9 15 23 triangle

Obtuse scalene triangle.

Sides: a = 9   b = 15   c = 23

Area: T = 38.05550587964
Perimeter: p = 47
Semiperimeter: s = 23.5

Angle ∠ A = α = 12.74548071461° = 12°44'41″ = 0.22224388472 rad
Angle ∠ B = β = 21.57327142774° = 21°34'22″ = 0.37765148927 rad
Angle ∠ C = γ = 145.6822478577° = 145°40'57″ = 2.54326389136 rad

Height: ha = 8.45766797325
Height: hb = 5.07440078395
Height: hc = 3.30991355475

Median: ma = 18.88878267675
Median: mb = 15.77218102956
Median: mc = 4.55552167896

Inradius: r = 1.61993642041
Circumradius: R = 20.39880764858

Vertex coordinates: A[23; 0] B[0; 0] C[8.37695652174; 3.30991355475]
Centroid: CG[10.45765217391; 1.10330451825]
Coordinates of the circumscribed circle: U[11.5; -16.84773002086]
Coordinates of the inscribed circle: I[8.5; 1.61993642041]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 167.2555192854° = 167°15'19″ = 0.22224388472 rad
∠ B' = β' = 158.4277285723° = 158°25'38″ = 0.37765148927 rad
∠ C' = γ' = 34.31875214235° = 34°19'3″ = 2.54326389136 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 = 9 ; ; b = 15 ; ; c = 23 ; ;

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

p = a+b+c = 9+15+23 = 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 * (23.5-9)(23.5-15)(23.5-23) } ; ; T = sqrt{ 1448.19 } = 38.06 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 38.06 }{ 9 } = 8.46 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 38.06 }{ 15 } = 5.07 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 38.06 }{ 23 } = 3.31 ; ;

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{ 9**2-15**2-23**2 }{ 2 * 15 * 23 } ) = 12° 44'41" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-9**2-23**2 }{ 2 * 9 * 23 } ) = 21° 34'22" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-9**2-15**2 }{ 2 * 15 * 9 } ) = 145° 40'57" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 38.06 }{ 23.5 } = 1.62 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 12° 44'41" } = 20.4 ; ;




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

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