Triangle calculator

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

You have entered side a, b and angle β.

Triangle has two solutions: a=39.8; b=27.5; c=20.13876781907 and a=39.8; b=27.5; c=41.1076526391.

#1 Obtuse scalene triangle.

Sides: a = 39.8   b = 27.5   c = 20.13876781907

Area: T = 255.9879684879
Perimeter: p = 87.43876781907
Semiperimeter: s = 43.71988390953

Angle ∠ A = α = 112.4111242703° = 112°24'40″ = 1.96219463014 rad
Angle ∠ B = β = 39.7° = 39°42' = 0.6932895713 rad
Angle ∠ C = γ = 27.88987572966° = 27°53'20″ = 0.48767506391 rad

Height: ha = 12.86333007477
Height: hb = 18.61767043549
Height: hc = 25.42329591371

Median: ma = 13.59769864844
Median: mb = 28.38552169528
Median: mc = 32.6921948233

Inradius: r = 5.8555134541
Circumradius: R = 21.5265818338

Vertex coordinates: A[20.13876781907; 0] B[0; 0] C[30.62221022909; 25.42329591371]
Centroid: CG[16.92199268272; 8.47443197124]
Coordinates of the circumscribed circle: U[10.06988390953; 19.02657545026]
Coordinates of the inscribed circle: I[16.21988390953; 5.8555134541]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 67.58987572966° = 67°35'20″ = 1.96219463014 rad
∠ B' = β' = 140.3° = 140°18' = 0.6932895713 rad
∠ C' = γ' = 152.1111242703° = 152°6'40″ = 0.48767506391 rad




How did we calculate this triangle?

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

a = 39.8 ; ; b = 27.5 ; ; beta = 39.7° ; ;

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

b**2 = a**2 + c**2 - 2a c cos beta ; ; ; ; 27.5**2 = 39.8**2 + c**2 - 2 * 39.8 * c * cos(39° 42') ; ; ; ; ; ; c**2 -61.244c +827.79 =0 ; ; a=1; b=-61.244; c=827.79 ; ; D = b**2 - 4ac = 61.244**2 - 4 * 1 * 827.79 = 439.692594849 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 61.24 ± sqrt{ 439.69 } }{ 2 } ; ; c_{1,2} = 30.62210229 ± 10.4844241002 ; ; c_{1} = 41.1065263902 ; ; c_{2} = 20.1376781898 ; ;
 ; ; (c -41.1065263902) (c -20.1376781898) = 0 ; ; ; ; c > 0 ; ; ; ; c = 41.107 ; ;


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 = 39.8 ; ; b = 27.5 ; ; c = 20.14 ; ;

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

p = a+b+c = 39.8+27.5+20.14 = 87.44 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 87.44 }{ 2 } = 43.72 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 43.72 * (43.72-39.8)(43.72-27.5)(43.72-20.14) } ; ; T = sqrt{ 65525.6 } = 255.98 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 255.98 }{ 39.8 } = 12.86 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 255.98 }{ 27.5 } = 18.62 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 255.98 }{ 20.14 } = 25.42 ; ;

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{ 39.8**2-27.5**2-20.14**2 }{ 2 * 27.5 * 20.14 } ) = 112° 24'40" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27.5**2-39.8**2-20.14**2 }{ 2 * 39.8 * 20.14 } ) = 39° 42' ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20.14**2-39.8**2-27.5**2 }{ 2 * 27.5 * 39.8 } ) = 27° 53'20" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 255.98 }{ 43.72 } = 5.86 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 39.8 }{ 2 * sin 112° 24'40" } = 21.53 ; ;





#2 Acute scalene triangle.

Sides: a = 39.8   b = 27.5   c = 41.1076526391

Area: T = 522.5254770354
Perimeter: p = 108.4076526391
Semiperimeter: s = 54.20332631955

Angle ∠ A = α = 67.58987572966° = 67°35'20″ = 1.18796463522 rad
Angle ∠ B = β = 39.7° = 39°42' = 0.6932895713 rad
Angle ∠ C = γ = 72.71112427034° = 72°42'40″ = 1.26990505884 rad

Height: ha = 26.25875261485
Height: hb = 38.00218014803
Height: hc = 25.42329591371

Median: ma = 28.75774034984
Median: mb = 38.05503712987
Median: mc = 27.3444256655

Inradius: r = 9.64400980227
Circumradius: R = 21.5265818338

Vertex coordinates: A[41.1076526391; 0] B[0; 0] C[30.62221022909; 25.42329591371]
Centroid: CG[23.9109542894; 8.47443197124]
Coordinates of the circumscribed circle: U[20.55332631955; 6.39772046345]
Coordinates of the inscribed circle: I[26.70332631955; 9.64400980227]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 112.4111242703° = 112°24'40″ = 1.18796463522 rad
∠ B' = β' = 140.3° = 140°18' = 0.6932895713 rad
∠ C' = γ' = 107.2898757297° = 107°17'20″ = 1.26990505884 rad

Calculate another triangle

How did we calculate this triangle?

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

a = 39.8 ; ; b = 27.5 ; ; beta = 39.7° ; ; : Nr. 1

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

b**2 = a**2 + c**2 - 2a c cos beta ; ; ; ; 27.5**2 = 39.8**2 + c**2 - 2 * 39.8 * c * cos(39° 42') ; ; ; ; ; ; c**2 -61.244c +827.79 =0 ; ; a=1; b=-61.244; c=827.79 ; ; D = b**2 - 4ac = 61.244**2 - 4 * 1 * 827.79 = 439.692594849 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 61.24 ± sqrt{ 439.69 } }{ 2 } ; ; c_{1,2} = 30.62210229 ± 10.4844241002 ; ; c_{1} = 41.1065263902 ; ; c_{2} = 20.1376781898 ; ; : Nr. 1
 ; ; (c -41.1065263902) (c -20.1376781898) = 0 ; ; ; ; c > 0 ; ; ; ; c = 41.107 ; ; : 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 = 39.8 ; ; b = 27.5 ; ; c = 41.11 ; ;

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

p = a+b+c = 39.8+27.5+41.11 = 108.41 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 108.41 }{ 2 } = 54.2 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 54.2 * (54.2-39.8)(54.2-27.5)(54.2-41.11) } ; ; T = sqrt{ 273032.14 } = 522.52 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 522.52 }{ 39.8 } = 26.26 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 522.52 }{ 27.5 } = 38 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 522.52 }{ 41.11 } = 25.42 ; ;

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{ 39.8**2-27.5**2-41.11**2 }{ 2 * 27.5 * 41.11 } ) = 67° 35'20" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27.5**2-39.8**2-41.11**2 }{ 2 * 39.8 * 41.11 } ) = 39° 42' ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 41.11**2-39.8**2-27.5**2 }{ 2 * 27.5 * 39.8 } ) = 72° 42'40" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 522.52 }{ 54.2 } = 9.64 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 39.8 }{ 2 * sin 67° 35'20" } = 21.53 ; ;




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