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=38; b=42; c=5.74663686918 and a=38; b=42; c=55.68773422442.

#1 Obtuse scalene triangle.

Sides: a = 38   b = 42   c = 5.74663686918

Area: T = 82.29992945061
Perimeter: p = 85.74663686918
Semiperimeter: s = 42.87331843459

Angle ∠ A = α = 43° = 0.75504915784 rad
Angle ∠ B = β = 131.0880449263° = 131°4'50″ = 2.28877854246 rad
Angle ∠ C = γ = 5.92195507372° = 5°55'10″ = 0.10333156506 rad

Height: ha = 4.33215418161
Height: hb = 3.91990140241
Height: hc = 28.64439311226

Median: ma = 23.18442700245
Median: mb = 17.24884891098
Median: mc = 39.94767747348

Inradius: r = 1.92195983634
Circumradius: R = 27.85993045272

Vertex coordinates: A[5.74663686918; 0] B[0; 0] C[-24.97704867762; 28.64439311226]
Centroid: CG[-6.40880393615; 9.54879770409]
Coordinates of the circumscribed circle: U[2.87331843459; 27.71107499078]
Coordinates of the inscribed circle: I[0.87331843459; 1.92195983634]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 137° = 0.75504915784 rad
∠ B' = β' = 48.92195507372° = 48°55'10″ = 2.28877854246 rad
∠ C' = γ' = 174.0880449263° = 174°4'50″ = 0.10333156506 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 = 38 ; ; b = 42 ; ; c = 5.75 ; ;

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

p = a+b+c = 38+42+5.75 = 85.75 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 85.75 }{ 2 } = 42.87 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 42.87 * (42.87-38)(42.87-42)(42.87-5.75) } ; ; T = sqrt{ 6773.17 } = 82.3 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 82.3 }{ 38 } = 4.33 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 82.3 }{ 42 } = 3.92 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 82.3 }{ 5.75 } = 28.64 ; ;

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{ 38**2-42**2-5.75**2 }{ 2 * 42 * 5.75 } ) = 43° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 42**2-38**2-5.75**2 }{ 2 * 38 * 5.75 } ) = 131° 4'50" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 5.75**2-38**2-42**2 }{ 2 * 42 * 38 } ) = 5° 55'10" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 82.3 }{ 42.87 } = 1.92 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 38 }{ 2 * sin 43° } = 27.86 ; ;





#2 Acute scalene triangle.

Sides: a = 38   b = 42   c = 55.68773422442

Area: T = 797.5522197823
Perimeter: p = 135.6877342244
Semiperimeter: s = 67.84436711221

Angle ∠ A = α = 43° = 0.75504915784 rad
Angle ∠ B = β = 48.92195507372° = 48°55'10″ = 0.8543807229 rad
Angle ∠ C = γ = 88.08804492628° = 88°4'50″ = 1.53772938463 rad

Height: ha = 41.97664314644
Height: hb = 37.97986760868
Height: hc = 28.64439311226

Median: ma = 45.51441740902
Median: mb = 42.79664956873
Median: mc = 28.78876705977

Inradius: r = 11.75657346858
Circumradius: R = 27.85993045272

Vertex coordinates: A[55.68773422442; 0] B[0; 0] C[24.97704867762; 28.64439311226]
Centroid: CG[26.88659430068; 9.54879770409]
Coordinates of the circumscribed circle: U[27.84436711221; 0.93331812149]
Coordinates of the inscribed circle: I[25.84436711221; 11.75657346858]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 137° = 0.75504915784 rad
∠ B' = β' = 131.0880449263° = 131°4'50″ = 0.8543807229 rad
∠ C' = γ' = 91.92195507372° = 91°55'10″ = 1.53772938463 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 = 38 ; ; b = 42 ; ; c = 55.69 ; ;

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

p = a+b+c = 38+42+55.69 = 135.69 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 135.69 }{ 2 } = 67.84 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 67.84 * (67.84-38)(67.84-42)(67.84-55.69) } ; ; T = sqrt{ 636089.51 } = 797.55 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 797.55 }{ 38 } = 41.98 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 797.55 }{ 42 } = 37.98 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 797.55 }{ 55.69 } = 28.64 ; ;

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{ 38**2-42**2-55.69**2 }{ 2 * 42 * 55.69 } ) = 43° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 42**2-38**2-55.69**2 }{ 2 * 38 * 55.69 } ) = 48° 55'10" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 55.69**2-38**2-42**2 }{ 2 * 42 * 38 } ) = 88° 4'50" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 797.55 }{ 67.84 } = 11.76 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 38 }{ 2 * sin 43° } = 27.86 ; ; : Nr. 1




Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.