Triangle calculator SSA

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Triangle has two solutions with side c=3.50771247279 and with side c=0.69328752721

#1 Acute scalene triangle.

Sides: a = 4.2   b = 3.9   c = 3.50771247279

Area: T = 6.37882441281
Perimeter: p = 11.60771247279
Semiperimeter: s = 5.8043562364

Angle ∠ A = α = 68.85105906375° = 68°51'2″ = 1.20216694986 rad
Angle ∠ B = β = 60° = 1.04771975512 rad
Angle ∠ C = γ = 51.14994093625° = 51°8'58″ = 0.89327256038 rad

Height: ha = 3.03772591086
Height: hb = 3.27108944247
Height: hc = 3.63773066959

Median: ma = 3.05769530465
Median: mb = 3.3421775266
Median: mc = 3.65437677862

Inradius: r = 1.09990222433
Circumradius: R = 2.25216660498

Vertex coordinates: A[3.50771247279; 0] B[0; 0] C[2.1; 3.63773066959]
Centroid: CG[1.8699041576; 1.21224355653]
Coordinates of the circumscribed circle: U[1.7543562364; 1.41224514277]
Coordinates of the inscribed circle: I[1.9043562364; 1.09990222433]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 111.1499409362° = 111°8'58″ = 1.20216694986 rad
∠ B' = β' = 120° = 1.04771975512 rad
∠ C' = γ' = 128.8510590638° = 128°51'2″ = 0.89327256038 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 = 4.2 ; ; b = 3.9 ; ; c = 3.51 ; ;

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

p = a+b+c = 4.2+3.9+3.51 = 11.61 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 11.61 }{ 2 } = 5.8 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 5.8 * (5.8-4.2)(5.8-3.9)(5.8-3.51) } ; ; T = sqrt{ 40.68 } = 6.38 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 6.38 }{ 4.2 } = 3.04 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 6.38 }{ 3.9 } = 3.27 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 6.38 }{ 3.51 } = 3.64 ; ;

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{ 4.2**2-3.9**2-3.51**2 }{ 2 * 3.9 * 3.51 } ) = 68° 51'2" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 3.9**2-4.2**2-3.51**2 }{ 2 * 4.2 * 3.51 } ) = 60° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 3.51**2-4.2**2-3.9**2 }{ 2 * 3.9 * 4.2 } ) = 51° 8'58" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 6.38 }{ 5.8 } = 1.1 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 4.2 }{ 2 * sin 68° 51'2" } = 2.25 ; ;





#2 Obtuse scalene triangle.

Sides: a = 4.2   b = 3.9   c = 0.69328752721

Area: T = 1.26600999332
Perimeter: p = 8.79328752721
Semiperimeter: s = 4.3966437636

Angle ∠ A = α = 111.1499409362° = 111°8'58″ = 1.9439923155 rad
Angle ∠ B = β = 60° = 1.04771975512 rad
Angle ∠ C = γ = 8.85105906375° = 8°51'2″ = 0.15444719474 rad

Height: ha = 0.66000475873
Height: hb = 0.6466205094
Height: hc = 3.63773066959

Median: ma = 1.85333855701
Median: mb = 2.29329321995
Median: mc = 4.03879426648

Inradius: r = 0.28766184028
Circumradius: R = 2.25216660498

Vertex coordinates: A[0.69328752721; 0] B[0; 0] C[2.1; 3.63773066959]
Centroid: CG[0.9310958424; 1.21224355653]
Coordinates of the circumscribed circle: U[0.3466437636; 2.22548552682]
Coordinates of the inscribed circle: I[0.4966437636; 0.28766184028]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 68.85105906375° = 68°51'2″ = 1.9439923155 rad
∠ B' = β' = 120° = 1.04771975512 rad
∠ C' = γ' = 171.1499409362° = 171°8'58″ = 0.15444719474 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 4.2 ; ; b = 3.9 ; ; beta = 60° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 3.9**2 = 4.2**2 + c**2 -2 * 3.9 * c * cos (60° ) ; ; ; ; c**2 -4.2c +2.43 =0 ; ; p=1; q=-4.2; r=2.43 ; ; D = q**2 - 4pr = 4.2**2 - 4 * 1 * 2.43 = 7.92 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 4.2 ± sqrt{ 7.92 } }{ 2 } ; ; c_{1,2} = 2.1 ± 1.40712472795 ; ; c_{1} = 3.50712472795 ; ; c_{2} = 0.692875272053 ; ;
 ; ; (c -3.50712472795) (c -0.692875272053) = 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 = 4.2 ; ; b = 3.9 ; ; c = 0.69 ; ;

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

p = a+b+c = 4.2+3.9+0.69 = 8.79 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 8.79 }{ 2 } = 4.4 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 4.4 * (4.4-4.2)(4.4-3.9)(4.4-0.69) } ; ; T = sqrt{ 1.59 } = 1.26 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1.26 }{ 4.2 } = 0.6 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1.26 }{ 3.9 } = 0.65 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1.26 }{ 0.69 } = 3.64 ; ;

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{ 4.2**2-3.9**2-0.69**2 }{ 2 * 3.9 * 0.69 } ) = 111° 8'58" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 3.9**2-4.2**2-0.69**2 }{ 2 * 4.2 * 0.69 } ) = 60° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 0.69**2-4.2**2-3.9**2 }{ 2 * 3.9 * 4.2 } ) = 8° 51'2" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1.26 }{ 4.4 } = 0.29 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 4.2 }{ 2 * sin 111° 8'58" } = 2.25 ; ;




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