Triangle calculator SSA

Right scalene triangle.

Sides: a = 210 b = 560 c = 519.1343894097

Area: T = 54509.05988802
Perimeter: p = 1289.13438941
Semiperimeter: s = 644.5676947048

Angle ∠ A = α = 22.0244312837° = 22°1'28″ = 0.38443967745 rad
Angle ∠ B = β = 90° = 1.57107963268 rad
Angle ∠ C = γ = 67.9765687163° = 67°58'32″ = 1.18663995523 rad

Height: h_a = 519.1343894097
Height: h_b = 194.6755210286
Height: h_c = 210

Median: m_a = 529.6466108265
Median: m_b = 280
Median: m_c = 333.8798720496

Inradius: r = 84.56769470483
Circumradius: R = 280

Vertex coordinates: A[519.1343894097; 0] B[0; 0] C[-0; 210]
Centroid: CG[173.0454631366; 70]
Coordinates of the circumscribed circle: U[259.5676947048; 105]
Coordinates of the inscribed circle: I[84.56769470483; 84.56769470483]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 157.9765687163° = 157°58'32″ = 0.38443967745 rad
∠ B' = β' = 90° = 1.57107963268 rad
∠ C' = γ' = 112.0244312837° = 112°1'28″ = 1.18663995523 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

$a = 210 ; ; b = 560 ; ; beta = 90° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 560**2 = 210**2 + c**2 -2 * 560 * c * cos (90° ) ; ; ; ; c**2 -269500 =0 ; ; p=1; q=-2.57175827821E-14; r=-269500 ; ; D = q**2 - 4pr = 0**2 - 4 * 1 * (-269500) = 1078000 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ ± sqrt{ 1078000 } }{ 2 } = fraction{ ± 140 sqrt{ 55 } }{ 2 } ; ; c_{1,2} = 1.2858791391E-14 ± 519.133894097 ; ;$
$c_{1} = 70 sqrt{ 55} = 519.133894097 ; ; c_{2} = - 70 sqrt{ 55} = -519.133894097 ; ; ; ; (c -519.133894097) (c +519.133894097) = 0 ; ; ; ; c>0 ; ;$

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.

$a = 210 ; ; b = 560 ; ; c = 519.13 ; ;$

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

$p = a+b+c = 210+560+519.13 = 1289.13 ; ;$

3. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 1289.13 }{ 2 } = 644.57 ; ;$

4. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 644.57 * (644.57-210)(644.57-560)(644.57-519.13) } ; ; T = sqrt{ 2971237500 } = 54509.06 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 54509.06 }{ 210 } = 519.13 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 54509.06 }{ 560 } = 194.68 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 54509.06 }{ 519.13 } = 210 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 210**2-560**2-519.13**2 }{ 2 * 560 * 519.13 } ) = 22° 1'28" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 560**2-210**2-519.13**2 }{ 2 * 210 * 519.13 } ) = 90° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 519.13**2-210**2-560**2 }{ 2 * 560 * 210 } ) = 67° 58'32" ; ;$

7. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 54509.06 }{ 644.57 } = 84.57 ; ;$

8. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 210 }{ 2 * sin 22° 1'28" } = 280 ; ;$

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