20 25 28 triangle

Acute scalene triangle.

Sides: a = 20   b = 25   c = 28

Area: T = 242.6311278899
Perimeter: p = 73
Semiperimeter: s = 36.5

Angle ∠ A = α = 43.88765152837° = 43°53'11″ = 0.76659641889 rad
Angle ∠ B = β = 60.05990533933° = 60°3'33″ = 1.04882282273 rad
Angle ∠ C = γ = 76.0544431323° = 76°3'16″ = 1.32774002373 rad

Height: ha = 24.26331278899
Height: hb = 19.41105023119
Height: hc = 17.33108056356

Median: ma = 24.58765817063
Median: mb = 20.87546257451
Median: mc = 17.79904468747

Inradius: r = 6.64774322986
Circumradius: R = 14.42551805286

Vertex coordinates: A[28; 0] B[0; 0] C[9.98221428571; 17.33108056356]
Centroid: CG[12.66107142857; 5.77769352119]
Coordinates of the circumscribed circle: U[14; 3.47664685074]
Coordinates of the inscribed circle: I[11.5; 6.64774322986]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 136.1133484716° = 136°6'49″ = 0.76659641889 rad
∠ B' = β' = 119.9410946607° = 119°56'27″ = 1.04882282273 rad
∠ C' = γ' = 103.9465568677° = 103°56'44″ = 1.32774002373 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 = 20 ; ; b = 25 ; ; c = 28 ; ;

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

p = a+b+c = 20+25+28 = 73 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 73 }{ 2 } = 36.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 36.5 * (36.5-20)(36.5-25)(36.5-28) } ; ; T = sqrt{ 58869.94 } = 242.63 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 242.63 }{ 20 } = 24.26 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 242.63 }{ 25 } = 19.41 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 242.63 }{ 28 } = 17.33 ; ;

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{ 20**2-25**2-28**2 }{ 2 * 25 * 28 } ) = 43° 53'11" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-20**2-28**2 }{ 2 * 20 * 28 } ) = 60° 3'33" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-20**2-25**2 }{ 2 * 25 * 20 } ) = 76° 3'16" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 242.63 }{ 36.5 } = 6.65 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 43° 53'11" } = 14.43 ; ;




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