6 15 20 triangle

Obtuse scalene triangle.

Sides: a = 6   b = 15   c = 20

Area: T = 28.59108639254
Perimeter: p = 41
Semiperimeter: s = 20.5

Angle ∠ A = α = 10.98881377331° = 10°59'17″ = 0.19217791821 rad
Angle ∠ B = β = 28.45879982553° = 28°27'29″ = 0.49766857681 rad
Angle ∠ C = γ = 140.5543864012° = 140°33'14″ = 2.45331277034 rad

Height: ha = 9.53302879751
Height: hb = 3.81221151901
Height: hc = 2.85990863925

Median: ma = 17.42112513902
Median: mb = 12.7188097342
Median: mc = 5.52326805086

Inradius: r = 1.3954676289
Circumradius: R = 15.73992935441

Vertex coordinates: A[20; 0] B[0; 0] C[5.275; 2.85990863925]
Centroid: CG[8.425; 0.95330287975]
Coordinates of the circumscribed circle: U[10; -12.15442322368]
Coordinates of the inscribed circle: I[5.5; 1.3954676289]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 169.0121862267° = 169°43″ = 0.19217791821 rad
∠ B' = β' = 151.5422001745° = 151°32'31″ = 0.49766857681 rad
∠ C' = γ' = 39.44661359884° = 39°26'46″ = 2.45331277034 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 = 6 ; ; b = 15 ; ; c = 20 ; ;

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

p = a+b+c = 6+15+20 = 41 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 41 }{ 2 } = 20.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 20.5 * (20.5-6)(20.5-15)(20.5-20) } ; ; T = sqrt{ 817.44 } = 28.59 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 28.59 }{ 6 } = 9.53 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 28.59 }{ 15 } = 3.81 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 28.59 }{ 20 } = 2.86 ; ;

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{ 6**2-15**2-20**2 }{ 2 * 15 * 20 } ) = 10° 59'17" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-6**2-20**2 }{ 2 * 6 * 20 } ) = 28° 27'29" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20**2-6**2-15**2 }{ 2 * 15 * 6 } ) = 140° 33'14" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 28.59 }{ 20.5 } = 1.39 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 10° 59'17" } = 15.74 ; ;




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

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