Triangle calculator SSA

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Triangle has two solutions with side c=3.67767673279 and with side c=0.5587554998

#1 Obtuse scalene triangle.

Sides: a = 2.3   b = 1.8   c = 3.67767673279

Area: T = 1.65221215643
Perimeter: p = 7.77767673279
Semiperimeter: s = 3.8888383664

Angle ∠ A = α = 29.95215535207° = 29°57'6″ = 0.5232753225 rad
Angle ∠ B = β = 23° = 0.4011425728 rad
Angle ∠ C = γ = 127.0488446479° = 127°2'54″ = 2.21774137006 rad

Height: ha = 1.43766274472
Height: hb = 1.8365690627
Height: hc = 0.89986815955

Median: ma = 2.65664655074
Median: mb = 2.93216051903
Median: mc = 0.94109280015

Inradius: r = 0.42548864585
Circumradius: R = 2.30333741987

Vertex coordinates: A[3.67767673279; 0] B[0; 0] C[2.11771611629; 0.89986815955]
Centroid: CG[1.9311309497; 0.32995605318]
Coordinates of the circumscribed circle: U[1.8388383664; -1.3887760139]
Coordinates of the inscribed circle: I[2.0888383664; 0.42548864585]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 150.0488446479° = 150°2'54″ = 0.5232753225 rad
∠ B' = β' = 157° = 0.4011425728 rad
∠ C' = γ' = 52.95215535207° = 52°57'6″ = 2.21774137006 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 = 2.3 ; ; b = 1.8 ; ; c = 3.68 ; ;

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

p = a+b+c = 2.3+1.8+3.68 = 7.78 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 7.78 }{ 2 } = 3.89 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 3.89 * (3.89-2.3)(3.89-1.8)(3.89-3.68) } ; ; T = sqrt{ 2.73 } = 1.65 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1.65 }{ 2.3 } = 1.44 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1.65 }{ 1.8 } = 1.84 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1.65 }{ 3.68 } = 0.9 ; ;

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{ 2.3**2-1.8**2-3.68**2 }{ 2 * 1.8 * 3.68 } ) = 29° 57'6" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 1.8**2-2.3**2-3.68**2 }{ 2 * 2.3 * 3.68 } ) = 23° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 3.68**2-2.3**2-1.8**2 }{ 2 * 1.8 * 2.3 } ) = 127° 2'54" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1.65 }{ 3.89 } = 0.42 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2.3 }{ 2 * sin 29° 57'6" } = 2.3 ; ;





#2 Obtuse scalene triangle.

Sides: a = 2.3   b = 1.8   c = 0.5587554998

Area: T = 0.25105322076
Perimeter: p = 4.6587554998
Semiperimeter: s = 2.3298777499

Angle ∠ A = α = 150.0488446479° = 150°2'54″ = 2.61988394286 rad
Angle ∠ B = β = 23° = 0.4011425728 rad
Angle ∠ C = γ = 6.95215535207° = 6°57'6″ = 0.12113274971 rad

Height: ha = 0.21878540935
Height: hb = 0.27883691195
Height: hc = 0.89986815955

Median: ma = 0.67330035571
Median: mb = 1.41108273416
Median: mc = 2.04662851967

Inradius: r = 0.10875809981
Circumradius: R = 2.30333741987

Vertex coordinates: A[0.5587554998; 0] B[0; 0] C[2.11771611629; 0.89986815955]
Centroid: CG[0.89215720536; 0.32995605318]
Coordinates of the circumscribed circle: U[0.2798777499; 2.28664417345]
Coordinates of the inscribed circle: I[0.5298777499; 0.10875809981]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 29.95215535207° = 29°57'6″ = 2.61988394286 rad
∠ B' = β' = 157° = 0.4011425728 rad
∠ C' = γ' = 173.0488446479° = 173°2'54″ = 0.12113274971 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 2.3 ; ; b = 1.8 ; ; beta = 23° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 1.8**2 = 2.3**2 + c**2 -2 * 1.8 * c * cos (23° ) ; ; ; ; c**2 -4.234c +2.05 =0 ; ; p=1; q=-4.23432232588; r=2.05 ; ; D = q**2 - 4pr = 4.234**2 - 4 * 1 * 2.05 = 9.72948555946 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 4.23 ± sqrt{ 9.73 } }{ 2 } ; ; c_{1,2} = 2.11716116294 ± 1.55960616499 ; ; c_{1} = 3.67676732793 ; ;
c_{2} = 0.557554997954 ; ; ; ; (c -3.67676732793) (c -0.557554997954) = 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 = 2.3 ; ; b = 1.8 ; ; c = 0.56 ; ;

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

p = a+b+c = 2.3+1.8+0.56 = 4.66 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 4.66 }{ 2 } = 2.33 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 2.33 * (2.33-2.3)(2.33-1.8)(2.33-0.56) } ; ; T = sqrt{ 0.06 } = 0.25 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 0.25 }{ 2.3 } = 0.22 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 0.25 }{ 1.8 } = 0.28 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 0.25 }{ 0.56 } = 0.9 ; ;

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{ 2.3**2-1.8**2-0.56**2 }{ 2 * 1.8 * 0.56 } ) = 150° 2'54" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 1.8**2-2.3**2-0.56**2 }{ 2 * 2.3 * 0.56 } ) = 23° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 0.56**2-2.3**2-1.8**2 }{ 2 * 1.8 * 2.3 } ) = 6° 57'6" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 0.25 }{ 2.33 } = 0.11 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2.3 }{ 2 * sin 150° 2'54" } = 2.3 ; ;




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