20 28 29 triangle

Acute scalene triangle.

Sides: a = 20   b = 28   c = 29

Area: T = 266.5466313987
Perimeter: p = 77
Semiperimeter: s = 38.5

Angle ∠ A = α = 41.03548543388° = 41°2'5″ = 0.71661933163 rad
Angle ∠ B = β = 66.7998528371° = 66°47'55″ = 1.16658542556 rad
Angle ∠ C = γ = 72.16766172902° = 72°10' = 1.26595450817 rad

Height: ha = 26.65546313987
Height: hb = 19.03990224276
Height: hc = 18.38325044129

Median: ma = 26.69326956301
Median: mb = 20.6033397778
Median: mc = 19.53884236826

Inradius: r = 6.92332808828
Circumradius: R = 15.23218744884

Vertex coordinates: A[29; 0] B[0; 0] C[7.87993103448; 18.38325044129]
Centroid: CG[12.29331034483; 6.1287501471]
Coordinates of the circumscribed circle: U[14.5; 4.66547615621]
Coordinates of the inscribed circle: I[10.5; 6.92332808828]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 138.9655145661° = 138°57'55″ = 0.71661933163 rad
∠ B' = β' = 113.2011471629° = 113°12'5″ = 1.16658542556 rad
∠ C' = γ' = 107.833338271° = 107°50' = 1.26595450817 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 = 28 ; ; c = 29 ; ;

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

p = a+b+c = 20+28+29 = 77 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 77 }{ 2 } = 38.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 38.5 * (38.5-20)(38.5-28)(38.5-29) } ; ; T = sqrt{ 71046.94 } = 266.55 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 266.55 }{ 20 } = 26.65 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 266.55 }{ 28 } = 19.04 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 266.55 }{ 29 } = 18.38 ; ;

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**2-28**2-29**2 }{ 2 * 28 * 29 } ) = 41° 2'5" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 28**2-20**2-29**2 }{ 2 * 20 * 29 } ) = 66° 47'55" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-20**2-28**2 }{ 2 * 28 * 20 } ) = 72° 10' ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 266.55 }{ 38.5 } = 6.92 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 41° 2'5" } = 15.23 ; ;




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

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