14 16 28 triangle

Obtuse scalene triangle.

Sides: a = 14   b = 16   c = 28

Area: T = 75.21997340421
Perimeter: p = 58
Semiperimeter: s = 29

Angle ∠ A = α = 19.61659069161° = 19°36'57″ = 0.34223621615 rad
Angle ∠ B = β = 22.56113280909° = 22°33'41″ = 0.39437694588 rad
Angle ∠ C = γ = 137.8232764993° = 137°49'22″ = 2.40554610333 rad

Height: ha = 10.74328191489
Height: hb = 9.43999667553
Height: hc = 5.37114095744

Median: ma = 21.70325344142
Median: mb = 20.64397674406
Median: mc = 5.47772255751

Inradius: r = 2.59330942773
Circumradius: R = 20.85111375735

Vertex coordinates: A[28; 0] B[0; 0] C[12.92985714286; 5.37114095744]
Centroid: CG[13.64328571429; 1.79904698581]
Coordinates of the circumscribed circle: U[14; -15.45221823089]
Coordinates of the inscribed circle: I[13; 2.59330942773]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 160.3844093084° = 160°23'3″ = 0.34223621615 rad
∠ B' = β' = 157.4398671909° = 157°26'19″ = 0.39437694588 rad
∠ C' = γ' = 42.17772350071° = 42°10'38″ = 2.40554610333 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 = 14 ; ; b = 16 ; ; c = 28 ; ;

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

p = a+b+c = 14+16+28 = 58 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 58 }{ 2 } = 29 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29 * (29-14)(29-16)(29-28) } ; ; T = sqrt{ 5655 } = 75.2 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 75.2 }{ 14 } = 10.74 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 75.2 }{ 16 } = 9.4 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 75.2 }{ 28 } = 5.37 ; ;

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{ 14**2-16**2-28**2 }{ 2 * 16 * 28 } ) = 19° 36'57" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 16**2-14**2-28**2 }{ 2 * 14 * 28 } ) = 22° 33'41" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-14**2-16**2 }{ 2 * 16 * 14 } ) = 137° 49'22" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 75.2 }{ 29 } = 2.59 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 19° 36'57" } = 20.85 ; ;




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

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