16 18 29 triangle

Obtuse scalene triangle.

Sides: a = 16   b = 18   c = 29

Area: T = 128.368836643
Perimeter: p = 63
Semiperimeter: s = 31.5

Angle ∠ A = α = 29.46111189642° = 29°27'40″ = 0.51441935272 rad
Angle ∠ B = β = 33.59545228553° = 33°35'40″ = 0.58663350345 rad
Angle ∠ C = γ = 116.944435818° = 116°56'40″ = 2.04110640919 rad

Height: ha = 16.04660458038
Height: hb = 14.26331518256
Height: hc = 8.85329907883

Median: ma = 22.77105950735
Median: mb = 21.62217483104
Median: mc = 8.93302855497

Inradius: r = 4.07551862359
Circumradius: R = 16.26656895781

Vertex coordinates: A[29; 0] B[0; 0] C[13.32875862069; 8.85329907883]
Centroid: CG[14.10991954023; 2.95109969294]
Coordinates of the circumscribed circle: U[14.5; -7.37703905901]
Coordinates of the inscribed circle: I[13.5; 4.07551862359]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 150.5398881036° = 150°32'20″ = 0.51441935272 rad
∠ B' = β' = 146.4055477145° = 146°24'20″ = 0.58663350345 rad
∠ C' = γ' = 63.05656418195° = 63°3'20″ = 2.04110640919 rad

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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 = 16 ; ; b = 18 ; ; c = 29 ; ;

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

p = a+b+c = 16+18+29 = 63 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 63 }{ 2 } = 31.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31.5 * (31.5-16)(31.5-18)(31.5-29) } ; ; T = sqrt{ 16478.44 } = 128.37 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 128.37 }{ 16 } = 16.05 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 128.37 }{ 18 } = 14.26 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 128.37 }{ 29 } = 8.85 ; ;

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{ 16**2-18**2-29**2 }{ 2 * 18 * 29 } ) = 29° 27'40" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 18**2-16**2-29**2 }{ 2 * 16 * 29 } ) = 33° 35'40" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-16**2-18**2 }{ 2 * 18 * 16 } ) = 116° 56'40" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 128.37 }{ 31.5 } = 4.08 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 29° 27'40" } = 16.27 ; ;




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