18 26 29 triangle

Acute scalene triangle.

Sides: a = 18   b = 26   c = 29

Area: T = 230.599908391
Perimeter: p = 73
Semiperimeter: s = 36.5

Angle ∠ A = α = 37.71102558131° = 37°42'37″ = 0.65881681257 rad
Angle ∠ B = β = 62.07701092129° = 62°4'12″ = 1.08333277728 rad
Angle ∠ C = γ = 80.2219634974° = 80°13'11″ = 1.4400096755 rad

Height: ha = 25.62221204344
Height: hb = 17.738839107
Height: hc = 15.90333850972

Median: ma = 26.02988301696
Median: mb = 20.33546994077
Median: mc = 17.02220445305

Inradius: r = 6.31877831208
Circumradius: R = 14.71438485655

Vertex coordinates: A[29; 0] B[0; 0] C[8.43110344828; 15.90333850972]
Centroid: CG[12.47770114943; 5.30111283657]
Coordinates of the circumscribed circle: U[14.5; 2.49994678653]
Coordinates of the inscribed circle: I[10.5; 6.31877831208]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 142.2989744187° = 142°17'23″ = 0.65881681257 rad
∠ B' = β' = 117.9329890787° = 117°55'48″ = 1.08333277728 rad
∠ C' = γ' = 99.7880365026° = 99°46'49″ = 1.4400096755 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 = 18 ; ; b = 26 ; ; c = 29 ; ;

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

p = a+b+c = 18+26+29 = 73 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 73 }{ 2 } = 36.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 36.5 * (36.5-18)(36.5-26)(36.5-29) } ; ; T = sqrt{ 53175.94 } = 230.6 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 230.6 }{ 18 } = 25.62 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 230.6 }{ 26 } = 17.74 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 230.6 }{ 29 } = 15.9 ; ;

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{ 18**2-26**2-29**2 }{ 2 * 26 * 29 } ) = 37° 42'37" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 26**2-18**2-29**2 }{ 2 * 18 * 29 } ) = 62° 4'12" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-18**2-26**2 }{ 2 * 26 * 18 } ) = 80° 13'11" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 230.6 }{ 36.5 } = 6.32 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 18 }{ 2 * sin 37° 42'37" } = 14.71 ; ;




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

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