100 210 160 triangle

Obtuse scalene triangle.

Sides: a = 100   b = 210   c = 160

Area: T = 7712.611142545
Perimeter: p = 470
Semiperimeter: s = 235

Angle ∠ A = α = 27.32880163439° = 27°19'41″ = 0.47769638632 rad
Angle ∠ B = β = 105.4044093699° = 105°24'15″ = 1.84396484801 rad
Angle ∠ C = γ = 47.26878899574° = 47°16'4″ = 0.82549803102 rad

Height: ha = 154.2522228509
Height: hb = 73.45334421472
Height: hc = 96.40876428181

Median: ma = 179.8611057486
Median: mb = 82.31103881658
Median: mc = 143.7011078632

Inradius: r = 32.8219623087
Circumradius: R = 108.9132526985

Vertex coordinates: A[160; 0] B[0; 0] C[-26.56325; 96.40876428181]
Centroid: CG[44.47991666667; 32.13658809394]
Coordinates of the circumscribed circle: U[80; 73.90549290256]
Coordinates of the inscribed circle: I[25; 32.8219623087]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 152.6721983656° = 152°40'19″ = 0.47769638632 rad
∠ B' = β' = 74.59659063013° = 74°35'45″ = 1.84396484801 rad
∠ C' = γ' = 132.7322110043° = 132°43'56″ = 0.82549803102 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 = 100 ; ; b = 210 ; ; c = 160 ; ;

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

p = a+b+c = 100+210+160 = 470 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 470 }{ 2 } = 235 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 235 * (235-100)(235-210)(235-160) } ; ; T = sqrt{ 59484375 } = 7712.61 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 7712.61 }{ 100 } = 154.25 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 7712.61 }{ 210 } = 73.45 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 7712.61 }{ 160 } = 96.41 ; ;

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{ 210**2+160**2-100**2 }{ 2 * 210 * 160 } ) = 27° 19'41" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 100**2+160**2-210**2 }{ 2 * 100 * 160 } ) = 105° 24'15" ; ; gamma = 180° - alpha - beta = 180° - 27° 19'41" - 105° 24'15" = 47° 16'4" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 7712.61 }{ 235 } = 32.82 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 100 }{ 2 * sin 27° 19'41" } = 108.91 ; ;

8. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 210**2+2 * 160**2 - 100**2 } }{ 2 } = 179.861 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 160**2+2 * 100**2 - 210**2 } }{ 2 } = 82.31 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 210**2+2 * 100**2 - 160**2 } }{ 2 } = 143.701 ; ;
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