14 22 23 triangle

Acute scalene triangle.

Sides: a = 14   b = 22   c = 23

Area: T = 149.3011498653
Perimeter: p = 59
Semiperimeter: s = 29.5

Angle ∠ A = α = 36.16658436388° = 36°9'57″ = 0.63112130483 rad
Angle ∠ B = β = 68.02436651379° = 68°1'25″ = 1.18772369259 rad
Angle ∠ C = γ = 75.81104912233° = 75°48'38″ = 1.32331426794 rad

Height: ha = 21.32987855218
Height: hb = 13.57328635139
Height: hc = 12.98327390133

Median: ma = 21.38992496362
Median: mb = 15.54402702679
Median: mc = 14.41435353054

Inradius: r = 5.06110677509
Circumradius: R = 11.86219037048

Vertex coordinates: A[23; 0] B[0; 0] C[5.23991304348; 12.98327390133]
Centroid: CG[9.41330434783; 4.32875796711]
Coordinates of the circumscribed circle: U[11.5; 2.90877069146]
Coordinates of the inscribed circle: I[7.5; 5.06110677509]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 143.8344156361° = 143°50'3″ = 0.63112130483 rad
∠ B' = β' = 111.9766334862° = 111°58'35″ = 1.18772369259 rad
∠ C' = γ' = 104.1989508777° = 104°11'22″ = 1.32331426794 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 = 14 ; ; b = 22 ; ; c = 23 ; ;

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

p = a+b+c = 14+22+23 = 59 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 59 }{ 2 } = 29.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29.5 * (29.5-14)(29.5-22)(29.5-23) } ; ; T = sqrt{ 22290.94 } = 149.3 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 149.3 }{ 14 } = 21.33 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 149.3 }{ 22 } = 13.57 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 149.3 }{ 23 } = 12.98 ; ;

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{ 14**2-22**2-23**2 }{ 2 * 22 * 23 } ) = 36° 9'57" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-14**2-23**2 }{ 2 * 14 * 23 } ) = 68° 1'25" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-14**2-22**2 }{ 2 * 22 * 14 } ) = 75° 48'38" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 149.3 }{ 29.5 } = 5.06 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 36° 9'57" } = 11.86 ; ;




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

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