Triangle calculator

Please enter what you know about the triangle:
Symbols definition of ABC triangle

You have entered side a, b and angle α.

Triangle has two solutions: a=50; b=60; c=14.14218130505 and a=50; b=60; c=77.78435201238.

#1 Obtuse scalene triangle.

Sides: a = 50   b = 60   c = 14.14218130505

Area: T = 272.7055466221
Perimeter: p = 124.1421813051
Semiperimeter: s = 62.07109065252

Angle ∠ A = α = 40° = 0.69881317008 rad
Angle ∠ B = β = 129.5255165368° = 129°31'31″ = 2.26106405999 rad
Angle ∠ C = γ = 10.47548346316° = 10°28'29″ = 0.18328203529 rad

Height: ha = 10.90882186488
Height: hb = 9.09901822074
Height: hc = 38.56772565812

Median: ma = 35.70770782644
Median: mb = 21.21330959121
Median: mc = 54.77222765723

Inradius: r = 4.39334506758
Circumradius: R = 38.89330956715

Vertex coordinates: A[14.14218130505; 0] B[0; 0] C[-31.82108535366; 38.56772565812]
Centroid: CG[-5.89330134954; 12.85657521937]
Coordinates of the circumscribed circle: U[7.07109065252; 38.24549365515]
Coordinates of the inscribed circle: I[2.07109065252; 4.39334506758]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 140° = 0.69881317008 rad
∠ B' = β' = 50.47548346316° = 50°28'29″ = 2.26106405999 rad
∠ C' = γ' = 169.5255165368° = 169°31'31″ = 0.18328203529 rad




How did we calculate this triangle?

Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.

a = 50 ; ; b = 60 ; ; c = 14.14 ; ;

1. The triangle circumference is the sum of the lengths of its three sides

p = a+b+c = 50+60+14.14 = 124.14 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 124.14 }{ 2 } = 62.07 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 62.07 * (62.07-50)(62.07-60)(62.07-14.14) } ; ; T = sqrt{ 74368.27 } = 272.71 ; ;

4. Calculate the heights of the triangle from its area.

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 272.71 }{ 50 } = 10.91 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 272.71 }{ 60 } = 9.09 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 272.71 }{ 14.14 } = 38.57 ; ;

5. Calculation of the inner angles of the triangle using a Law of Cosines

a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 50**2-60**2-14.14**2 }{ 2 * 60 * 14.14 } ) = 40° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 60**2-50**2-14.14**2 }{ 2 * 50 * 14.14 } ) = 129° 31'31" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 14.14**2-50**2-60**2 }{ 2 * 60 * 50 } ) = 10° 28'29" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 272.71 }{ 62.07 } = 4.39 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 50 }{ 2 * sin 40° } = 38.89 ; ;





#2 Acute scalene triangle.

Sides: a = 50   b = 60   c = 77.78435201238

Area: T = 1499.94884892
Perimeter: p = 187.7843520124
Semiperimeter: s = 93.89217600619

Angle ∠ A = α = 40° = 0.69881317008 rad
Angle ∠ B = β = 50.47548346316° = 50°28'29″ = 0.88109520537 rad
Angle ∠ C = γ = 89.52551653684° = 89°31'31″ = 1.56325088991 rad

Height: ha = 59.9987939568
Height: hb = 49.99882829734
Height: hc = 38.56772565812

Median: ma = 64.80884716794
Median: mb = 58.09659379081
Median: mc = 39.21100879786

Inradius: r = 15.97552941921
Circumradius: R = 38.89330956715

Vertex coordinates: A[77.78435201238; 0] B[0; 0] C[31.82108535366; 38.56772565812]
Centroid: CG[36.53547912201; 12.85657521937]
Coordinates of the circumscribed circle: U[38.89217600619; 0.32223200297]
Coordinates of the inscribed circle: I[33.89217600619; 15.97552941921]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 140° = 0.69881317008 rad
∠ B' = β' = 129.5255165368° = 129°31'31″ = 0.88109520537 rad
∠ C' = γ' = 90.47548346316° = 90°28'29″ = 1.56325088991 rad

Calculate another triangle

How did we calculate this triangle?

Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.

a = 50 ; ; b = 60 ; ; c = 77.78 ; ;

1. The triangle circumference is the sum of the lengths of its three sides

p = a+b+c = 50+60+77.78 = 187.78 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 187.78 }{ 2 } = 93.89 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 93.89 * (93.89-50)(93.89-60)(93.89-77.78) } ; ; T = sqrt{ 2249845.47 } = 1499.95 ; ;

4. Calculate the heights of the triangle from its area.

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1499.95 }{ 50 } = 60 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1499.95 }{ 60 } = 50 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1499.95 }{ 77.78 } = 38.57 ; ;

5. Calculation of the inner angles of the triangle using a Law of Cosines

a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 50**2-60**2-77.78**2 }{ 2 * 60 * 77.78 } ) = 40° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 60**2-50**2-77.78**2 }{ 2 * 50 * 77.78 } ) = 50° 28'29" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 77.78**2-50**2-60**2 }{ 2 * 60 * 50 } ) = 89° 31'31" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1499.95 }{ 93.89 } = 15.98 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 50 }{ 2 * sin 40° } = 38.89 ; ; : Nr. 1




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