16 28 29 triangle

Acute scalene triangle.

Sides: a = 16   b = 28   c = 29

Area: T = 218.4055442927
Perimeter: p = 73
Semiperimeter: s = 36.5

Angle ∠ A = α = 32.54438173506° = 32°32'38″ = 0.56879967639 rad
Angle ∠ B = β = 70.28884838181° = 70°17'19″ = 1.22767654689 rad
Angle ∠ C = γ = 77.16876988313° = 77°10'4″ = 1.34768304208 rad

Height: ha = 27.30106803658
Height: hb = 15.66003887805
Height: hc = 15.06224443398

Median: ma = 27.35987280406
Median: mb = 18.77549833555
Median: mc = 17.65997159068

Inradius: r = 5.98437107651
Circumradius: R = 14.87114242488

Vertex coordinates: A[29; 0] B[0; 0] C[5.39765517241; 15.06224443398]
Centroid: CG[11.46655172414; 5.02108147799]
Coordinates of the circumscribed circle: U[14.5; 3.30329167695]
Coordinates of the inscribed circle: I[8.5; 5.98437107651]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 147.4566182649° = 147°27'22″ = 0.56879967639 rad
∠ B' = β' = 109.7121516182° = 109°42'41″ = 1.22767654689 rad
∠ C' = γ' = 102.8322301169° = 102°49'56″ = 1.34768304208 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 = 16 ; ; b = 28 ; ; c = 29 ; ;

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

p = a+b+c = 16+28+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 * (36.5-16)(36.5-28)(36.5-29) } ; ; T = sqrt{ 47700.94 } = 218.41 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 218.41 }{ 16 } = 27.3 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 218.41 }{ 28 } = 15.6 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 218.41 }{ 29 } = 15.06 ; ;

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{ 16**2-28**2-29**2 }{ 2 * 28 * 29 } ) = 32° 32'38" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 28**2-16**2-29**2 }{ 2 * 16 * 29 } ) = 70° 17'19" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-16**2-28**2 }{ 2 * 28 * 16 } ) = 77° 10'4" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 218.41 }{ 36.5 } = 5.98 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 32° 32'38" } = 14.87 ; ;




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

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