11 27 28 triangle

Acute scalene triangle.

Sides: a = 11   b = 27   c = 28

Area: T = 147.5880486515
Perimeter: p = 66
Semiperimeter: s = 33

Angle ∠ A = α = 22.98109198075° = 22°58'51″ = 0.40110927158 rad
Angle ∠ B = β = 73.3988450401° = 73°23'54″ = 1.28110446254 rad
Angle ∠ C = γ = 83.62106297916° = 83°37'14″ = 1.45994553125 rad

Height: ha = 26.833281573
Height: hb = 10.932188789
Height: hc = 10.54114633225

Median: ma = 26.94990259564
Median: mb = 16.43992822228
Median: mc = 15.13327459504

Inradius: r = 4.4722135955
Circumradius: R = 14.08772282582

Vertex coordinates: A[28; 0] B[0; 0] C[3.14328571429; 10.54114633225]
Centroid: CG[10.3810952381; 3.51438211075]
Coordinates of the circumscribed circle: U[14; 1.56552475842]
Coordinates of the inscribed circle: I[6; 4.4722135955]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 157.0199080193° = 157°1'9″ = 0.40110927158 rad
∠ B' = β' = 106.6021549599° = 106°36'6″ = 1.28110446254 rad
∠ C' = γ' = 96.37993702084° = 96°22'46″ = 1.45994553125 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 = 11 ; ; b = 27 ; ; c = 28 ; ;

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

p = a+b+c = 11+27+28 = 66 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 66 }{ 2 } = 33 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33 * (33-11)(33-27)(33-28) } ; ; T = sqrt{ 21780 } = 147.58 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 147.58 }{ 11 } = 26.83 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 147.58 }{ 27 } = 10.93 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 147.58 }{ 28 } = 10.54 ; ;

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{ 11**2-27**2-28**2 }{ 2 * 27 * 28 } ) = 22° 58'51" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-11**2-28**2 }{ 2 * 11 * 28 } ) = 73° 23'54" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-11**2-27**2 }{ 2 * 27 * 11 } ) = 83° 37'14" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 147.58 }{ 33 } = 4.47 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 22° 58'51" } = 14.09 ; ;




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

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