Triangle Calculator
Any three values give every side, angle, height, median, radius and centre — with steps.
Solution
This is the last solution worked out. Fix the input above to update it.
Sides, angles, area
Heights, medians, bisectors and radii
Centres
Step by step
About the Triangle Calculator
Type any three of a triangle’s sides and angles — at least one of them a side — and the calculator works out the rest with the law of sines and the law of cosines, showing each step. It recognises every case: three sides (SSS), two sides and the angle between them (SAS), two sides and another angle (SSA, where it gives both triangles when two fit), and two angles with a side (ASA and AAS). You can also start from a right triangle with the right angle fixed, or from the coordinates of the three corners.
You get all three sides and angles, the area, perimeter, the three heights, medians and angle bisectors, the inradius, circumradius and exradii, and the triangle’s type. The centroid, incentre, circumcentre, orthocentre and nine-point centre are given as coordinates, and a drawing to scale can show the incircle, circumcircle, medians, altitudes, angle bisectors and the Euler line.
How to use it
- Choose “Sides and angles”, “Right triangle” or “Corner coordinates”.
- Type three values (two for a right triangle) and leave the others empty — side a is opposite angle A, b opposite B, c opposite C. Angles can be degrees, degrees–minutes–seconds or radians.
- Read the solution, the tables and the steps. If two triangles fit (the SSA ambiguous case), switch between them above the result.
- Tick what to show on the drawing — incircle, circumcircle, medians, altitudes, bisectors or the Euler line — and copy or download the full solution.
Examples
a = 13, b = 14, c = 15
area 84, R = 8.125, r = 4
Heron: s = 21, area = √(21 × 8 × 7 × 6) = 84; R = abc/(4 × area) = 2730/336; r = area/s.
b = 5, c = 7, A = 60°
a = √39 ≈ 6.2450
Law of cosines: a² = 25 + 49 − 2 × 5 × 7 × cos 60° = 39.
a = 6, b = 10, A = 30°
B = 56.4427° or 123.5573°
sin B = 10 × sin 30° / 6 = 0.8333. Both angles fit, giving c = 11.9769 or c = 5.3436.
A = 30°, B = 45°, a = 10
C = 105°, b = 14.1421, c = 19.3185
a / sin A = 20, so b = 20 sin 45° and c = 20 sin 105°.
A (0, 0), B (6, 0), C (2, 4)
area 12, circumcentre (3, 1), orthocentre (2, 2)
Area = ½ |6 × 4 − 0 × 2| = 12; R = √10 ≈ 3.1623.
Common uses
- Checking trigonometry homework on the laws of sines and cosines, including the ambiguous case.
- Surveying and carpentry: finding a distance or an angle you cannot measure directly from three you can.
- Finding where a triangle’s centres are — for a circle through three points (the circumcircle) or the largest circle inside a triangle (the incircle).
- Teaching: showing how medians, altitudes and bisectors meet, and that O, G, N and H always lie on one line.
The laws of sines and cosines
With side a opposite angle A and so on, the law of sines says a / sin A = b / sin B = c / sin C = 2R, where R is the circumradius. The law of cosines says c² = a² + b² − 2ab cos C (and likewise for the other sides); for C = 90° it is Pythagoras’ theorem. SSS and SAS are solved with the law of cosines, which never leaves any doubt about an angle. ASA and AAS first use the angle sum (A + B + C = 180°) and then the law of sines.
The ambiguous case (SSA)
When you know two sides and an angle that is not between them, the law of sines gives sin of the second angle, and two angles in a triangle share each sine: θ and 180° − θ. If both fit with the known angle, two different triangles have your values; if sin would be more than 1, the opposite side is too short and no triangle exists; if it is exactly 1, there is one right triangle. The calculator checks every possibility and shows each triangle that exists.
Lines and centres of a triangle
- Height (altitude) to side a: h_a = 2 × area / a. Median to a: m_a = ½√(2b² + 2c² − a²). Bisector of A: t_a = √(bc((b + c)² − a²)) / (b + c).
- Inradius r = area / s and circumradius R = abc / (4 × area), with s the semi-perimeter; exradius r_a = area / (s − a).
- The centroid (where the medians meet), incentre (bisectors), circumcentre (perpendicular bisectors) and orthocentre (altitudes) are X(2), X(1), X(3) and X(4) in Clark Kimberling’s Encyclopedia of Triangle Centers; the nine-point centre X(5) is halfway between the circumcentre and the orthocentre.
- The circumcentre O, centroid G, nine-point centre N and orthocentre H lie on the Euler line, with OG : GH = 1 : 2 and OH² = 9R² − (a² + b² + c²).
Limitations
- Flat (Euclidean) triangles only — not triangles on a sphere, such as long routes on the Earth.
- Exactly three values are used (two in the right-triangle mode). Extra values are not checked against each other, so clear them instead.
- Calculations use double-precision numbers (about 15 significant digits); results are rounded to the decimal places you choose. Exact forms (√39, cos A = 5/7) are shown where the inputs allow.
- Without coordinates, the triangle is placed with A at (0, 0) and B on the positive x-axis, so the centres are given in that frame.
Privacy
Everything happens in your browser. What you enter or open here is not uploaded or stored by MySmartCoPilot.
Frequently asked questions
What do SSS, SAS, ASA, AAS and SSA mean?
They say which three parts you know, in order around the triangle: S for a side, A for an angle. SSS is three sides; SAS two sides and the angle between them; ASA two angles and the side between them; AAS two angles and a side that is not between them; SSA two sides and an angle that is not between them. All of these fix the triangle except SSA, which can have two solutions, and AAA, which fixes only the shape.
Why does the calculator give two triangles?
You entered two sides and an angle that is not between them (SSA). The law of sines gives the sine of another angle, and both θ and 180° − θ have that sine. When both angles leave room for the third, two triangles match your values — for example a = 6, b = 10, A = 30° gives B = 56.44° or B = 123.56°.
How do I find the area of a triangle from three sides?
Use Heron’s formula: s = (a + b + c)/2 and area = √(s(s − a)(s − b)(s − c)). For sides 13, 14 and 15, s = 21 and the area is √(21 × 8 × 7 × 6) = 84. With two sides and the angle between them, area = ½ab sin C.
How do I find a missing angle when I know all three sides?
Use the law of cosines: cos A = (b² + c² − a²)/(2bc), then A = arccos of that. For a 5-6-7 triangle, cos A = 60/84 = 5/7, so A ≈ 44.42°. Find a second angle the same way and the third from A + B + C = 180°.
What is the difference between the centroid, incentre, circumcentre and orthocentre?
The centroid is where the three medians cross — the balance point. The incentre is where the angle bisectors cross, the centre of the largest circle inside. The circumcentre is the centre of the circle through all three corners; it lies outside an obtuse triangle and on the hypotenuse of a right triangle. The orthocentre is where the altitudes cross; it is the right-angle corner in a right triangle.
How do I enter angles in degrees, minutes and seconds?
Type them with the symbols, such as 36° 52′ 12″ or 36°52'12", in degree mode. Decimal degrees (36.87) work too. In radian mode you can type expressions such as pi/6, 2pi/3 or 0.75.