, an ambiguous case with two answers, and an area from two sides and an angle.

Every sheet number prints its own questions, and the second page is the answer key, worked by the same solver as the calculator. Print on A4 or US Letter.