Triangle calculator SSA

Triangle has two solutions with side c=104.3221603047 and with side c=48.8877285577

#1 Acute scalene triangle.

Sides: a = 100 b = 70 c = 104.3221603047

Area: T = 3352.832169306
Perimeter: p = 274.3221603047
Semiperimeter: s = 137.1610801523

Angle ∠ A = α = 66.67441765214° = 66°40'27″ = 1.16436839064 rad
Angle ∠ B = β = 40° = 0.69881317008 rad
Angle ∠ C = γ = 73.32658234786° = 73°19'33″ = 1.28797770464 rad

Height: h_a = 67.05766338611
Height: h_b = 95.79551912302
Height: h_c = 64.27987609687

Median: m_a = 73.42768236486
Median: m_b = 96.00325959603
Median: m_c = 68.77695483804

Inradius: r = 24.4454532664
Circumradius: R = 54.45503339401

Vertex coordinates: A[104.3221603047; 0] B[0; 0] C[76.60444443119; 64.27987609687]
Centroid: CG[60.30986824529; 21.42662536562]
Coordinates of the circumscribed circle: U[52.16108015234; 15.62333687349]
Coordinates of the inscribed circle: I[67.16108015234; 24.4454532664]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 113.3265823479° = 113°19'33″ = 1.16436839064 rad
∠ B' = β' = 140° = 0.69881317008 rad
∠ C' = γ' = 106.6744176521° = 106°40'27″ = 1.28797770464 rad

How did we calculate this triangle?

1. Use Law of Cosines

$a = 100 ; ; b = 70 ; ; beta = 40° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 70**2 = 100**2 + c**2 -2 * 100 * c * cos (40° ) ; ; ; ; c**2 -153.209c +5100 =0 ; ; p=1; q=-153.209; r=5100 ; ; D = q**2 - 4pr = 153.209**2 - 4 * 1 * 5100 = 3072.96355334 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 153.21 ± sqrt{ 3072.96 } }{ 2 } ; ; c_{1,2} = 76.60444431 ± 27.7171587349 ; ; c_{1} = 104.321603045 ; ;$
$c_{2} = 48.8872855751 ; ; ; ; (c -104.321603045) (c -48.8872855751) = 0 ; ; ; ; c>0 ; ;$

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.

$a = 100 ; ; b = 70 ; ; c = 104.32 ; ;$

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

$p = a+b+c = 100+70+104.32 = 274.32 ; ;$

3. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 274.32 }{ 2 } = 137.16 ; ;$

4. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 137.16 * (137.16-100)(137.16-70)(137.16-104.32) } ; ; T = sqrt{ 11241480.36 } = 3352.83 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 3352.83 }{ 100 } = 67.06 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 3352.83 }{ 70 } = 95.8 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 3352.83 }{ 104.32 } = 64.28 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 100**2-70**2-104.32**2 }{ 2 * 70 * 104.32 } ) = 66° 40'27" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 70**2-100**2-104.32**2 }{ 2 * 100 * 104.32 } ) = 40° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 104.32**2-100**2-70**2 }{ 2 * 70 * 100 } ) = 73° 19'33" ; ;$

7. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 3352.83 }{ 137.16 } = 24.44 ; ;$

8. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 100 }{ 2 * sin 66° 40'27" } = 54.45 ; ;$

#2 Obtuse scalene triangle.

Sides: a = 100 b = 70 c = 48.8877285577

Area: T = 1571.207707201
Perimeter: p = 218.8877285577
Semiperimeter: s = 109.4443642789

Angle ∠ A = α = 113.3265823479° = 113°19'33″ = 1.97879087472 rad
Angle ∠ B = β = 40° = 0.69881317008 rad
Angle ∠ C = γ = 26.67441765214° = 26°40'27″ = 0.46655522056 rad

Height: h_a = 31.42441414401
Height: h_b = 44.89216306287
Height: h_c = 64.27987609687

Median: m_a = 33.83876025384
Median: m_b = 70.49881088083
Median: m_c = 82.78798787582

Inradius: r = 14.3566311906
Circumradius: R = 54.45503339401

Vertex coordinates: A[48.8877285577; 0] B[0; 0] C[76.60444443119; 64.27987609687]
Centroid: CG[41.83105766296; 21.42662536562]
Coordinates of the circumscribed circle: U[24.44436427885; 48.65553922337]
Coordinates of the inscribed circle: I[39.44436427885; 14.3566311906]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 66.67441765214° = 66°40'27″ = 1.97879087472 rad
∠ B' = β' = 140° = 0.69881317008 rad
∠ C' = γ' = 153.3265823479° = 153°19'33″ = 0.46655522056 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

$a = 100 ; ; b = 70 ; ; beta = 40° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 70**2 = 100**2 + c**2 -2 * 100 * c * cos (40° ) ; ; ; ; c**2 -153.209c +5100 =0 ; ; p=1; q=-153.209; r=5100 ; ; D = q**2 - 4pr = 153.209**2 - 4 * 1 * 5100 = 3072.96355334 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 153.21 ± sqrt{ 3072.96 } }{ 2 } ; ; c_{1,2} = 76.60444431 ± 27.7171587349 ; ; c_{1} = 104.321603045 ; ; : Nr. 1$
$c_{2} = 48.8872855751 ; ; ; ; (c -104.321603045) (c -48.8872855751) = 0 ; ; ; ; c>0 ; ; : Nr. 1$

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.

$a = 100 ; ; b = 70 ; ; c = 48.89 ; ;$

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

$p = a+b+c = 100+70+48.89 = 218.89 ; ;$

3. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 218.89 }{ 2 } = 109.44 ; ;$

4. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 109.44 * (109.44-100)(109.44-70)(109.44-48.89) } ; ; T = sqrt{ 2468691.66 } = 1571.21 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1571.21 }{ 100 } = 31.42 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1571.21 }{ 70 } = 44.89 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1571.21 }{ 48.89 } = 64.28 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 100**2-70**2-48.89**2 }{ 2 * 70 * 48.89 } ) = 113° 19'33" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 70**2-100**2-48.89**2 }{ 2 * 100 * 48.89 } ) = 40° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 48.89**2-100**2-70**2 }{ 2 * 70 * 100 } ) = 26° 40'27" ; ;$

7. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 1571.21 }{ 109.44 } = 14.36 ; ;$

8. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 100 }{ 2 * sin 113° 19'33" } = 54.45 ; ;$

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