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=155; b=63; c=99.77106663709 and a=155; b=63; c=201.0211008775.

#1 Obtuse scalene triangle.

Sides: a = 155   b = 63   c = 99.77106663709

Area: T = 1870.595492686
Perimeter: p = 317.7710666371
Semiperimeter: s = 158.8855333186

Angle ∠ A = α = 143.4732776509° = 143°28'22″ = 2.50440723371 rad
Angle ∠ B = β = 14° = 0.24443460953 rad
Angle ∠ C = γ = 22.52772234906° = 22°31'38″ = 0.39331742212 rad

Height: ha = 24.13767087337
Height: hb = 59.38439659321
Height: hc = 37.49878938179

Median: ma = 30.90986223253
Median: mb = 126.4810602995
Median: mc = 107.2777460508

Inradius: r = 11.77332385322
Circumradius: R = 130.2077313075

Vertex coordinates: A[99.77106663709; 0] B[0; 0] C[150.3965837573; 37.49878938179]
Centroid: CG[83.38988346479; 12.49992979393]
Coordinates of the circumscribed circle: U[49.88553331855; 120.2722182616]
Coordinates of the inscribed circle: I[95.88553331855; 11.77332385322]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 36.52772234906° = 36°31'38″ = 2.50440723371 rad
∠ B' = β' = 166° = 0.24443460953 rad
∠ C' = γ' = 157.4732776509° = 157°28'22″ = 0.39331742212 rad




How did we calculate this triangle?

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

a = 155 ; ; b = 63 ; ; beta = 14° ; ;

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

b**2 = a**2 + c**2 - 2a c cos beta ; ; ; ; 63**2 = 155**2 + c**2 - 2 * 155 * c * cos 14° ; ; ; ; ; ; c**2 -300.792c +20056 =0 ; ; p=1; q=-300.792; r=20056 ; ; D = q**2 - 4pr = 300.792**2 - 4 * 1 * 20056 = 10251.6318369 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 300.79 ± sqrt{ 10251.63 } }{ 2 } ; ; c_{1,2} = 150.39583757 ± 50.6251712019 ; ; c_{1} = 201.021008772 ; ; c_{2} = 99.7706663681 ; ; ; ; text{ Factored form: } ; ; (c -201.021008772) (c -99.7706663681) = 0 ; ; ; ; c > 0 ; ; ; ; c = 201.021 ; ;


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 = 155 ; ; b = 63 ; ; c = 99.77 ; ;

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

p = a+b+c = 155+63+99.77 = 317.77 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 317.77 }{ 2 } = 158.89 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 158.89 * (158.89-155)(158.89-63)(158.89-99.77) } ; ; T = sqrt{ 3499125.38 } = 1870.59 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1870.59 }{ 155 } = 24.14 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1870.59 }{ 63 } = 59.38 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1870.59 }{ 99.77 } = 37.5 ; ;

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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 63**2+99.77**2-155**2 }{ 2 * 63 * 99.77 } ) = 143° 28'22" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 155**2+99.77**2-63**2 }{ 2 * 155 * 99.77 } ) = 14° ; ; gamma = 180° - alpha - beta = 180° - 143° 28'22" - 14° = 22° 31'38" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1870.59 }{ 158.89 } = 11.77 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 155 }{ 2 * sin 143° 28'22" } = 130.21 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 63**2+2 * 99.77**2 - 155**2 } }{ 2 } = 30.909 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 99.77**2+2 * 155**2 - 63**2 } }{ 2 } = 126.481 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 63**2+2 * 155**2 - 99.77**2 } }{ 2 } = 107.277 ; ;







#2 Obtuse scalene triangle.

Sides: a = 155   b = 63   c = 201.0211008775

Area: T = 3768.93222211
Perimeter: p = 419.0211008775
Semiperimeter: s = 209.5110504387

Angle ∠ A = α = 36.52772234906° = 36°31'38″ = 0.63875203165 rad
Angle ∠ B = β = 14° = 0.24443460953 rad
Angle ∠ C = γ = 129.4732776509° = 129°28'22″ = 2.26597262418 rad

Height: ha = 48.63113834981
Height: hb = 119.649864194
Height: hc = 37.49878938179

Median: ma = 127.2122314594
Median: mb = 176.7065894028
Median: mc = 62.40770389284

Inradius: r = 17.98992279489
Circumradius: R = 130.2077313075

Vertex coordinates: A[201.0211008775; 0] B[0; 0] C[150.3965837573; 37.49878938179]
Centroid: CG[117.1398948782; 12.49992979393]
Coordinates of the circumscribed circle: U[100.5110504387; -82.77442887977]
Coordinates of the inscribed circle: I[146.5110504387; 17.98992279489]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 143.4732776509° = 143°28'22″ = 0.63875203165 rad
∠ B' = β' = 166° = 0.24443460953 rad
∠ C' = γ' = 50.52772234906° = 50°31'38″ = 2.26597262418 rad

Calculate another triangle

How did we calculate this triangle?

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

a = 155 ; ; b = 63 ; ; beta = 14° ; ; : Nr. 1

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

b**2 = a**2 + c**2 - 2a c cos beta ; ; ; ; 63**2 = 155**2 + c**2 - 2 * 155 * c * cos 14° ; ; ; ; ; ; c**2 -300.792c +20056 =0 ; ; p=1; q=-300.792; r=20056 ; ; D = q**2 - 4pr = 300.792**2 - 4 * 1 * 20056 = 10251.6318369 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 300.79 ± sqrt{ 10251.63 } }{ 2 } ; ; c_{1,2} = 150.39583757 ± 50.6251712019 ; ; c_{1} = 201.021008772 ; ; c_{2} = 99.7706663681 ; ; ; ; text{ Factored form: } ; ; (c -201.021008772) (c -99.7706663681) = 0 ; ; ; ; c > 0 ; ; ; ; c = 201.021 ; ; : 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 = 155 ; ; b = 63 ; ; c = 201.02 ; ;

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

p = a+b+c = 155+63+201.02 = 419.02 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 419.02 }{ 2 } = 209.51 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 209.51 * (209.51-155)(209.51-63)(209.51-201.02) } ; ; T = sqrt{ 14204850.09 } = 3768.93 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 3768.93 }{ 155 } = 48.63 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 3768.93 }{ 63 } = 119.65 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 3768.93 }{ 201.02 } = 37.5 ; ;

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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 63**2+201.02**2-155**2 }{ 2 * 63 * 201.02 } ) = 36° 31'38" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 155**2+201.02**2-63**2 }{ 2 * 155 * 201.02 } ) = 14° ; ; gamma = 180° - alpha - beta = 180° - 36° 31'38" - 14° = 129° 28'22" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 3768.93 }{ 209.51 } = 17.99 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 155 }{ 2 * sin 36° 31'38" } = 130.21 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 63**2+2 * 201.02**2 - 155**2 } }{ 2 } = 127.212 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 201.02**2+2 * 155**2 - 63**2 } }{ 2 } = 176.706 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 63**2+2 * 155**2 - 201.02**2 } }{ 2 } = 62.407 ; ;
Calculate another triangle

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

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