6 14 19 triangle

Obtuse scalene triangle.

Sides: a = 6 b = 14 c = 19

Area: T = 26.9066086672
Perimeter: p = 39
Semiperimeter: s = 19.5

Angle ∠ A = α = 11.67215712088° = 11°40'18″ = 0.20437073465 rad
Angle ∠ B = β = 28.16765785968° = 28°10' = 0.49215995355 rad
Angle ∠ C = γ = 140.1621850194° = 140°9'43″ = 2.44662857716 rad

Height: h_a = 8.96986955573
Height: h_b = 3.84437266674
Height: h_c = 2.83222196497

Median: m_a = 16.41664551594
Median: m_b = 12.22770192606
Median: m_c = 5.07444457825

Inradius: r = 1.38797993165
Circumradius: R = 14.82993583108

Vertex coordinates: A[19; 0] B[0; 0] C[5.28994736842; 2.83222196497]
Centroid: CG[8.09664912281; 0.94440732166]
Coordinates of the circumscribed circle: U[9.5; -11.38768287029]
Coordinates of the inscribed circle: I[5.5; 1.38797993165]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 168.3288428791° = 168°19'42″ = 0.20437073465 rad
∠ B' = β' = 151.8333421403° = 151°50' = 0.49215995355 rad
∠ C' = γ' = 39.83881498056° = 39°50'17″ = 2.44662857716 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.

$a = 6 ; ; b = 14 ; ; c = 19 ; ;$

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

$p = a+b+c = 6+14+19 = 39 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 39 }{ 2 } = 19.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 19.5 * (19.5-6)(19.5-14)(19.5-19) } ; ; T = sqrt{ 723.94 } = 26.91 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 26.91 }{ 6 } = 8.97 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 26.91 }{ 14 } = 3.84 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 26.91 }{ 19 } = 2.83 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 6**2-14**2-19**2 }{ 2 * 14 * 19 } ) = 11° 40'18" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 14**2-6**2-19**2 }{ 2 * 6 * 19 } ) = 28° 10' ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 19**2-6**2-14**2 }{ 2 * 14 * 6 } ) = 140° 9'43" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 26.91 }{ 19.5 } = 1.38 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 11° 40'18" } = 14.83 ; ;$

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