Triangle calculator SSA

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Triangle has two solutions with side c=43.94443174473 and with side c=11.62883521894

#1 Acute scalene triangle.

Sides: a = 40   b = 33   c = 43.94443174473

Area: T = 632.2187930772
Perimeter: p = 116.9444317447
Semiperimeter: s = 58.47221587237

Angle ∠ A = α = 60.68333509748° = 60°41' = 1.05991242757 rad
Angle ∠ B = β = 46° = 0.80328514559 rad
Angle ∠ C = γ = 73.31766490252° = 73°19' = 1.2879616922 rad

Height: ha = 31.61108965386
Height: hb = 38.31662382286
Height: hc = 28.77435920135

Median: ma = 33.3177435645
Median: mb = 38.64332596704
Median: mc = 29.35551399421

Inradius: r = 10.81222898927
Circumradius: R = 22.93876992518

Vertex coordinates: A[43.94443174473; 0] B[0; 0] C[27.78663348184; 28.77435920135]
Centroid: CG[23.91102174219; 9.59111973378]
Coordinates of the circumscribed circle: U[21.97221587237; 6.58550047826]
Coordinates of the inscribed circle: I[25.47221587237; 10.81222898927]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 119.3176649025° = 119°19' = 1.05991242757 rad
∠ B' = β' = 134° = 0.80328514559 rad
∠ C' = γ' = 106.6833350975° = 106°41' = 1.2879616922 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 = 40 ; ; b = 33 ; ; c = 43.94 ; ;

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

p = a+b+c = 40+33+43.94 = 116.94 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 116.94 }{ 2 } = 58.47 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 58.47 * (58.47-40)(58.47-33)(58.47-43.94) } ; ; T = sqrt{ 399699.51 } = 632.22 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 632.22 }{ 40 } = 31.61 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 632.22 }{ 33 } = 38.32 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 632.22 }{ 43.94 } = 28.77 ; ;

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{ 40**2-33**2-43.94**2 }{ 2 * 33 * 43.94 } ) = 60° 41' ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 33**2-40**2-43.94**2 }{ 2 * 40 * 43.94 } ) = 46° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 43.94**2-40**2-33**2 }{ 2 * 33 * 40 } ) = 73° 19' ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 632.22 }{ 58.47 } = 10.81 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 40 }{ 2 * sin 60° 41' } = 22.94 ; ;





#2 Obtuse scalene triangle.

Sides: a = 40   b = 33   c = 11.62883521894

Area: T = 167.2954730844
Perimeter: p = 84.62883521894
Semiperimeter: s = 42.31441760947

Angle ∠ A = α = 119.3176649025° = 119°19' = 2.08224683779 rad
Angle ∠ B = β = 46° = 0.80328514559 rad
Angle ∠ C = γ = 14.68333509748° = 14°41' = 0.25662728197 rad

Height: ha = 8.36547365422
Height: hb = 10.13990745966
Height: hc = 28.77435920135

Median: ma = 14.5643972237
Median: mb = 24.43999853959
Median: mc = 36.20435268495

Inradius: r = 3.95436331859
Circumradius: R = 22.93876992518

Vertex coordinates: A[11.62883521894; 0] B[0; 0] C[27.78663348184; 28.77435920135]
Centroid: CG[13.13882290026; 9.59111973378]
Coordinates of the circumscribed circle: U[5.81441760947; 22.18985872309]
Coordinates of the inscribed circle: I[9.31441760947; 3.95436331859]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 60.68333509748° = 60°41' = 2.08224683779 rad
∠ B' = β' = 134° = 0.80328514559 rad
∠ C' = γ' = 165.3176649025° = 165°19' = 0.25662728197 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 40 ; ; b = 33 ; ; beta = 46° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 33**2 = 40**2 + c**2 -2 * 33 * c * cos (46° ) ; ; ; ; c**2 -55.573c +511 =0 ; ; p=1; q=-55.5726696367; r=511 ; ; D = q**2 - 4pr = 55.573**2 - 4 * 1 * 511 = 1044.32161055 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 55.57 ± sqrt{ 1044.32 } }{ 2 } ; ; c_{1,2} = 27.7863348184 ± 16.157982629 ; ; c_{1} = 43.9443174473 ; ;
c_{2} = 11.6283521894 ; ; ; ; (c -43.9443174473) (c -11.6283521894) = 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 = 40 ; ; b = 33 ; ; c = 11.63 ; ;

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

p = a+b+c = 40+33+11.63 = 84.63 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 84.63 }{ 2 } = 42.31 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 42.31 * (42.31-40)(42.31-33)(42.31-11.63) } ; ; T = sqrt{ 27987.53 } = 167.29 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 167.29 }{ 40 } = 8.36 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 167.29 }{ 33 } = 10.14 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 167.29 }{ 11.63 } = 28.77 ; ;

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{ 40**2-33**2-11.63**2 }{ 2 * 33 * 11.63 } ) = 119° 19' ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 33**2-40**2-11.63**2 }{ 2 * 40 * 11.63 } ) = 46° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 11.63**2-40**2-33**2 }{ 2 * 33 * 40 } ) = 14° 41' ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 167.29 }{ 42.31 } = 3.95 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 40 }{ 2 * sin 119° 19' } = 22.94 ; ;




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