14 20 23 triangle

Acute scalene triangle.

Sides: a = 14   b = 20   c = 23

Area: T = 138.9944379383
Perimeter: p = 57
Semiperimeter: s = 28.5

Angle ∠ A = α = 37.18801691374° = 37°10'49″ = 0.64989163679 rad
Angle ∠ B = β = 59.69113221687° = 59°41'29″ = 1.04218101067 rad
Angle ∠ C = γ = 83.12985086939° = 83°7'43″ = 1.4510866179 rad

Height: ha = 19.85663399118
Height: hb = 13.89994379383
Height: hc = 12.08664677724

Median: ma = 20.38438171106
Median: mb = 16.2021851746
Median: mc = 12.87443931896

Inradius: r = 4.87769957678
Circumradius: R = 11.58332021924

Vertex coordinates: A[23; 0] B[0; 0] C[7.06552173913; 12.08664677724]
Centroid: CG[10.02217391304; 4.02988225908]
Coordinates of the circumscribed circle: U[11.5; 1.38658474052]
Coordinates of the inscribed circle: I[8.5; 4.87769957678]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 142.8219830863° = 142°49'11″ = 0.64989163679 rad
∠ B' = β' = 120.3098677831° = 120°18'31″ = 1.04218101067 rad
∠ C' = γ' = 96.87114913061° = 96°52'17″ = 1.4510866179 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 = 20 ; ; c = 23 ; ;

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

p = a+b+c = 14+20+23 = 57 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 57 }{ 2 } = 28.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 28.5 * (28.5-14)(28.5-20)(28.5-23) } ; ; T = sqrt{ 19319.44 } = 138.99 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 138.99 }{ 14 } = 19.86 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 138.99 }{ 20 } = 13.9 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 138.99 }{ 23 } = 12.09 ; ;

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**2-20**2-23**2 }{ 2 * 20 * 23 } ) = 37° 10'49" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-14**2-23**2 }{ 2 * 14 * 23 } ) = 59° 41'29" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-14**2-20**2 }{ 2 * 20 * 14 } ) = 83° 7'43" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 138.99 }{ 28.5 } = 4.88 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 37° 10'49" } = 11.58 ; ;




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

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