20 23 29 triangle

Acute scalene triangle.

Sides: a = 20   b = 23   c = 29

Area: T = 228.945540834
Perimeter: p = 72
Semiperimeter: s = 36

Angle ∠ A = α = 43.35331348587° = 43°21'11″ = 0.75766549443 rad
Angle ∠ B = β = 52.13657227104° = 52°8'9″ = 0.91099400192 rad
Angle ∠ C = γ = 84.5111142431° = 84°30'40″ = 1.475499769 rad

Height: ha = 22.8954540834
Height: hb = 19.90882963774
Height: hc = 15.78993385062

Median: ma = 24.18767732449
Median: mb = 22.0966379794
Median: mc = 15.94552187191

Inradius: r = 6.36595946761
Circumradius: R = 14.56767913769

Vertex coordinates: A[29; 0] B[0; 0] C[12.2765862069; 15.78993385062]
Centroid: CG[13.75986206897; 5.26331128354]
Coordinates of the circumscribed circle: U[14.5; 1.39333452621]
Coordinates of the inscribed circle: I[13; 6.36595946761]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 136.6476865141° = 136°38'49″ = 0.75766549443 rad
∠ B' = β' = 127.864427729° = 127°51'51″ = 0.91099400192 rad
∠ C' = γ' = 95.4898857569° = 95°29'20″ = 1.475499769 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 = 20 ; ; b = 23 ; ; c = 29 ; ;

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

p = a+b+c = 20+23+29 = 72 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 72 }{ 2 } = 36 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 36 * (36-20)(36-23)(36-29) } ; ; T = sqrt{ 52416 } = 228.95 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 228.95 }{ 20 } = 22.89 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 228.95 }{ 23 } = 19.91 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 228.95 }{ 29 } = 15.79 ; ;

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{ 20**2-23**2-29**2 }{ 2 * 23 * 29 } ) = 43° 21'11" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-20**2-29**2 }{ 2 * 20 * 29 } ) = 52° 8'9" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-20**2-23**2 }{ 2 * 23 * 20 } ) = 84° 30'40" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 228.95 }{ 36 } = 6.36 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 43° 21'11" } = 14.57 ; ;




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

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