19 20 28 triangle

Obtuse scalene triangle.

Sides: a = 19   b = 20   c = 28

Area: T = 189.9132973491
Perimeter: p = 67
Semiperimeter: s = 33.5

Angle ∠ A = α = 42.70878707931° = 42°42'28″ = 0.74553929619 rad
Angle ∠ B = β = 45.55879131999° = 45°33'28″ = 0.79551355857 rad
Angle ∠ C = γ = 91.7344216007° = 91°44'3″ = 1.60110641061 rad

Height: ha = 19.99108393148
Height: hb = 18.9911297349
Height: hc = 13.56552123922

Median: ma = 22.43997767846
Median: mb = 21.73770651193
Median: mc = 13.58330777072

Inradius: r = 5.66990439848
Circumradius: R = 14.00664154181

Vertex coordinates: A[28; 0] B[0; 0] C[13.30435714286; 13.56552123922]
Centroid: CG[13.76878571429; 4.52217374641]
Coordinates of the circumscribed circle: U[14; -0.42438783613]
Coordinates of the inscribed circle: I[13.5; 5.66990439848]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 137.2922129207° = 137°17'32″ = 0.74553929619 rad
∠ B' = β' = 134.44220868° = 134°26'32″ = 0.79551355857 rad
∠ C' = γ' = 88.2665783993° = 88°15'57″ = 1.60110641061 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 = 19 ; ; b = 20 ; ; c = 28 ; ;

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

p = a+b+c = 19+20+28 = 67 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 67 }{ 2 } = 33.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33.5 * (33.5-19)(33.5-20)(33.5-28) } ; ; T = sqrt{ 36066.94 } = 189.91 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 189.91 }{ 19 } = 19.99 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 189.91 }{ 20 } = 18.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 189.91 }{ 28 } = 13.57 ; ;

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{ 19**2-20**2-28**2 }{ 2 * 20 * 28 } ) = 42° 42'28" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-19**2-28**2 }{ 2 * 19 * 28 } ) = 45° 33'28" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-19**2-20**2 }{ 2 * 20 * 19 } ) = 91° 44'3" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 189.91 }{ 33.5 } = 5.67 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 19 }{ 2 * sin 42° 42'28" } = 14.01 ; ;




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

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