Triangle calculator SSA

Triangle has two solutions with side c=112.6298611148 and with side c=71.2166451293

#1 Acute scalene triangle.

Sides: a = 105.1 b = 55 c = 112.6298611148

Area: T = 2869.411046718
Perimeter: p = 272.7298611148
Semiperimeter: s = 136.3644305574

Angle ∠ A = α = 67.88545444666° = 67°53'4″ = 1.18548088122 rad
Angle ∠ B = β = 29° = 0.50661454831 rad
Angle ∠ C = γ = 83.11554555334° = 83°6'56″ = 1.45106383584 rad

Height: h_a = 54.60334341994
Height: h_b = 104.3422198806
Height: h_c = 50.95334910879

Median: m_a = 71.36994579251
Median: m_b = 105.4010934647
Median: m_c = 62.16327218495

Inradius: r = 21.04222401603
Circumradius: R = 56.72332968398

Vertex coordinates: A[112.6298611148; 0] B[0; 0] C[91.92325312204; 50.95334910879]
Centroid: CG[68.18437141227; 16.98444970293]
Coordinates of the circumscribed circle: U[56.31443055738; 6.79993670373]
Coordinates of the inscribed circle: I[81.36443055738; 21.04222401603]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 112.1155455533° = 112°6'56″ = 1.18548088122 rad
∠ B' = β' = 151° = 0.50661454831 rad
∠ C' = γ' = 96.88545444666° = 96°53'4″ = 1.45106383584 rad

How did we calculate this triangle?

1. Use Law of Cosines

$a = 105.1 ; ; b = 55 ; ; beta = 29° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 55**2 = 105.1**2 + c**2 -2 * 105.1 * c * cos (29° ) ; ; ; ; c**2 -183.845c +8021.01 =0 ; ; p=1; q=-183.845; r=8021.01 ; ; D = q**2 - 4pr = 183.845**2 - 4 * 1 * 8021.01 = 1714.96698383 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 183.85 ± sqrt{ 1714.97 } }{ 2 } ; ; c_{1,2} = 91.92253122 ± 20.7060799273 ; ;$
$c_{1} = 112.628611147 ; ; c_{2} = 71.2164512927 ; ; ; ; (c -112.628611147) (c -71.2164512927) = 0 ; ; ; ; c>0 ; ;$

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.

$a = 105.1 ; ; b = 55 ; ; c = 112.63 ; ;$

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

$p = a+b+c = 105.1+55+112.63 = 272.73 ; ;$

3. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 272.73 }{ 2 } = 136.36 ; ;$

4. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 136.36 * (136.36-105.1)(136.36-55)(136.36-112.63) } ; ; T = sqrt{ 8233516.43 } = 2869.41 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2869.41 }{ 105.1 } = 54.6 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2869.41 }{ 55 } = 104.34 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2869.41 }{ 112.63 } = 50.95 ; ;$

6. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 105.1**2-55**2-112.63**2 }{ 2 * 55 * 112.63 } ) = 67° 53'4" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 55**2-105.1**2-112.63**2 }{ 2 * 105.1 * 112.63 } ) = 29° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 112.63**2-105.1**2-55**2 }{ 2 * 55 * 105.1 } ) = 83° 6'56" ; ;$

7. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 2869.41 }{ 136.36 } = 21.04 ; ;$

8. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 105.1 }{ 2 * sin 67° 53'4" } = 56.72 ; ;$

#2 Obtuse scalene triangle.

Sides: a = 105.1 b = 55 c = 71.2166451293

Area: T = 1814.363340814
Perimeter: p = 231.3166451293
Semiperimeter: s = 115.6588225647

Angle ∠ A = α = 112.1155455533° = 112°6'56″ = 1.95767838414 rad
Angle ∠ B = β = 29° = 0.50661454831 rad
Angle ∠ C = γ = 38.88545444666° = 38°53'4″ = 0.67986633291 rad

Height: h_a = 34.52664207067
Height: h_b = 65.97768512049
Height: h_c = 50.95334910879

Median: m_a = 35.87332346937
Median: m_b = 85.45655233287
Median: m_c = 75.94444485549

Inradius: r = 15.6877283788
Circumradius: R = 56.72332968398

Vertex coordinates: A[71.2166451293; 0] B[0; 0] C[91.92325312204; 50.95334910879]
Centroid: CG[54.38796608378; 16.98444970293]
Coordinates of the circumscribed circle: U[35.60882256465; 44.15441240506]
Coordinates of the inscribed circle: I[60.65882256465; 15.6877283788]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 67.88545444666° = 67°53'4″ = 1.95767838414 rad
∠ B' = β' = 151° = 0.50661454831 rad
∠ C' = γ' = 141.1155455533° = 141°6'56″ = 0.67986633291 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

$a = 105.1 ; ; b = 55 ; ; beta = 29° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 55**2 = 105.1**2 + c**2 -2 * 105.1 * c * cos (29° ) ; ; ; ; c**2 -183.845c +8021.01 =0 ; ; p=1; q=-183.845; r=8021.01 ; ; D = q**2 - 4pr = 183.845**2 - 4 * 1 * 8021.01 = 1714.96698383 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 183.85 ± sqrt{ 1714.97 } }{ 2 } ; ; c_{1,2} = 91.92253122 ± 20.7060799273 ; ; : Nr. 1$
$c_{1} = 112.628611147 ; ; c_{2} = 71.2164512927 ; ; ; ; (c -112.628611147) (c -71.2164512927) = 0 ; ; ; ; c>0 ; ; : Nr. 1$

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.

$a = 105.1 ; ; b = 55 ; ; c = 71.22 ; ;$

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

$p = a+b+c = 105.1+55+71.22 = 231.32 ; ;$

3. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 231.32 }{ 2 } = 115.66 ; ;$

4. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 115.66 * (115.66-105.1)(115.66-55)(115.66-71.22) } ; ; T = sqrt{ 3291914.58 } = 1814.36 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1814.36 }{ 105.1 } = 34.53 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1814.36 }{ 55 } = 65.98 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1814.36 }{ 71.22 } = 50.95 ; ;$

6. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 105.1**2-55**2-71.22**2 }{ 2 * 55 * 71.22 } ) = 112° 6'56" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 55**2-105.1**2-71.22**2 }{ 2 * 105.1 * 71.22 } ) = 29° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 71.22**2-105.1**2-55**2 }{ 2 * 55 * 105.1 } ) = 38° 53'4" ; ;$

7. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 1814.36 }{ 115.66 } = 15.69 ; ;$

8. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 105.1 }{ 2 * sin 112° 6'56" } = 56.72 ; ;$

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