Triangle calculator

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

You have entered side b, c and angle γ.

Triangle has two solutions: a=113.5244019669; b=500; c=400 and a=792.7843767367; b=500; c=400.

#1 Obtuse scalene triangle.

Sides: a = 113.5244019669   b = 500   c = 400

Area: T = 11994.33109646
Perimeter: p = 1013.524401967
Semiperimeter: s = 506.7622009835

Angle ∠ A = α = 6.88988308305° = 6°53'20″ = 0.12202327796 rad
Angle ∠ B = β = 148.1111169169° = 148°6'40″ = 2.5855027561 rad
Angle ∠ C = γ = 25° = 0.4366332313 rad

Height: ha = 211.309913087
Height: hb = 47.97773238584
Height: hc = 59.9721654823

Median: ma = 449.1977144069
Median: mb = 154.7388009296
Median: mc = 302.3976844429

Inradius: r = 23.66985677534
Circumradius: R = 473.244031663

Vertex coordinates: A[400; 0] B[0; 0] C[-96.39903711977; 59.9721654823]
Centroid: CG[101.2033209601; 19.99105516077]
Coordinates of the circumscribed circle: U[200; 428.9011384102]
Coordinates of the inscribed circle: I[6.76220098346; 23.66985677534]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 173.1111169169° = 173°6'40″ = 0.12202327796 rad
∠ B' = β' = 31.88988308305° = 31°53'20″ = 2.5855027561 rad
∠ C' = γ' = 155° = 0.4366332313 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 = 113.52 ; ; b = 500 ; ; c = 400 ; ;

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

p = a+b+c = 113.52+500+400 = 1013.52 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1013.52 }{ 2 } = 506.76 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 506.76 * (506.76-113.52)(506.76-500)(506.76-400) } ; ; T = sqrt{ 143863975.29 } = 11994.33 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 11994.33 }{ 113.52 } = 211.31 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 11994.33 }{ 500 } = 47.98 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 11994.33 }{ 400 } = 59.97 ; ;

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{ 113.52**2-500**2-400**2 }{ 2 * 500 * 400 } ) = 6° 53'20" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 500**2-113.52**2-400**2 }{ 2 * 113.52 * 400 } ) = 148° 6'40" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 400**2-113.52**2-500**2 }{ 2 * 500 * 113.52 } ) = 25° ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 11994.33 }{ 506.76 } = 23.67 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 113.52 }{ 2 * sin 6° 53'20" } = 473.24 ; ;





#2 Obtuse scalene triangle.

Sides: a = 792.7843767367   b = 500   c = 400

Area: T = 83761.22444253
Perimeter: p = 1692.784376737
Semiperimeter: s = 846.3921883684

Angle ∠ A = α = 123.1111169169° = 123°6'40″ = 2.1498695248 rad
Angle ∠ B = β = 31.88988308305° = 31°53'20″ = 0.55765650926 rad
Angle ∠ C = γ = 25° = 0.4366332313 rad

Height: ha = 211.309913087
Height: hb = 335.0454897701
Height: hc = 418.8066122126

Median: ma = 218.8800078953
Median: mb = 575.9880078562
Median: mc = 631.8654740986

Inradius: r = 98.96326980598
Circumradius: R = 473.244031663

Vertex coordinates: A[400; 0] B[0; 0] C[673.1332627252; 418.8066122126]
Centroid: CG[357.7110875751; 139.6022040709]
Coordinates of the circumscribed circle: U[200; 428.9011384102]
Coordinates of the inscribed circle: I[346.3921883684; 98.96326980598]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 56.88988308305° = 56°53'20″ = 2.1498695248 rad
∠ B' = β' = 148.1111169169° = 148°6'40″ = 0.55765650926 rad
∠ C' = γ' = 155° = 0.4366332313 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 = 792.78 ; ; b = 500 ; ; c = 400 ; ;

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

p = a+b+c = 792.78+500+400 = 1692.78 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1692.78 }{ 2 } = 846.39 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 846.39 * (846.39-792.78)(846.39-500)(846.39-400) } ; ; T = sqrt{ 7015942717.22 } = 83761.22 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 83761.22 }{ 792.78 } = 211.31 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 83761.22 }{ 500 } = 335.04 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 83761.22 }{ 400 } = 418.81 ; ;

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{ 792.78**2-500**2-400**2 }{ 2 * 500 * 400 } ) = 123° 6'40" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 500**2-792.78**2-400**2 }{ 2 * 792.78 * 400 } ) = 31° 53'20" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 400**2-792.78**2-500**2 }{ 2 * 500 * 792.78 } ) = 25° ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 83761.22 }{ 846.39 } = 98.96 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 792.78 }{ 2 * sin 123° 6'40" } = 473.24 ; ;




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