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=125; b=150; c=28.21333475737 and a=125; b=150; c=243.6798988537.

#1 Obtuse scalene triangle.

Sides: a = 125   b = 150   c = 28.21333475737

Area: T = 894.261069321
Perimeter: p = 303.2133347574
Semiperimeter: s = 151.6076673787

Angle ∠ A = α = 25° = 0.4366332313 rad
Angle ∠ B = β = 149.5266359102° = 149°31'35″ = 2.61097272848 rad
Angle ∠ C = γ = 5.47436408984° = 5°28'25″ = 0.09655330557 rad

Height: ha = 14.30881710914
Height: hb = 11.92334759095
Height: hc = 63.39327392611

Median: ma = 87.9877195038
Median: mb = 50.84877776373
Median: mc = 137.3444463866

Inradius: r = 5.89985575692
Circumradius: R = 147.8887598947

Vertex coordinates: A[28.21333475737; 0] B[0; 0] C[-107.7332820482; 63.39327392611]
Centroid: CG[-26.50664909694; 21.1310913087]
Coordinates of the circumscribed circle: U[14.10766737868; 147.2133259175]
Coordinates of the inscribed circle: I[1.60766737868; 5.89985575692]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155° = 0.4366332313 rad
∠ B' = β' = 30.47436408984° = 30°28'25″ = 2.61097272848 rad
∠ C' = γ' = 174.5266359102° = 174°31'35″ = 0.09655330557 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 = 125 ; ; b = 150 ; ; c = 28.21 ; ;

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

p = a+b+c = 125+150+28.21 = 303.21 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 303.21 }{ 2 } = 151.61 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 151.61 * (151.61-125)(151.61-150)(151.61-28.21) } ; ; T = sqrt{ 799702.19 } = 894.26 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 894.26 }{ 125 } = 14.31 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 894.26 }{ 150 } = 11.92 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 894.26 }{ 28.21 } = 63.39 ; ;

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{ 125**2-150**2-28.21**2 }{ 2 * 150 * 28.21 } ) = 25° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 150**2-125**2-28.21**2 }{ 2 * 125 * 28.21 } ) = 149° 31'35" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28.21**2-125**2-150**2 }{ 2 * 150 * 125 } ) = 5° 28'25" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 894.26 }{ 151.61 } = 5.9 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 125 }{ 2 * sin 25° } = 147.89 ; ;





#2 Obtuse scalene triangle.

Sides: a = 125   b = 150   c = 243.6798988537

Area: T = 7723.739929188
Perimeter: p = 518.6798988537
Semiperimeter: s = 259.3399494269

Angle ∠ A = α = 25° = 0.4366332313 rad
Angle ∠ B = β = 30.47436408984° = 30°28'25″ = 0.53218653687 rad
Angle ∠ C = γ = 124.5266359102° = 124°31'35″ = 2.17333949718 rad

Height: ha = 123.587982867
Height: hb = 102.9833190558
Height: hc = 63.39327392611

Median: ma = 192.4410834355
Median: mb = 178.5421941087
Median: mc = 64.943334174

Inradius: r = 29.78223488615
Circumradius: R = 147.8887598947

Vertex coordinates: A[243.6798988537; 0] B[0; 0] C[107.7332820482; 63.39327392611]
Centroid: CG[117.1377269673; 21.1310913087]
Coordinates of the circumscribed circle: U[121.8399494269; -83.82105199141]
Coordinates of the inscribed circle: I[109.3399494269; 29.78223488615]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155° = 0.4366332313 rad
∠ B' = β' = 149.5266359102° = 149°31'35″ = 0.53218653687 rad
∠ C' = γ' = 55.47436408984° = 55°28'25″ = 2.17333949718 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 = 125 ; ; b = 150 ; ; c = 243.68 ; ;

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

p = a+b+c = 125+150+243.68 = 518.68 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 518.68 }{ 2 } = 259.34 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 259.34 * (259.34-125)(259.34-150)(259.34-243.68) } ; ; T = sqrt{ 59656148.65 } = 7723.74 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 7723.74 }{ 125 } = 123.58 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 7723.74 }{ 150 } = 102.98 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 7723.74 }{ 243.68 } = 63.39 ; ;

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{ 125**2-150**2-243.68**2 }{ 2 * 150 * 243.68 } ) = 25° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 150**2-125**2-243.68**2 }{ 2 * 125 * 243.68 } ) = 30° 28'25" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 243.68**2-125**2-150**2 }{ 2 * 150 * 125 } ) = 124° 31'35" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 7723.74 }{ 259.34 } = 29.78 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 125 }{ 2 * sin 25° } = 147.89 ; ; : Nr. 1




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