Triangle calculator SSA

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Triangle has two solutions with side c=3.92200877706 and with side c=0.12549972013

#1 Obtuse scalene triangle.

Sides: a = 2.5   b = 2.4   c = 3.92200877706

Area: T = 2.88802122241
Perimeter: p = 8.82200877706
Semiperimeter: s = 4.41100438853

Angle ∠ A = α = 37.75442772144° = 37°45'15″ = 0.65989364441 rad
Angle ∠ B = β = 36° = 0.62883185307 rad
Angle ∠ C = γ = 106.2465722786° = 106°14'45″ = 1.85443376788 rad

Height: ha = 2.30441697793
Height: hb = 2.44001768534
Height: hc = 1.46994631307

Median: ma = 33.0001740057
Median: mb = 3.06108077471
Median: mc = 1.47107916126

Inradius: r = 0.65331028486
Circumradius: R = 2.042156194

Vertex coordinates: A[3.92200877706; 0] B[0; 0] C[2.02325424859; 1.46994631307]
Centroid: CG[1.98108767522; 0.49898210436]
Coordinates of the circumscribed circle: U[1.96600438853; -0.57111419462]
Coordinates of the inscribed circle: I[2.01100438853; 0.65331028486]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 142.2465722786° = 142°14'45″ = 0.65989364441 rad
∠ B' = β' = 144° = 0.62883185307 rad
∠ C' = γ' = 73.75442772144° = 73°45'15″ = 1.85443376788 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 = 2.5 ; ; b = 2.4 ; ; c = 3.92 ; ;

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

p = a+b+c = 2.5+2.4+3.92 = 8.82 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 8.82 }{ 2 } = 4.41 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 4.41 * (4.41-2.5)(4.41-2.4)(4.41-3.92) } ; ; T = sqrt{ 8.3 } = 2.88 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2.88 }{ 2.5 } = 2.3 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2.88 }{ 2.4 } = 2.4 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2.88 }{ 3.92 } = 1.47 ; ;

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{ 2.5**2-2.4**2-3.92**2 }{ 2 * 2.4 * 3.92 } ) = 37° 45'15" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 2.4**2-2.5**2-3.92**2 }{ 2 * 2.5 * 3.92 } ) = 36° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 3.92**2-2.5**2-2.4**2 }{ 2 * 2.4 * 2.5 } ) = 106° 14'45" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 2.88 }{ 4.41 } = 0.65 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2.5 }{ 2 * sin 37° 45'15" } = 2.04 ; ;





#2 Obtuse scalene triangle.

Sides: a = 2.5   b = 2.4   c = 0.12549972013

Area: T = 0.09218393893
Perimeter: p = 5.02549972013
Semiperimeter: s = 2.51224986006

Angle ∠ A = α = 142.2465722786° = 142°14'45″ = 2.48326562095 rad
Angle ∠ B = β = 36° = 0.62883185307 rad
Angle ∠ C = γ = 1.75442772144° = 1°45'15″ = 0.03106179134 rad

Height: ha = 0.07334715115
Height: hb = 0.07765328245
Height: hc = 1.46994631307

Median: ma = 1.15112220247
Median: mb = 1.30110811466
Median: mc = 2.45497130291

Inradius: r = 0.03765530111
Circumradius: R = 2.042156194

Vertex coordinates: A[0.12549972013; 0] B[0; 0] C[2.02325424859; 1.46994631307]
Centroid: CG[0.71658465624; 0.49898210436]
Coordinates of the circumscribed circle: U[0.06224986006; 2.04106050769]
Coordinates of the inscribed circle: I[0.11224986006; 0.03765530111]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 37.75442772144° = 37°45'15″ = 2.48326562095 rad
∠ B' = β' = 144° = 0.62883185307 rad
∠ C' = γ' = 178.2465722786° = 178°14'45″ = 0.03106179134 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 2.5 ; ; b = 2.4 ; ; beta = 36° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 2.4**2 = 2.5**2 + c**2 -2 * 2.4 * c * cos (36° ) ; ; ; ; c**2 -4.045c +0.49 =0 ; ; p=1; q=-4.04508497187; r=0.49 ; ; D = q**2 - 4pr = 4.045**2 - 4 * 1 * 0.49 = 14.4027124297 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 4.05 ± sqrt{ 14.4 } }{ 2 } ; ; c_{1,2} = 2.02254248594 ± 1.89754528468 ; ; c_{1} = 3.92008777062 ; ;
c_{2} = 0.124997201255 ; ; ; ; (c -3.92008777062) (c -0.124997201255) = 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 = 2.5 ; ; b = 2.4 ; ; c = 0.12 ; ;

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

p = a+b+c = 2.5+2.4+0.12 = 5.02 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 5.02 }{ 2 } = 2.51 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 2.51 * (2.51-2.5)(2.51-2.4)(2.51-0.12) } ; ; T = sqrt{ 0.01 } = 0.09 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 0.09 }{ 2.5 } = 0.07 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 0.09 }{ 2.4 } = 0.08 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 0.09 }{ 0.12 } = 1.47 ; ;

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{ 2.5**2-2.4**2-0.12**2 }{ 2 * 2.4 * 0.12 } ) = 142° 14'45" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 2.4**2-2.5**2-0.12**2 }{ 2 * 2.5 * 0.12 } ) = 36° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 0.12**2-2.5**2-2.4**2 }{ 2 * 2.4 * 2.5 } ) = 1° 45'15" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 0.09 }{ 2.51 } = 0.04 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2.5 }{ 2 * sin 142° 14'45" } = 2.04 ; ;




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