Triangle calculator SSA

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Triangle has two solutions with side c=122.0411264747 and with side c=41.78991441111

#1 Acute scalene triangle.

Sides: a = 100   b = 70   c = 122.0411264747

Area: T = 35009.99968606
Perimeter: p = 292.0411264747
Semiperimeter: s = 146.0210632373

Angle ∠ A = α = 55.02442676241° = 55°1'27″ = 0.96603546385 rad
Angle ∠ B = β = 35° = 0.61108652382 rad
Angle ∠ C = γ = 89.97657323759° = 89°58'33″ = 1.57703727769 rad

Height: ha = 709.9999937212
Height: hb = 100.9999910303
Height: hc = 57.35876436351

Median: ma = 86.00660181062
Median: mb = 105.9344107588
Median: mc = 61.04549213674

Inradius: r = 23.96992133171
Circumradius: R = 61.02106378467

Vertex coordinates: A[122.0411264747; 0] B[0; 0] C[81.91552044289; 57.35876436351]
Centroid: CG[67.98554897252; 19.1199214545]
Coordinates of the circumscribed circle: U[61.02106323734; 0.02658452869]
Coordinates of the inscribed circle: I[76.02106323734; 23.96992133171]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 124.9765732376° = 124°58'33″ = 0.96603546385 rad
∠ B' = β' = 145° = 0.61108652382 rad
∠ C' = γ' = 90.02442676241° = 90°1'27″ = 1.57703727769 rad




How did we calculate this triangle?

1. Use Law of Cosines

a = 100 ; ; b = 70 ; ; beta = 35° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 70**2 = 100**2 + c**2 -2 * 100 * c * cos (35° ) ; ; ; ; c**2 -163.83c +5100 =0 ; ; p=1; q=-163.83; r=5100 ; ; D = q**2 - 4pr = 163.83**2 - 4 * 1 * 5100 = 6440.40286651 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 163.83 ± sqrt{ 6440.4 } }{ 2 } ; ; c_{1,2} = 81.91520443 ± 40.1260603178 ; ; c_{1} = 122.041264748 ; ;
c_{2} = 41.7891441122 ; ; ; ; (c -122.041264748) (c -41.7891441122) = 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 = 100 ; ; b = 70 ; ; c = 122.04 ; ;

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

p = a+b+c = 100+70+122.04 = 292.04 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 292.04 }{ 2 } = 146.02 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 146.02 * (146.02-100)(146.02-70)(146.02-122.04) } ; ; T = sqrt{ 12249997.8 } = 3500 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 3500 }{ 100 } = 70 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 3500 }{ 70 } = 100 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 3500 }{ 122.04 } = 57.36 ; ;

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{ 100**2-70**2-122.04**2 }{ 2 * 70 * 122.04 } ) = 55° 1'27" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 70**2-100**2-122.04**2 }{ 2 * 100 * 122.04 } ) = 35° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 122.04**2-100**2-70**2 }{ 2 * 70 * 100 } ) = 89° 58'33" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 3500 }{ 146.02 } = 23.97 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 100 }{ 2 * sin 55° 1'27" } = 61.02 ; ;





#2 Obtuse scalene triangle.

Sides: a = 100   b = 70   c = 41.78991441111

Area: T = 1198.463341787
Perimeter: p = 211.7899144111
Semiperimeter: s = 105.8954572056

Angle ∠ A = α = 124.9765732376° = 124°58'33″ = 2.1811238015 rad
Angle ∠ B = β = 35° = 0.61108652382 rad
Angle ∠ C = γ = 20.02442676241° = 20°1'27″ = 0.34994894003 rad

Height: ha = 23.96992683574
Height: hb = 34.24218119391
Height: hc = 57.35876436351

Median: ma = 28.69108745557
Median: mb = 68.1777461692
Median: mc = 83.74661453359

Inradius: r = 11.31875150964
Circumradius: R = 61.02106378467

Vertex coordinates: A[41.78991441111; 0] B[0; 0] C[81.91552044289; 57.35876436351]
Centroid: CG[41.23547828467; 19.1199214545]
Coordinates of the circumscribed circle: U[20.89545720555; 57.33217983482]
Coordinates of the inscribed circle: I[35.89545720555; 11.31875150964]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 55.02442676241° = 55°1'27″ = 2.1811238015 rad
∠ B' = β' = 145° = 0.61108652382 rad
∠ C' = γ' = 159.9765732376° = 159°58'33″ = 0.34994894003 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 100 ; ; b = 70 ; ; beta = 35° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 70**2 = 100**2 + c**2 -2 * 100 * c * cos (35° ) ; ; ; ; c**2 -163.83c +5100 =0 ; ; p=1; q=-163.83; r=5100 ; ; D = q**2 - 4pr = 163.83**2 - 4 * 1 * 5100 = 6440.40286651 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 163.83 ± sqrt{ 6440.4 } }{ 2 } ; ; c_{1,2} = 81.91520443 ± 40.1260603178 ; ; c_{1} = 122.041264748 ; ; : Nr. 1
c_{2} = 41.7891441122 ; ; ; ; (c -122.041264748) (c -41.7891441122) = 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 = 100 ; ; b = 70 ; ; c = 41.79 ; ;

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

p = a+b+c = 100+70+41.79 = 211.79 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 211.79 }{ 2 } = 105.89 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 105.89 * (105.89-100)(105.89-70)(105.89-41.79) } ; ; T = sqrt{ 1436314.56 } = 1198.46 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1198.46 }{ 100 } = 23.97 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1198.46 }{ 70 } = 34.24 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1198.46 }{ 41.79 } = 57.36 ; ;

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{ 100**2-70**2-41.79**2 }{ 2 * 70 * 41.79 } ) = 124° 58'33" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 70**2-100**2-41.79**2 }{ 2 * 100 * 41.79 } ) = 35° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 41.79**2-100**2-70**2 }{ 2 * 70 * 100 } ) = 20° 1'27" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1198.46 }{ 105.89 } = 11.32 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 100 }{ 2 * sin 124° 58'33" } = 61.02 ; ;




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