16 18 28 triangle

Obtuse scalene triangle.

Sides: a = 16   b = 18   c = 28

Area: T = 134.6666254125
Perimeter: p = 62
Semiperimeter: s = 31

Angle ∠ A = α = 32.30325452092° = 32°18'9″ = 0.56437857707 rad
Angle ∠ B = β = 36.95550748363° = 36°57'18″ = 0.64549877312 rad
Angle ∠ C = γ = 110.7422379954° = 110°44'33″ = 1.93328191517 rad

Height: ha = 16.83332817656
Height: hb = 14.9632917125
Height: hc = 9.61990181518

Median: ma = 22.13659436212
Median: mb = 20.95223268398
Median: mc = 9.69553597148

Inradius: r = 4.34440727137
Circumradius: R = 14.97703428903

Vertex coordinates: A[28; 0] B[0; 0] C[12.78657142857; 9.61990181518]
Centroid: CG[13.59552380952; 3.20663393839]
Coordinates of the circumscribed circle: U[14; -5.30219964403]
Coordinates of the inscribed circle: I[13; 4.34440727137]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 147.6977454791° = 147°41'51″ = 0.56437857707 rad
∠ B' = β' = 143.0454925164° = 143°2'42″ = 0.64549877312 rad
∠ C' = γ' = 69.25876200455° = 69°15'27″ = 1.93328191517 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 = 18 ; ; c = 28 ; ;

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

p = a+b+c = 16+18+28 = 62 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 62 }{ 2 } = 31 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31 * (31-16)(31-18)(31-28) } ; ; T = sqrt{ 18135 } = 134.67 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 134.67 }{ 16 } = 16.83 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 134.67 }{ 18 } = 14.96 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 134.67 }{ 28 } = 9.62 ; ;

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**2-18**2-28**2 }{ 2 * 18 * 28 } ) = 32° 18'9" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 18**2-16**2-28**2 }{ 2 * 16 * 28 } ) = 36° 57'18" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-16**2-18**2 }{ 2 * 18 * 16 } ) = 110° 44'33" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 134.67 }{ 31 } = 4.34 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 32° 18'9" } = 14.97 ; ;




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

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