20 29 30 triangle

Acute scalene triangle.

Sides: a = 20   b = 29   c = 30

Area: T = 277.1876647406
Perimeter: p = 79
Semiperimeter: s = 39.5

Angle ∠ A = α = 39.58441402624° = 39°35'3″ = 0.69108735792 rad
Angle ∠ B = β = 67.51113753665° = 67°30'41″ = 1.17882957827 rad
Angle ∠ C = γ = 72.90444843711° = 72°54'16″ = 1.27224232917 rad

Height: ha = 27.71986647406
Height: hb = 19.11663205107
Height: hc = 18.4799109827

Median: ma = 27.75878817636
Median: mb = 20.97702169755
Median: mc = 19.88771818014

Inradius: r = 7.01773834786
Circumradius: R = 15.69333966362

Vertex coordinates: A[30; 0] B[0; 0] C[7.65; 18.4799109827]
Centroid: CG[12.55; 6.16597032757]
Coordinates of the circumscribed circle: U[15; 4.61333174594]
Coordinates of the inscribed circle: I[10.5; 7.01773834786]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 140.4165859738° = 140°24'57″ = 0.69108735792 rad
∠ B' = β' = 112.4898624634° = 112°29'19″ = 1.17882957827 rad
∠ C' = γ' = 107.0965515629° = 107°5'44″ = 1.27224232917 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 = 29 ; ; c = 30 ; ;

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

p = a+b+c = 20+29+30 = 79 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 79 }{ 2 } = 39.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 39.5 * (39.5-20)(39.5-29)(39.5-30) } ; ; T = sqrt{ 76832.44 } = 277.19 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 277.19 }{ 20 } = 27.72 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 277.19 }{ 29 } = 19.12 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 277.19 }{ 30 } = 18.48 ; ;

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**2-29**2-30**2 }{ 2 * 29 * 30 } ) = 39° 35'3" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-20**2-30**2 }{ 2 * 20 * 30 } ) = 67° 30'41" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-20**2-29**2 }{ 2 * 29 * 20 } ) = 72° 54'16" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 277.19 }{ 39.5 } = 7.02 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 39° 35'3" } = 15.69 ; ;




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

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