14 18 20 triangle

Acute scalene triangle.

Sides: a = 14 b = 18 c = 20

Area: T = 122.3766468326
Perimeter: p = 52
Semiperimeter: s = 26

Angle ∠ A = α = 42.83334280661° = 42°50' = 0.74875843497 rad
Angle ∠ B = β = 60.9410718932° = 60°56'27″ = 1.06436161939 rad
Angle ∠ C = γ = 76.2265853002° = 76°13'33″ = 1.333039211 rad

Height: h_a = 17.4822352618
Height: h_b = 13.59773853696
Height: h_c = 12.23876468326

Median: m_a = 17.6921806013
Median: m_b = 14.73109198627
Median: m_c = 12.64991106407

Inradius: r = 4.70767872433
Circumradius: R = 10.29660970948

Vertex coordinates: A[20; 0] B[0; 0] C[6.8; 12.23876468326]
Centroid: CG[8.93333333333; 4.07992156109]
Coordinates of the circumscribed circle: U[10; 2.45114516892]
Coordinates of the inscribed circle: I[8; 4.70767872433]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 137.1676571934° = 137°10' = 0.74875843497 rad
∠ B' = β' = 119.0599281068° = 119°3'33″ = 1.06436161939 rad
∠ C' = γ' = 103.7744146998° = 103°46'27″ = 1.333039211 rad

Calculate another triangle

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.

$a = 14 ; ; b = 18 ; ; c = 20 ; ;$

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

$p = a+b+c = 14+18+20 = 52 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 52 }{ 2 } = 26 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 26 * (26-14)(26-18)(26-20) } ; ; T = sqrt{ 14976 } = 122.38 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 122.38 }{ 14 } = 17.48 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 122.38 }{ 18 } = 13.6 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 122.38 }{ 20 } = 12.24 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 14**2-18**2-20**2 }{ 2 * 18 * 20 } ) = 42° 50' ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 18**2-14**2-20**2 }{ 2 * 14 * 20 } ) = 60° 56'27" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20**2-14**2-18**2 }{ 2 * 18 * 14 } ) = 76° 13'33" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 122.38 }{ 26 } = 4.71 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 42° 50' } = 10.3 ; ;$

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