10 19 21 triangle

Acute scalene triangle.

Sides: a = 10 b = 19 c = 21

Area: T = 94.86883298051
Perimeter: p = 50
Semiperimeter: s = 25

Angle ∠ A = α = 28.3943894835° = 28°23'38″ = 0.49655669523 rad
Angle ∠ B = β = 64.62330664748° = 64°37'23″ = 1.12878852827 rad
Angle ∠ C = γ = 86.98330386902° = 86°58'59″ = 1.51881404185 rad

Height: h_a = 18.9743665961
Height: h_b = 9.98661399795
Height: h_c = 9.03550790291

Median: m_a = 19.39107194297
Median: m_b = 13.42657215821
Median: m_c = 10.96658560997

Inradius: r = 3.79547331922
Circumradius: R = 10.51545732201

Vertex coordinates: A[21; 0] B[0; 0] C[4.28657142857; 9.03550790291]
Centroid: CG[8.42985714286; 3.01216930097]
Coordinates of the circumscribed circle: U[10.5; 0.55333985905]
Coordinates of the inscribed circle: I[6; 3.79547331922]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 151.6066105165° = 151°36'22″ = 0.49655669523 rad
∠ B' = β' = 115.3776933525° = 115°22'37″ = 1.12878852827 rad
∠ C' = γ' = 93.01769613098° = 93°1'1″ = 1.51881404185 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.

$a = 10 ; ; b = 19 ; ; c = 21 ; ;$

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

$p = a+b+c = 10+19+21 = 50 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 50 }{ 2 } = 25 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 25 * (25-10)(25-19)(25-21) } ; ; T = sqrt{ 9000 } = 94.87 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 94.87 }{ 10 } = 18.97 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 94.87 }{ 19 } = 9.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 94.87 }{ 21 } = 9.04 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 10**2-19**2-21**2 }{ 2 * 19 * 21 } ) = 28° 23'38" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 19**2-10**2-21**2 }{ 2 * 10 * 21 } ) = 64° 37'23" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 21**2-10**2-19**2 }{ 2 * 19 * 10 } ) = 86° 58'59" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 94.87 }{ 25 } = 3.79 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 28° 23'38" } = 10.51 ; ;$

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