50 15.5 35 triangle

Obtuse scalene triangle.

Sides: a = 50   b = 15.5   c = 35

Area: T = 81.59325232099
Perimeter: p = 100.5
Semiperimeter: s = 50.25

Angle ∠ A = α = 162.4944224811° = 162°29'39″ = 2.83660592384 rad
Angle ∠ B = β = 5.35105243676° = 5°21'2″ = 0.09333842669 rad
Angle ∠ C = γ = 12.1555250821° = 12°9'19″ = 0.21221491482 rad

Height: ha = 3.26437009284
Height: hb = 10.52880675109
Height: hc = 4.66224298977

Median: ma = 10.37442469606
Median: mb = 42.45551233657
Median: mc = 32.61770967439

Inradius: r = 1.62437318052
Circumradius: R = 83.11111691761

Vertex coordinates: A[35; 0] B[0; 0] C[49.78221428571; 4.66224298977]
Centroid: CG[28.26107142857; 1.55441432992]
Coordinates of the circumscribed circle: U[17.5; 81.24878703833]
Coordinates of the inscribed circle: I[34.75; 1.62437318052]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 17.50657751886° = 17°30'21″ = 2.83660592384 rad
∠ B' = β' = 174.6499475632° = 174°38'58″ = 0.09333842669 rad
∠ C' = γ' = 167.8454749179° = 167°50'41″ = 0.21221491482 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 = 50 ; ; b = 15.5 ; ; c = 35 ; ;

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

p = a+b+c = 50+15.5+35 = 100.5 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 100.5 }{ 2 } = 50.25 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 50.25 * (50.25-50)(50.25-15.5)(50.25-35) } ; ; T = sqrt{ 6657.34 } = 81.59 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 81.59 }{ 50 } = 3.26 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 81.59 }{ 15.5 } = 10.53 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 81.59 }{ 35 } = 4.66 ; ;

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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 15.5**2+35**2-50**2 }{ 2 * 15.5 * 35 } ) = 162° 29'39" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 50**2+35**2-15.5**2 }{ 2 * 50 * 35 } ) = 5° 21'2" ; ; gamma = 180° - alpha - beta = 180° - 162° 29'39" - 5° 21'2" = 12° 9'19" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 81.59 }{ 50.25 } = 1.62 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 50 }{ 2 * sin 162° 29'39" } = 83.11 ; ;

8. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 15.5**2+2 * 35**2 - 50**2 } }{ 2 } = 10.374 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 35**2+2 * 50**2 - 15.5**2 } }{ 2 } = 42.455 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 15.5**2+2 * 50**2 - 35**2 } }{ 2 } = 32.617 ; ;
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