420 293 512 triangle

Acute scalene triangle.

Sides: a = 420   b = 293   c = 512

Area: T = 61529.99444006
Perimeter: p = 1225
Semiperimeter: s = 612.5

Angle ∠ A = α = 55.11660809659° = 55°6'58″ = 0.96219570837 rad
Angle ∠ B = β = 34.90883626268° = 34°54'30″ = 0.60992658643 rad
Angle ∠ C = γ = 89.97655564073° = 89°58'32″ = 1.57703697056 rad

Height: ha = 2932.999973336
Height: hb = 4209.999961779
Height: hc = 240.3521540627

Median: ma = 360.4121570292
Median: mb = 444.7588080309
Median: mc = 256.1032518535

Inradius: r = 100.4577133715
Circumradius: R = 2566.000023297

Vertex coordinates: A[512; 0] B[0; 0] C[344.4298710938; 240.3521540627]
Centroid: CG[285.4766236979; 80.11771802091]
Coordinates of the circumscribed circle: U[256; 0.1099215027]
Coordinates of the inscribed circle: I[319.5; 100.4577133715]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 124.8843919034° = 124°53'2″ = 0.96219570837 rad
∠ B' = β' = 145.0921637373° = 145°5'30″ = 0.60992658643 rad
∠ C' = γ' = 90.02444435927° = 90°1'28″ = 1.57703697056 rad

Calculate another triangle




How did we calculate this triangle?

a = 420 ; ; b = 293 ; ; c = 512 ; ;

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

p = a+b+c = 420+293+512 = 1225 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1225 }{ 2 } = 612.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 612.5 * (612.5-420)(612.5-293)(612.5-512) } ; ; T = sqrt{ 3785940210.94 } = 61529.99 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 61529.99 }{ 420 } = 293 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 61529.99 }{ 293 } = 420 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 61529.99 }{ 512 } = 240.35 ; ;

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{ 420**2-293**2-512**2 }{ 2 * 293 * 512 } ) = 55° 6'58" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 293**2-420**2-512**2 }{ 2 * 420 * 512 } ) = 34° 54'30" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 512**2-420**2-293**2 }{ 2 * 293 * 420 } ) = 89° 58'32" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 61529.99 }{ 612.5 } = 100.46 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 420 }{ 2 * sin 55° 6'58" } = 256 ; ;




Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.