Triangle calculator SSA

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Triangle has two solutions with side c=69.82553913618 and with side c=30.63435554772

#1 Obtuse scalene triangle.

Sides: a = 58   b = 35   c = 69.82553913618

Area: T = 1012.468817475
Perimeter: p = 162.8255391362
Semiperimeter: s = 81.41326956809

Angle ∠ A = α = 55.952226763° = 55°57'8″ = 0.97765512941 rad
Angle ∠ B = β = 30° = 0.52435987756 rad
Angle ∠ C = γ = 94.048773237° = 94°2'52″ = 1.64114425839 rad

Height: ha = 34.91326956809
Height: hb = 57.85553242712
Height: hc = 29

Median: ma = 47.00331130821
Median: mb = 61.75438876461
Median: mc = 32.79663973676

Inradius: r = 12.43662443262
Circumradius: R = 35

Vertex coordinates: A[69.82553913618; 0] B[0; 0] C[50.22994734195; 29]
Centroid: CG[40.01882882604; 9.66766666667]
Coordinates of the circumscribed circle: U[34.91326956809; -2.47105627485]
Coordinates of the inscribed circle: I[46.41326956809; 12.43662443262]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 124.048773237° = 124°2'52″ = 0.97765512941 rad
∠ B' = β' = 150° = 0.52435987756 rad
∠ C' = γ' = 85.952226763° = 85°57'8″ = 1.64114425839 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 = 58 ; ; b = 35 ; ; c = 69.83 ; ;

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

p = a+b+c = 58+35+69.83 = 162.83 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 162.83 }{ 2 } = 81.41 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 81.41 * (81.41-58)(81.41-35)(81.41-69.83) } ; ; T = sqrt{ 1025091.8 } = 1012.47 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1012.47 }{ 58 } = 34.91 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1012.47 }{ 35 } = 57.86 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1012.47 }{ 69.83 } = 29 ; ;

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{ 58**2-35**2-69.83**2 }{ 2 * 35 * 69.83 } ) = 55° 57'8" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 35**2-58**2-69.83**2 }{ 2 * 58 * 69.83 } ) = 30° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 69.83**2-58**2-35**2 }{ 2 * 35 * 58 } ) = 94° 2'52" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1012.47 }{ 81.41 } = 12.44 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 58 }{ 2 * sin 55° 57'8" } = 35 ; ;





#2 Obtuse scalene triangle.

Sides: a = 58   b = 35   c = 30.63435554772

Area: T = 444.187655442
Perimeter: p = 123.6343555477
Semiperimeter: s = 61.81767777386

Angle ∠ A = α = 124.048773237° = 124°2'52″ = 2.16550413595 rad
Angle ∠ B = β = 30° = 0.52435987756 rad
Angle ∠ C = γ = 25.952226763° = 25°57'8″ = 0.45329525185 rad

Height: ha = 15.31767777386
Height: hb = 25.3822088824
Height: hc = 29

Median: ma = 15.51547465525
Median: mb = 42.95329668427
Median: mc = 45.38660806824

Inradius: r = 7.18655339387
Circumradius: R = 35

Vertex coordinates: A[30.63435554772; 0] B[0; 0] C[50.22994734195; 29]
Centroid: CG[26.95443429656; 9.66766666667]
Coordinates of the circumscribed circle: U[15.31767777386; 31.47105627485]
Coordinates of the inscribed circle: I[26.81767777386; 7.18655339387]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 55.952226763° = 55°57'8″ = 2.16550413595 rad
∠ B' = β' = 150° = 0.52435987756 rad
∠ C' = γ' = 154.048773237° = 154°2'52″ = 0.45329525185 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 58 ; ; b = 35 ; ; beta = 30° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 35**2 = 58**2 + c**2 -2 * 35 * c * cos (30° ) ; ; ; ; c**2 -100.459c +2139 =0 ; ; p=1; q=-100.458946839; r=2139 ; ; D = q**2 - 4pr = 100.459**2 - 4 * 1 * 2139 = 1536 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 100.46 ± sqrt{ 1536 } }{ 2 } ; ; c_{1,2} = 50.2294734195 ± 19.5959179423 ; ; c_{1} = 69.8253913618 ; ;
c_{2} = 30.6335554772 ; ; ; ; (c -69.8253913618) (c -30.6335554772) = 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 = 58 ; ; b = 35 ; ; c = 30.63 ; ;

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

p = a+b+c = 58+35+30.63 = 123.63 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 123.63 }{ 2 } = 61.82 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 61.82 * (61.82-58)(61.82-35)(61.82-30.63) } ; ; T = sqrt{ 197301.7 } = 444.19 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 444.19 }{ 58 } = 15.32 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 444.19 }{ 35 } = 25.38 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 444.19 }{ 30.63 } = 29 ; ;

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{ 58**2-35**2-30.63**2 }{ 2 * 35 * 30.63 } ) = 124° 2'52" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 35**2-58**2-30.63**2 }{ 2 * 58 * 30.63 } ) = 30° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30.63**2-58**2-35**2 }{ 2 * 35 * 58 } ) = 25° 57'8" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 444.19 }{ 61.82 } = 7.19 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 58 }{ 2 * sin 124° 2'52" } = 35 ; ;




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