19 20 22 triangle

Acute scalene triangle.

Sides: a = 19   b = 20   c = 22

Area: T = 176.9310600802
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 53.53659579233° = 53°32'9″ = 0.9344378734 rad
Angle ∠ B = β = 57.83992380024° = 57°50'21″ = 1.009948514 rad
Angle ∠ C = γ = 68.62548040744° = 68°37'29″ = 1.19877287796 rad

Height: ha = 18.62442737686
Height: hb = 17.69330600802
Height: hc = 16.08546000729

Median: ma = 18.75549993335
Median: mb = 17.95882849961
Median: mc = 16.10990036936

Inradius: r = 5.8011003305
Circumradius: R = 11.8132541135

Vertex coordinates: A[22; 0] B[0; 0] C[10.11436363636; 16.08546000729]
Centroid: CG[10.70545454545; 5.36215333576]
Coordinates of the circumscribed circle: U[11; 4.30553603873]
Coordinates of the inscribed circle: I[10.5; 5.8011003305]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 126.4644042077° = 126°27'51″ = 0.9344378734 rad
∠ B' = β' = 122.1610761998° = 122°9'39″ = 1.009948514 rad
∠ C' = γ' = 111.3755195926° = 111°22'31″ = 1.19877287796 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 = 22 ; ;

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

p = a+b+c = 19+20+22 = 61 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 61 }{ 2 } = 30.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30.5 * (30.5-19)(30.5-20)(30.5-22) } ; ; T = sqrt{ 31304.44 } = 176.93 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 176.93 }{ 19 } = 18.62 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 176.93 }{ 20 } = 17.69 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 176.93 }{ 22 } = 16.08 ; ;

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**2-22**2 }{ 2 * 20 * 22 } ) = 53° 32'9" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-19**2-22**2 }{ 2 * 19 * 22 } ) = 57° 50'21" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-19**2-20**2 }{ 2 * 20 * 19 } ) = 68° 37'29" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 176.93 }{ 30.5 } = 5.8 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 19 }{ 2 * sin 53° 32'9" } = 11.81 ; ;




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

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