Missing sides, angles, area, perimeter, heights, inradius and circumradius of any triangle from three measurements, with both SSA answers.
By AdminDrafted with AI writing toolsUpdated Checked examples: 8
The triangle has sides 7, 8 and 9 cm with angles 48.1897°, 58.4119° and 73.3985°; it is Scalene, acute and encloses 26.8328 cm².
Δ is the area and s is the semi-perimeter.
Three compatible measurements can determine a triangle when at least one is a side; the SSA case may have two solutions. The law of cosines, c² = a² + b² − 2ab cos C, solves three sides (SSS) and two sides with the angle between them (SAS). The law of sines, a/sin A = b/sin B = c/sin C, solves two angles and a side (ASA, AAS). The area comes from ½ab sin C, and the results add the perimeter, all three heights, the inradius and the circumradius.
The default sides of 7, 8 and 9 cm give angles of 48.19°, 58.41° and 73.40° and an area of 26.8328 cm². Surveying and roof framing use the same two laws to find a length that cannot be measured directly from ones that can.
Two sides and an angle that is not between them (SSA) can fit two triangles, one or none. With a = 6, b = 8 and A = 35°, both B = 49.89° and B = 130.11° work, so both triangles are drawn and compared in a table.
Use the law of cosines, c² = a² + b² − 2ab cos C, for three sides or two sides and the angle between them. Use the law of sines, a/sin A = b/sin B = c/sin C, for two angles and a side. Sides of 5 and 7 cm at 49° leave a third side of 5.298667 cm.
Lowercase a, b and c are sides, and A, B and C are the angles opposite them, so side a faces angle A.
The angles of a triangle always add to 180°. With A = 40° and B = 60°, C = 80°; if the side between them is c = 10 cm, then a = 10 sin 40°/sin 80° ≈ 6.527036 cm and b = 10 sin 60°/sin 80° ≈ 8.793852 cm.
Two angles fix a triangle's shape but not its size, so at least one side is always needed. The largest angle sits opposite the longest side.
From two sides and the angle between them, area = ½ab sin C. The 5 cm and 7 cm sides at 49° enclose ½ × 5 × 7 × sin 49° ≈ 13.207418 cm².
From three sides, Heron's formula needs no angles: with s = (a + b + c)/2, area = √(s(s − a)(s − b)(s − c)). Sides of 7, 8 and 9 cm give s = 12 and √(12 × 5 × 4 × 3) = √720 ≈ 26.832816 cm².
When a base and its perpendicular height are known, area = ½bh; the area and perimeter calculator has that form.
An equilateral triangle with side x has three 60° angles, height (√3/2)x ≈ 0.866x and area (√3/4)x² ≈ 0.433x². A side of 5 cm gives a height of about 4.330127 cm and an area of about 10.825318 cm².
| Side, cm | Height, cm | Area, cm² | Perimeter, cm |
|---|---|---|---|
| 1 | 0.866025 | 0.433013 | 3 |
| 2 | 1.732051 | 1.732051 | 6 |
| 3 | 2.598076 | 3.897114 | 9 |
| 4 | 3.464102 | 6.928203 | 12 |
| 5 | 4.330127 | 10.825318 | 15 |
| 6 | 5.196152 | 15.588457 | 18 |
| 8 | 6.928203 | 27.712813 | 24 |
| 10 | 8.660254 | 43.30127 | 30 |
| 12 | 10.392305 | 62.353829 | 36 |
Doubling the side doubles the perimeter and multiplies the area by four: 10 cm sides enclose 43.30127 cm², four times the 5 cm triangle.
With two equal legs L and a base b, the height from the apex cuts the base in half: h = √(L² − (b/2)²), and area = bh/2. Legs of 5 cm on a 6 cm base give h = √(25 − 9) = 4 cm and an area of 12 cm².
The two base angles are equal, about 53.130102° each in that triangle, which leaves about 73.739795° at the apex.
| Equal leg, cm | Base, cm | Height, cm | Area, cm² | Perimeter, cm |
|---|---|---|---|---|
| 5 | 6 | 4 | 12 | 16 |
| 5 | 8 | 3 | 12 | 18 |
| 10 | 12 | 8 | 48 | 32 |
| 13 | 10 | 12 | 60 | 36 |
| 13 | 24 | 5 | 60 | 50 |
| 17 | 16 | 15 | 120 | 50 |
Every height in this chart is a whole number because each half-triangle is a Pythagorean triple, such as 3-4-5 or 5-12-13. The base must be shorter than twice the leg; at exactly 2L the triangle flattens to zero area.
Sides are in centimeters, angles in degrees and areas in square centimeters.
| Mode | Entered measurements | Side c, cm | Area, cm² | Solutions |
|---|---|---|---|---|
| SSS | a=7, b=8, c=9 | 9 | 26.832816 | 1 |
| SAS | a=5, b=7, C=49° | 5.298667 | 13.207418 | 1 |
| ASA | A=40°, B=60°, c=10 | 10 | 28.262897 | 1 |
| AAS | A=30°, B=45°, a=10 | 19.318517 | 68.30127 | 1 |
| SSA | A=35°, a=6, b=8 | 10.419047 | 23.904479 | 2 |
For an acute angle A, compare side a with the height h = b sin A, the distance from corner C to the line that side c lies on:
When A is 90° or more, a triangle exists only if a is longer than b, and then there is one.
For a = 6 cm, b = 8 cm and A = 35°, h = 8 sin 35° ≈ 4.588611 cm. Since 4.59 < 6 < 8, two triangles fit: sin B = 8 sin 35°/6 ≈ 0.764769, so B ≈ 49.886408° or its supplement, 130.113592°.
| Triangle | Angle B, ° | Angle C, ° | Side c, cm | Area, cm² |
|---|---|---|---|---|
| 1 | 49.886408 | 95.113592 | 10.419047 | 23.904479 |
| 2 | 130.113592 | 14.886408 | 2.687386 | 6.165685 |
Both triangles fit every given measurement, so both are answers unless the problem adds a condition, such as B being obtuse.
Shorten a to 4 cm and it falls below h, so no triangle closes. Lengthen it to 9 cm and only one fits, with B ≈ 30.653554°.
When one angle is 90°, the right triangle calculator solves it from any two values.
Checked against: Right triangle: area ½·3·4, r = (3 + 4 − 5)/2, R = hypotenuse/2
Checked against: sqrt(3)/4*25, 5/sqrt(3)
Checked against: Law of cosines and ½ab·sin C
Checked against: C = 80°, law of sines a = 10·sin40/sin80
With two sides and the angle between them, use the law of cosines: c² = a² + b² − 2ab cos C. For a = 5, b = 7 and C = 49°, c = √(74 − 70 cos 49°) ≈ 5.2987. With two angles and any side, use the law of sines, a/sin A = b/sin B. When C = 90° the cosine term is zero and the law of cosines becomes Pythagoras' theorem.
It is the SSA case: two sides and an angle opposite one of them can fit two different triangles. For an acute angle A, find the height h = b sin A. If a < h there is no triangle, if a = h one right triangle, if h < a < b two triangles, and if a ≥ b one. For a = 6, b = 8 and A = 35°, h = 4.589, so B is either 49.89° or 130.11°.
Rearrange the law of cosines: cos A = (b² + c² − a²)/(2bc), and the same pattern for B and C. For sides 7, 8 and 9, cos A = (64 + 81 − 49)/144 = 0.6667, so A = 48.19°; B = 58.41° and C = 73.40° follow the same way, and the three add up to 180°. The largest angle is always opposite the longest side.
Take half the product of two sides and the sine of the angle between them: Area = ½ab sin C. Sides 7 and 8 with C = 73.40° give ½ × 7 × 8 × 0.9583 ≈ 26.83. From three sides alone, Heron's formula gives the same answer: the semi-perimeter is 12, and √(12 × 5 × 4 × 3) = √720 ≈ 26.83.
Compare the square of the longest side c with the sum of the squares of the other two. If c² < a² + b² the triangle is acute, if they are equal it has a right angle, and if c² is larger it is obtuse. Sides 7, 8 and 9 give 81 < 113, so acute; 3, 4 and 5 give 25 = 25, right; 4, 5 and 8 give 64 > 41, obtuse.
This calculator includes 8 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.