11 15 23 triangle

Obtuse scalene triangle.

Sides: a = 11   b = 15   c = 23

Area: T = 68.65326583608
Perimeter: p = 49
Semiperimeter: s = 24.5

Angle ∠ A = α = 23.45223604058° = 23°27'8″ = 0.40993209064 rad
Angle ∠ B = β = 32.8688227068° = 32°52'6″ = 0.57436587816 rad
Angle ∠ C = γ = 123.6799412526° = 123°40'46″ = 2.15986129655 rad

Height: ha = 12.48223015201
Height: hb = 9.15436877814
Height: hc = 5.97697963792

Median: ma = 18.62112244495
Median: mb = 16.39435963108
Median: mc = 6.38435726674

Inradius: r = 2.80221493208
Circumradius: R = 13.8219566826

Vertex coordinates: A[23; 0] B[0; 0] C[9.23991304348; 5.97697963792]
Centroid: CG[10.74663768116; 1.99899321264]
Coordinates of the circumscribed circle: U[11.5; -7.66435779672]
Coordinates of the inscribed circle: I[9.5; 2.80221493208]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 156.5487639594° = 156°32'52″ = 0.40993209064 rad
∠ B' = β' = 147.1321772932° = 147°7'54″ = 0.57436587816 rad
∠ C' = γ' = 56.32105874737° = 56°19'14″ = 2.15986129655 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 = 11 ; ; b = 15 ; ; c = 23 ; ;

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

p = a+b+c = 11+15+23 = 49 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 49 }{ 2 } = 24.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 24.5 * (24.5-11)(24.5-15)(24.5-23) } ; ; T = sqrt{ 4713.19 } = 68.65 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 68.65 }{ 11 } = 12.48 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 68.65 }{ 15 } = 9.15 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 68.65 }{ 23 } = 5.97 ; ;

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{ 11**2-15**2-23**2 }{ 2 * 15 * 23 } ) = 23° 27'8" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-11**2-23**2 }{ 2 * 11 * 23 } ) = 32° 52'6" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-11**2-15**2 }{ 2 * 15 * 11 } ) = 123° 40'46" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 68.65 }{ 24.5 } = 2.8 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 23° 27'8" } = 13.82 ; ;




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

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