Triangle calculator SSA

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Triangle has two solutions with side c=47.58774152732 and with side c=16.23545442711

#1 Acute scalene triangle.

Sides: a = 39.1   b = 27.5   c = 47.58774152732

Area: T = 537.6010576973
Perimeter: p = 114.1877415273
Semiperimeter: s = 57.09437076366

Angle ∠ A = α = 55.2466134099° = 55°14'46″ = 0.9644226939 rad
Angle ∠ B = β = 35.3° = 35°18' = 0.6166101226 rad
Angle ∠ C = γ = 89.4543865901° = 89°27'14″ = 1.56112644886 rad

Height: ha = 27.49987507403
Height: hb = 39.09882237798
Height: hc = 22.59442331134

Median: ma = 33.58987413606
Median: mb = 41.32334019194
Median: mc = 24.0088112731

Inradius: r = 9.41661090465
Circumradius: R = 23.79547885773

Vertex coordinates: A[47.58774152732; 0] B[0; 0] C[31.91109797721; 22.59442331134]
Centroid: CG[26.49994650151; 7.53114110378]
Coordinates of the circumscribed circle: U[23.79437076366; 0.22768046399]
Coordinates of the inscribed circle: I[29.59437076366; 9.41661090465]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 124.7543865901° = 124°45'14″ = 0.9644226939 rad
∠ B' = β' = 144.7° = 144°42' = 0.6166101226 rad
∠ C' = γ' = 90.5466134099° = 90°32'46″ = 1.56112644886 rad




How did we calculate this triangle?

1. Use Law of Cosines

a = 39.1 ; ; b = 27.5 ; ; beta = 35° 18' ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 27.5**2 = 39.1**2 + c**2 -2 * 39.1 * c * cos (35° 18') ; ; ; ; c**2 -63.822c +772.56 =0 ; ; p=1; q=-63.822; r=772.56 ; ; D = q**2 - 4pr = 63.822**2 - 4 * 1 * 772.56 = 983.00252007 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 63.82 ± sqrt{ 983 } }{ 2 } ; ; c_{1,2} = 31.91097977 ± 15.676435501 ; ; c_{1} = 47.587415271 ; ;
c_{2} = 16.234544269 ; ; ; ; (c -47.587415271) (c -16.234544269) = 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 = 39.1 ; ; b = 27.5 ; ; c = 47.59 ; ;

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

p = a+b+c = 39.1+27.5+47.59 = 114.19 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 114.19 }{ 2 } = 57.09 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 57.09 * (57.09-39.1)(57.09-27.5)(57.09-47.59) } ; ; T = sqrt{ 289014.38 } = 537.6 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 537.6 }{ 39.1 } = 27.5 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 537.6 }{ 27.5 } = 39.1 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 537.6 }{ 47.59 } = 22.59 ; ;

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{ 39.1**2-27.5**2-47.59**2 }{ 2 * 27.5 * 47.59 } ) = 55° 14'46" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27.5**2-39.1**2-47.59**2 }{ 2 * 39.1 * 47.59 } ) = 35° 18' ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 47.59**2-39.1**2-27.5**2 }{ 2 * 27.5 * 39.1 } ) = 89° 27'14" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 537.6 }{ 57.09 } = 9.42 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 39.1 }{ 2 * sin 55° 14'46" } = 23.79 ; ;





#2 Obtuse scalene triangle.

Sides: a = 39.1   b = 27.5   c = 16.23545442711

Area: T = 183.4043538876
Perimeter: p = 82.83545442711
Semiperimeter: s = 41.41772721356

Angle ∠ A = α = 124.7543865901° = 124°45'14″ = 2.17773657146 rad
Angle ∠ B = β = 35.3° = 35°18' = 0.6166101226 rad
Angle ∠ C = γ = 19.9466134099° = 19°56'46″ = 0.34881257131 rad

Height: ha = 9.38112551855
Height: hb = 13.3388439191
Height: hc = 22.59442331134

Median: ma = 11.30105625455
Median: mb = 26.59217790651
Median: mc = 32.81221912264

Inradius: r = 4.42881897242
Circumradius: R = 23.79547885773

Vertex coordinates: A[16.23545442711; 0] B[0; 0] C[31.91109797721; 22.59442331134]
Centroid: CG[16.04985080144; 7.53114110378]
Coordinates of the circumscribed circle: U[8.11772721356; 22.36774284735]
Coordinates of the inscribed circle: I[13.91772721356; 4.42881897242]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 55.2466134099° = 55°14'46″ = 2.17773657146 rad
∠ B' = β' = 144.7° = 144°42' = 0.6166101226 rad
∠ C' = γ' = 160.0543865901° = 160°3'14″ = 0.34881257131 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 39.1 ; ; b = 27.5 ; ; beta = 35° 18' ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 27.5**2 = 39.1**2 + c**2 -2 * 39.1 * c * cos (35° 18') ; ; ; ; c**2 -63.822c +772.56 =0 ; ; p=1; q=-63.822; r=772.56 ; ; D = q**2 - 4pr = 63.822**2 - 4 * 1 * 772.56 = 983.00252007 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 63.82 ± sqrt{ 983 } }{ 2 } ; ; c_{1,2} = 31.91097977 ± 15.676435501 ; ; c_{1} = 47.587415271 ; ; : Nr. 1
c_{2} = 16.234544269 ; ; ; ; (c -47.587415271) (c -16.234544269) = 0 ; ; ; ; c>0 ; ; : 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 = 39.1 ; ; b = 27.5 ; ; c = 16.23 ; ;

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

p = a+b+c = 39.1+27.5+16.23 = 82.83 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 82.83 }{ 2 } = 41.42 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 41.42 * (41.42-39.1)(41.42-27.5)(41.42-16.23) } ; ; T = sqrt{ 33636.86 } = 183.4 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 183.4 }{ 39.1 } = 9.38 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 183.4 }{ 27.5 } = 13.34 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 183.4 }{ 16.23 } = 22.59 ; ;

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{ 39.1**2-27.5**2-16.23**2 }{ 2 * 27.5 * 16.23 } ) = 124° 45'14" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27.5**2-39.1**2-16.23**2 }{ 2 * 39.1 * 16.23 } ) = 35° 18' ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 16.23**2-39.1**2-27.5**2 }{ 2 * 27.5 * 39.1 } ) = 19° 56'46" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 183.4 }{ 41.42 } = 4.43 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 39.1 }{ 2 * sin 124° 45'14" } = 23.79 ; ;




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