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Triangle calculator: solve SSS, SAS, ASA, AAS and SSA

Missing sides, angles, area, perimeter, heights, inradius and circumradius of any triangle from three measurements, with both SSA answers.

Updated Checked against 8 worked examples

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Area
cm²
Area: 26.8328 cm²
Shown to up to 4 decimal places, half-up
Side a
7cm
Side b
8cm
Side c
9cm
Angle A
48.1897°
Angle B
58.4119°
Angle C
73.3985°
Perimeter
24cm
Inradius
2.2361cm
Circumradius
4.6957cm
Height onto a
7.6665cm
Height onto b
6.7082cm
Height onto c
5.9628cm
Type
Scalene, acute
Number of triangles
1

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.

The triangle, to scale

c = 9 cma = 7 cmb = 8 cm48.19°58.41°73.4°ABC
How it's calculated S

  1. Law of cosines for each angle

    cos⁡A=b2+c2−a22bc=0.666667⇒A=48.189685∘cos⁡B=a2+c2−b22ac=0.52381⇒B=58.411864∘cos⁡C=a2+b2−c22ab=0.285714⇒C=73.39845∘\begin{gathered} \cos A = \frac{b^2 + c^2 - a^2}{2bc} = 0.666667 \Rightarrow A = 48.189685^\circ \\[6pt] \cos B = \frac{a^2 + c^2 - b^2}{2ac} = 0.52381 \Rightarrow B = 58.411864^\circ \\[6pt] \cos C = \frac{a^2 + b^2 - c^2}{2ab} = 0.285714 \Rightarrow C = 73.39845^\circ \end{gathered}

  2. Area

    Area=12absin⁡C=12×7×8×sin⁡73.39845∘=26.832816 cm2\text{Area} = \tfrac{1}{2}ab\sin C = \tfrac{1}{2} \times 7 \times 8 \times \sin 73.39845^\circ = 26.832816\,\text{cm}^{2}

  3. Inradius and circumradius

    r=Areas=26.83281612=2.236068 cmR=a2sin⁡A=4.695743 cm\begin{gathered} r = \frac{\text{Area}}{s} = \frac{26.832816}{12} = 2.236068\,\text{cm} \\[6pt] R = \frac{a}{2\sin A} = 4.695743\,\text{cm} \end{gathered}

    s is the semi-perimeter.

  4. Heights

    ha=2 Areaa=7.666519 cmhb=2 Areab=6.708204 cm,hc=2 Areac=5.962848 cm\begin{gathered} h_a = \frac{2\,\text{Area}}{a} = 7.666519\,\text{cm} \\[6pt] h_b = \frac{2\,\text{Area}}{b} = 6.708204\,\text{cm},\quad h_c = \frac{2\,\text{Area}}{c} = 5.962848\,\text{cm} \end{gathered}

About the triangle calculator

Three measurements fix a triangle as long as at least one of them is a side. 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.

Worked examples

SSS 3-4-5

I know
SSS
Side a (opposite A)
3 cm
Side b (opposite B)
4 cm
Side c (opposite C)
5 cm
Show results in
Centimetres (cm)
Area
6 cm²
Angle C
90 °
Angle A
36.869898 °
Angle B
53.130102 °
Inradius
1 cm
Circumradius
2.5 cm
Type
Scalene, right
Number of triangles
1

Checked against: Right triangle: area ½·3·4, r = (3 + 4 − 5)/2, R = hypotenuse/2; Python 3.8 math: degrees(acos(0.8)), degrees(acos(0.6))

Equilateral, side 5

I know
SSS
Side a (opposite A)
5 cm
Side b (opposite B)
5 cm
Side c (opposite C)
5 cm
Show results in
Centimetres (cm)
Area
10.825318 cm²
Angle A
60 °
Angle B
60 °
Angle C
60 °
Type
Equilateral
Circumradius
2.886751 cm

Checked against: Python 3.8 math: sqrt(3)/4*25, 5/sqrt(3)

SAS a = 5, b = 7, C = 49°

I know
SAS
Side a (opposite A)
5 cm
Side b (opposite B)
7 cm
Angle C (between a and b)
49 °
Show results in
Centimetres (cm)
Side c
5.298667 cm
Angle A
45.411694 °
Angle B
85.588306 °
Area
13.207418 cm²

Checked against: Python 3.8 math: law of cosines and ½ab·sin C

ASA A = 40°, c = 10, B = 60°

I know
ASA
Side c (opposite C)
10 cm
Angle A
40 °
Angle B
60 °
Show results in
Centimetres (cm)
Angle C
80 °
Side a
6.527036 cm
Side b
8.793852 cm
Area
28.262897 cm²

Checked against: Python 3.8 math: C = 80°, law of sines a = 10·sin40/sin80

Questions

How do you find a missing side of a triangle that is not right-angled?

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.

What is the ambiguous case of the law of sines?

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°.

How do you find the angles of a triangle from its three sides?

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.

How do you find the area of a triangle without the height?

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.

How can you tell if a triangle is acute, right or obtuse from its sides?

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.

How accurate is the triangle calculator?

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 8 worked examples whose answers come from independent sources; for example, “SSS 3-4-5” is checked against Right triangle: area ½·3·4, r = (3 + 4 − 5)/2, R = hypotenuse/2; Python 3.8 math: degrees(acos(0.8)), degrees(acos(0.6)).

Where does the method come from?

Weisstein, E. W. “Law of Sines”, “Law of Cosines”, “Triangle” — MathWorld; OpenStax Precalculus 2e, §8.1 Non-right Triangles: Law of Sines (ambiguous case).

About this calculator

asin⁡A=bsin⁡B=csin⁡C=2R,c2=a2+b2−2abcos⁡CArea=12absin⁡C,r=Areas\begin{gathered} \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,\qquad c^2 = a^2 + b^2 - 2ab\cos C \\ \text{Area} = \tfrac{1}{2}ab\sin C,\qquad r = \frac{\text{Area}}{s} \end{gathered}

Sources

  1. Weisstein, E. W. “Law of Sines”, “Law of Cosines”, “Triangle” — MathWorld
  2. OpenStax Precalculus 2e, §8.1 Non-right Triangles: Law of Sines (ambiguous case)

Checked against references

8 worked examples with independently sourced answers ship with this calculator. They run in the test suite; you can run them here too.

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