16 26 30 triangle

Acute scalene triangle.

Sides: a = 16   b = 26   c = 30

Area: T = 207.8466096908
Perimeter: p = 72
Semiperimeter: s = 36

Angle ∠ A = α = 32.2044227504° = 32°12'15″ = 0.5622069803 rad
Angle ∠ B = β = 60° = 1.04771975512 rad
Angle ∠ C = γ = 87.7965772496° = 87°47'45″ = 1.53223252994 rad

Height: ha = 25.98107621135
Height: hb = 15.98881613006
Height: hc = 13.85664064606

Median: ma = 26.90772480941
Median: mb = 20.22437484162
Median: mc = 15.52441746963

Inradius: r = 5.77435026919
Circumradius: R = 15.01111069989

Vertex coordinates: A[30; 0] B[0; 0] C[8; 13.85664064606]
Centroid: CG[12.66766666667; 4.61988021535]
Coordinates of the circumscribed circle: U[15; 0.57773502692]
Coordinates of the inscribed circle: I[10; 5.77435026919]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 147.7965772496° = 147°47'45″ = 0.5622069803 rad
∠ B' = β' = 120° = 1.04771975512 rad
∠ C' = γ' = 92.2044227504° = 92°12'15″ = 1.53223252994 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 = 26 ; ; c = 30 ; ;

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

p = a+b+c = 16+26+30 = 72 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 72 }{ 2 } = 36 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 36 * (36-16)(36-26)(36-30) } ; ; T = sqrt{ 43200 } = 207.85 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 207.85 }{ 16 } = 25.98 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 207.85 }{ 26 } = 15.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 207.85 }{ 30 } = 13.86 ; ;

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**2-26**2-30**2 }{ 2 * 26 * 30 } ) = 32° 12'15" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 26**2-16**2-30**2 }{ 2 * 16 * 30 } ) = 60° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-16**2-26**2 }{ 2 * 26 * 16 } ) = 87° 47'45" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 207.85 }{ 36 } = 5.77 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 32° 12'15" } = 15.01 ; ;




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

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