13 15 19 triangle

Acute scalene triangle.

Sides: a = 13   b = 15   c = 19

Area: T = 97.15503345336
Perimeter: p = 47
Semiperimeter: s = 23.5

Angle ∠ A = α = 42.98110733896° = 42°58'52″ = 0.75501612467 rad
Angle ∠ B = β = 51.87328349633° = 51°52'22″ = 0.90553517625 rad
Angle ∠ C = γ = 85.14660916471° = 85°8'46″ = 1.48660796444 rad

Height: ha = 14.94662053129
Height: hb = 12.95333779378
Height: hc = 10.22663510035

Median: ma = 15.83550876221
Median: mb = 14.44881832768
Median: mc = 10.33219891599

Inradius: r = 4.13440567887
Circumradius: R = 9.53441925938

Vertex coordinates: A[19; 0] B[0; 0] C[8.02663157895; 10.22663510035]
Centroid: CG[9.00987719298; 3.40987836678]
Coordinates of the circumscribed circle: U[9.5; 0.80767393733]
Coordinates of the inscribed circle: I[8.5; 4.13440567887]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 137.019892661° = 137°1'8″ = 0.75501612467 rad
∠ B' = β' = 128.1277165037° = 128°7'38″ = 0.90553517625 rad
∠ C' = γ' = 94.85439083529° = 94°51'14″ = 1.48660796444 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 = 13 ; ; b = 15 ; ; c = 19 ; ;

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

p = a+b+c = 13+15+19 = 47 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 47 }{ 2 } = 23.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 23.5 * (23.5-13)(23.5-15)(23.5-19) } ; ; T = sqrt{ 9438.19 } = 97.15 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 97.15 }{ 13 } = 14.95 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 97.15 }{ 15 } = 12.95 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 97.15 }{ 19 } = 10.23 ; ;

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{ 13**2-15**2-19**2 }{ 2 * 15 * 19 } ) = 42° 58'52" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-13**2-19**2 }{ 2 * 13 * 19 } ) = 51° 52'22" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 19**2-13**2-15**2 }{ 2 * 15 * 13 } ) = 85° 8'46" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 97.15 }{ 23.5 } = 4.13 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 13 }{ 2 * sin 42° 58'52" } = 9.53 ; ;




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

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