Triangle calculator SSA

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Triangle has two solutions with side c=79.27883767368 and with side c=11.35224019669

#1 Obtuse scalene triangle.

Sides: a = 50   b = 40   c = 79.27883767368

Area: T = 837.6122244253
Perimeter: p = 169.2788376737
Semiperimeter: s = 84.63991883684

Angle ∠ A = α = 31.88988308305° = 31°53'20″ = 0.55765650926 rad
Angle ∠ B = β = 25° = 0.4366332313 rad
Angle ∠ C = γ = 123.1111169169° = 123°6'40″ = 2.1498695248 rad

Height: ha = 33.50444897701
Height: hb = 41.88106122126
Height: hc = 21.1310913087

Median: ma = 57.59880078562
Median: mb = 63.18664740986
Median: mc = 21.88800078953

Inradius: r = 9.8966269806
Circumradius: R = 47.32440316631

Vertex coordinates: A[79.27883767368; 0] B[0; 0] C[45.31553893518; 21.1310913087]
Centroid: CG[41.53112553629; 7.04436376957]
Coordinates of the circumscribed circle: U[39.63991883684; -25.85114742006]
Coordinates of the inscribed circle: I[44.63991883684; 9.8966269806]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 148.1111169169° = 148°6'40″ = 0.55765650926 rad
∠ B' = β' = 155° = 0.4366332313 rad
∠ C' = γ' = 56.88988308305° = 56°53'20″ = 2.1498695248 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 = 50 ; ; b = 40 ; ; c = 79.28 ; ;

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

p = a+b+c = 50+40+79.28 = 169.28 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 169.28 }{ 2 } = 84.64 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 84.64 * (84.64-50)(84.64-40)(84.64-79.28) } ; ; T = sqrt{ 701594.27 } = 837.61 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 837.61 }{ 50 } = 33.5 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 837.61 }{ 40 } = 41.88 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 837.61 }{ 79.28 } = 21.13 ; ;

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{ 50**2-40**2-79.28**2 }{ 2 * 40 * 79.28 } ) = 31° 53'20" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 40**2-50**2-79.28**2 }{ 2 * 50 * 79.28 } ) = 25° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 79.28**2-50**2-40**2 }{ 2 * 40 * 50 } ) = 123° 6'40" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 837.61 }{ 84.64 } = 9.9 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 50 }{ 2 * sin 31° 53'20" } = 47.32 ; ;





#2 Obtuse scalene triangle.

Sides: a = 50   b = 40   c = 11.35224019669

Area: T = 119.9433309646
Perimeter: p = 101.3522401967
Semiperimeter: s = 50.67662009835

Angle ∠ A = α = 148.1111169169° = 148°6'40″ = 2.5855027561 rad
Angle ∠ B = β = 25° = 0.4366332313 rad
Angle ∠ C = γ = 6.88988308305° = 6°53'20″ = 0.12202327796 rad

Height: ha = 4.79877323858
Height: hb = 5.99771654823
Height: hc = 21.1310913087

Median: ma = 15.47438009296
Median: mb = 30.2439684443
Median: mc = 44.92197144069

Inradius: r = 2.36768567753
Circumradius: R = 47.32440316631

Vertex coordinates: A[11.35224019669; 0] B[0; 0] C[45.31553893518; 21.1310913087]
Centroid: CG[18.88992637729; 7.04436376957]
Coordinates of the circumscribed circle: U[5.67662009835; 46.98223872876]
Coordinates of the inscribed circle: I[10.67662009835; 2.36768567753]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 31.88988308305° = 31°53'20″ = 2.5855027561 rad
∠ B' = β' = 155° = 0.4366332313 rad
∠ C' = γ' = 173.1111169169° = 173°6'40″ = 0.12202327796 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 50 ; ; b = 40 ; ; beta = 25° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 40**2 = 50**2 + c**2 -2 * 40 * c * cos (25° ) ; ; ; ; c**2 -90.631c +900 =0 ; ; p=1; q=-90.6307787037; r=900 ; ; D = q**2 - 4pr = 90.631**2 - 4 * 1 * 900 = 4613.93804843 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 90.63 ± sqrt{ 4613.94 } }{ 2 } ; ; c_{1,2} = 45.3153893518 ± 33.9629873849 ; ; c_{1} = 79.2783767368 ; ;
c_{2} = 11.3524019669 ; ; ; ; (c -79.2783767368) (c -11.3524019669) = 0 ; ; ; ; c>0 ; ;


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 = 40 ; ; c = 11.35 ; ;

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

p = a+b+c = 50+40+11.35 = 101.35 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 101.35 }{ 2 } = 50.68 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 50.68 * (50.68-50)(50.68-40)(50.68-11.35) } ; ; T = sqrt{ 14386.4 } = 119.94 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 119.94 }{ 50 } = 4.8 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 119.94 }{ 40 } = 6 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 119.94 }{ 11.35 } = 21.13 ; ;

6. 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**2-40**2-11.35**2 }{ 2 * 40 * 11.35 } ) = 148° 6'40" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 40**2-50**2-11.35**2 }{ 2 * 50 * 11.35 } ) = 25° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 11.35**2-50**2-40**2 }{ 2 * 40 * 50 } ) = 6° 53'20" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 119.94 }{ 50.68 } = 2.37 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 50 }{ 2 * sin 148° 6'40" } = 47.32 ; ;




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