16 19 23 triangle

Acute scalene triangle.

Sides: a = 16   b = 19   c = 23

Area: T = 150.3999468084
Perimeter: p = 58
Semiperimeter: s = 29

Angle ∠ A = α = 43.49878286118° = 43°29'52″ = 0.75991803267 rad
Angle ∠ B = β = 54.82442119874° = 54°49'27″ = 0.9576863009 rad
Angle ∠ C = γ = 81.67879594008° = 81°40'41″ = 1.42655493179 rad

Height: ha = 18.87999335105
Height: hb = 15.83215229562
Height: hc = 13.0788214616

Median: ma = 19.51992212959
Median: mb = 17.38553386507
Median: mc = 13.27659180474

Inradius: r = 5.18661885546
Circumradius: R = 11.62223815301

Vertex coordinates: A[23; 0] B[0; 0] C[9.21773913043; 13.0788214616]
Centroid: CG[10.73991304348; 4.3599404872]
Coordinates of the circumscribed circle: U[11.5; 1.68221868004]
Coordinates of the inscribed circle: I[10; 5.18661885546]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 136.5022171388° = 136°30'8″ = 0.75991803267 rad
∠ B' = β' = 125.1765788013° = 125°10'33″ = 0.9576863009 rad
∠ C' = γ' = 98.32220405992° = 98°19'19″ = 1.42655493179 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 = 16 ; ; b = 19 ; ; c = 23 ; ;

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

p = a+b+c = 16+19+23 = 58 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 58 }{ 2 } = 29 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29 * (29-16)(29-19)(29-23) } ; ; T = sqrt{ 22620 } = 150.4 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 150.4 }{ 16 } = 18.8 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 150.4 }{ 19 } = 15.83 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 150.4 }{ 23 } = 13.08 ; ;

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{ 16**2-19**2-23**2 }{ 2 * 19 * 23 } ) = 43° 29'52" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 19**2-16**2-23**2 }{ 2 * 16 * 23 } ) = 54° 49'27" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-16**2-19**2 }{ 2 * 19 * 16 } ) = 81° 40'41" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 150.4 }{ 29 } = 5.19 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 43° 29'52" } = 11.62 ; ;




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

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