Triangle calculator SSA

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Triangle has two solutions with side c=322.8711376228 and with side c=23.53987852859

#1 Obtuse scalene triangle.

Sides: a = 200   b = 180   c = 322.8711376228

Area: T = 16143.56988114
Perimeter: p = 702.8711376228
Semiperimeter: s = 351.4365688114

Angle ∠ A = α = 33.74989885959° = 33°44'56″ = 0.58990309702 rad
Angle ∠ B = β = 30° = 0.52435987756 rad
Angle ∠ C = γ = 116.2511011404° = 116°15'4″ = 2.02989629078 rad

Height: ha = 161.4365688114
Height: hb = 179.3732986793
Height: hc = 100

Median: ma = 241.5011475759
Median: mb = 253.0287592949
Median: mc = 100.6990211059

Inradius: r = 45.9366054184
Circumradius: R = 180

Vertex coordinates: A[322.8711376228; 0] B[0; 0] C[173.2055080757; 100]
Centroid: CG[165.3598818995; 33.33333333333]
Coordinates of the circumscribed circle: U[161.4365688114; -79.61548139682]
Coordinates of the inscribed circle: I[171.4365688114; 45.9366054184]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 146.2511011404° = 146°15'4″ = 0.58990309702 rad
∠ B' = β' = 150° = 0.52435987756 rad
∠ C' = γ' = 63.74989885959° = 63°44'56″ = 2.02989629078 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 = 200 ; ; b = 180 ; ; c = 322.87 ; ;

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

p = a+b+c = 200+180+322.87 = 702.87 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 702.87 }{ 2 } = 351.44 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 351.44 * (351.44-200)(351.44-180)(351.44-322.87) } ; ; T = sqrt{ 260614813.97 } = 16143.57 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 16143.57 }{ 200 } = 161.44 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 16143.57 }{ 180 } = 179.37 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 16143.57 }{ 322.87 } = 100 ; ;

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{ 200**2-180**2-322.87**2 }{ 2 * 180 * 322.87 } ) = 33° 44'56" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 180**2-200**2-322.87**2 }{ 2 * 200 * 322.87 } ) = 30° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 322.87**2-200**2-180**2 }{ 2 * 180 * 200 } ) = 116° 15'4" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 16143.57 }{ 351.44 } = 45.94 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 200 }{ 2 * sin 33° 44'56" } = 180 ; ;





#2 Obtuse scalene triangle.

Sides: a = 200   b = 180   c = 23.53987852859

Area: T = 1176.93992643
Perimeter: p = 403.5398785286
Semiperimeter: s = 201.7699392643

Angle ∠ A = α = 146.2511011404° = 146°15'4″ = 2.55325616834 rad
Angle ∠ B = β = 30° = 0.52435987756 rad
Angle ∠ C = γ = 3.74989885959° = 3°44'56″ = 0.06554321946 rad

Height: ha = 11.7699392643
Height: hb = 13.07771029366
Height: hc = 100

Median: ma = 80.48800422861
Median: mb = 110.3549613531
Median: mc = 189.8998608201

Inradius: r = 5.83330911784
Circumradius: R = 180

Vertex coordinates: A[23.53987852859; 0] B[0; 0] C[173.2055080757; 100]
Centroid: CG[65.58112886809; 33.33333333333]
Coordinates of the circumscribed circle: U[11.7699392643; 179.6154813968]
Coordinates of the inscribed circle: I[21.7699392643; 5.83330911784]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 33.74989885959° = 33°44'56″ = 2.55325616834 rad
∠ B' = β' = 150° = 0.52435987756 rad
∠ C' = γ' = 176.2511011404° = 176°15'4″ = 0.06554321946 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 200 ; ; b = 180 ; ; beta = 30° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 180**2 = 200**2 + c**2 -2 * 180 * c * cos (30° ) ; ; ; ; c**2 -346.41c +7600 =0 ; ; p=1; q=-346.410161514; r=7600 ; ; D = q**2 - 4pr = 346.41**2 - 4 * 1 * 7600 = 89600 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 346.41 ± sqrt{ 89600 } }{ 2 } ; ; c_{1,2} = 173.205080757 ± 149.666295471 ; ; c_{1} = 322.871376228 ; ;
c_{2} = 23.5387852859 ; ; ; ; (c -322.871376228) (c -23.5387852859) = 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 = 200 ; ; b = 180 ; ; c = 23.54 ; ;

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

p = a+b+c = 200+180+23.54 = 403.54 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 403.54 }{ 2 } = 201.77 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 201.77 * (201.77-200)(201.77-180)(201.77-23.54) } ; ; T = sqrt{ 1385186.03 } = 1176.94 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1176.94 }{ 200 } = 11.77 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1176.94 }{ 180 } = 13.08 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1176.94 }{ 23.54 } = 100 ; ;

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{ 200**2-180**2-23.54**2 }{ 2 * 180 * 23.54 } ) = 146° 15'4" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 180**2-200**2-23.54**2 }{ 2 * 200 * 23.54 } ) = 30° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23.54**2-200**2-180**2 }{ 2 * 180 * 200 } ) = 3° 44'56" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1176.94 }{ 201.77 } = 5.83 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 200 }{ 2 * sin 146° 15'4" } = 180 ; ;




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