18 28 29 triangle

Acute scalene triangle.

Sides: a = 18   b = 28   c = 29

Area: T = 242.999884259
Perimeter: p = 75
Semiperimeter: s = 37.5

Angle ∠ A = α = 36.76439249588° = 36°45'50″ = 0.64216515365 rad
Angle ∠ B = β = 68.59659541323° = 68°35'45″ = 1.19772252532 rad
Angle ∠ C = γ = 74.64401209089° = 74°38'24″ = 1.30327158639 rad

Height: ha = 276.9998713989
Height: hb = 17.3577060185
Height: hc = 16.75985408683

Median: ma = 27.04662566726
Median: mb = 19.66596032513
Median: mc = 18.54404962177

Inradius: r = 6.48799691357
Circumradius: R = 15.03771086589

Vertex coordinates: A[29; 0] B[0; 0] C[6.56989655172; 16.75985408683]
Centroid: CG[11.85663218391; 5.58661802894]
Coordinates of the circumscribed circle: U[14.5; 3.98330436626]
Coordinates of the inscribed circle: I[9.5; 6.48799691357]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 143.2366075041° = 143°14'10″ = 0.64216515365 rad
∠ B' = β' = 111.4044045868° = 111°24'15″ = 1.19772252532 rad
∠ C' = γ' = 105.3659879091° = 105°21'36″ = 1.30327158639 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 = 28 ; ; c = 29 ; ;

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

p = a+b+c = 18+28+29 = 75 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 75 }{ 2 } = 37.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 37.5 * (37.5-18)(37.5-28)(37.5-29) } ; ; T = sqrt{ 59048.44 } = 243 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 243 }{ 18 } = 27 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 243 }{ 28 } = 17.36 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 243 }{ 29 } = 16.76 ; ;

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**2-28**2-29**2 }{ 2 * 28 * 29 } ) = 36° 45'50" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 28**2-18**2-29**2 }{ 2 * 18 * 29 } ) = 68° 35'45" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-18**2-28**2 }{ 2 * 28 * 18 } ) = 74° 38'24" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 243 }{ 37.5 } = 6.48 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 18 }{ 2 * sin 36° 45'50" } = 15.04 ; ;




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

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