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?

1. Input data entered: side a, b and angle α.

a = 50 ; ; b = 60 ; ; alpha = 40° ; ;

2. From angle α, b and side a we calculate c - by using the law of cosines and quadratic equation:

a**2 = b**2 + c**2 - 2b c cos alpha ; ; ; ; 50**2 = 60**2 + c**2 - 2 * 60 * c * cos(40° ) ; ; ; ; ; ; c**2 -91.925c +1100 =0 ; ; a=1; b=-91.925; c=1100 ; ; D = b**2 - 4ac = 91.925**2 - 4 * 1 * 1100 = 4050.2668792 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 91.93 ± sqrt{ 4050.27 } }{ 2 } ; ; c_{1,2} = 45.96266659 ± 31.8208535366 ; ; c_{1} = 77.7835201266 ; ; c_{2} = 14.1418130534 ; ; ; ;
(c -77.7835201266) (c -14.1418130534) = 0 ; ; ; ; c > 0 ; ; ; ; c = 77.784 ; ;


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 ; ;

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

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

4. Semiperimeter of the triangle

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

5. 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 ; ;

6. 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 ; ;

7. 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" ; ;

8. Inradius

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

9. 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?

1. Input data entered: side a, b and angle α.

a = 50 ; ; b = 60 ; ; alpha = 40° ; ; : Nr. 1

2. From angle α, b and side a we calculate c - by using the law of cosines and quadratic equation:

a**2 = b**2 + c**2 - 2b c cos alpha ; ; ; ; 50**2 = 60**2 + c**2 - 2 * 60 * c * cos(40° ) ; ; ; ; ; ; c**2 -91.925c +1100 =0 ; ; a=1; b=-91.925; c=1100 ; ; D = b**2 - 4ac = 91.925**2 - 4 * 1 * 1100 = 4050.2668792 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 91.93 ± sqrt{ 4050.27 } }{ 2 } ; ; c_{1,2} = 45.96266659 ± 31.8208535366 ; ; c_{1} = 77.7835201266 ; ; c_{2} = 14.1418130534 ; ; ; ; : Nr. 1
(c -77.7835201266) (c -14.1418130534) = 0 ; ; ; ; c > 0 ; ; ; ; c = 77.784 ; ; : Nr. 1


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 ; ;

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

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

4. Semiperimeter of the triangle

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

5. 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 ; ;

6. 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 ; ;

7. 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" ; ;

8. Inradius

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

9. Circumradius

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

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