15 23 27 triangle

Acute scalene triangle.

Sides: a = 15   b = 23   c = 27

Area: T = 172.3876738179
Perimeter: p = 65
Semiperimeter: s = 32.5

Angle ∠ A = α = 33.72438564576° = 33°43'26″ = 0.58985923317 rad
Angle ∠ B = β = 58.35325320481° = 58°21'9″ = 1.01884438111 rad
Angle ∠ C = γ = 87.92436114943° = 87°55'25″ = 1.53545565108 rad

Height: ha = 22.98548984239
Height: hb = 14.9990151146
Height: hc = 12.76993880133

Median: ma = 23.93221958875
Median: mb = 18.56774446276
Median: mc = 13.9555285737

Inradius: r = 5.30442073286
Circumradius: R = 13.50988697924

Vertex coordinates: A[27; 0] B[0; 0] C[7.87703703704; 12.76993880133]
Centroid: CG[11.62334567901; 4.25664626711]
Coordinates of the circumscribed circle: U[13.5; 0.48994518041]
Coordinates of the inscribed circle: I[9.5; 5.30442073286]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 146.2766143542° = 146°16'34″ = 0.58985923317 rad
∠ B' = β' = 121.6477467952° = 121°38'51″ = 1.01884438111 rad
∠ C' = γ' = 92.07663885057° = 92°4'35″ = 1.53545565108 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 = 15 ; ; b = 23 ; ; c = 27 ; ;

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

p = a+b+c = 15+23+27 = 65 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 65 }{ 2 } = 32.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 32.5 * (32.5-15)(32.5-23)(32.5-27) } ; ; T = sqrt{ 29717.19 } = 172.39 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 172.39 }{ 15 } = 22.98 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 172.39 }{ 23 } = 14.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 172.39 }{ 27 } = 12.77 ; ;

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{ 15**2-23**2-27**2 }{ 2 * 23 * 27 } ) = 33° 43'26" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-15**2-27**2 }{ 2 * 15 * 27 } ) = 58° 21'9" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-15**2-23**2 }{ 2 * 23 * 15 } ) = 87° 55'25" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 172.39 }{ 32.5 } = 5.3 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 33° 43'26" } = 13.51 ; ;




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

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