2 27 28 triangle

Obtuse scalene triangle.

Sides: a = 2   b = 27   c = 28

Area: T = 23.87999474789
Perimeter: p = 57
Semiperimeter: s = 28.5

Angle ∠ A = α = 3.61098918931° = 3°36'36″ = 0.06330044992 rad
Angle ∠ B = β = 58.21114653664° = 58°12'41″ = 1.01659817331 rad
Angle ∠ C = γ = 118.1798642741° = 118°10'43″ = 2.06326064214 rad

Height: ha = 23.87999474789
Height: hb = 1.76329590725
Height: hc = 1.76999962485

Median: ma = 27.48663602538
Median: mb = 14.55216322109
Median: mc = 13.05875648572

Inradius: r = 0.83550858765
Circumradius: R = 15.88223879899

Vertex coordinates: A[28; 0] B[0; 0] C[1.05435714286; 1.76999962485]
Centroid: CG[9.68545238095; 0.56766654162]
Coordinates of the circumscribed circle: U[14; -7.55000165508]
Coordinates of the inscribed circle: I[1.5; 0.83550858765]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 176.3990108107° = 176°23'24″ = 0.06330044992 rad
∠ B' = β' = 121.7898534634° = 121°47'19″ = 1.01659817331 rad
∠ C' = γ' = 61.82113572595° = 61°49'17″ = 2.06326064214 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 = 2 ; ; b = 27 ; ; c = 28 ; ;

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

p = a+b+c = 2+27+28 = 57 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 57 }{ 2 } = 28.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 28.5 * (28.5-2)(28.5-27)(28.5-28) } ; ; T = sqrt{ 566.44 } = 23.8 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 23.8 }{ 2 } = 23.8 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 23.8 }{ 27 } = 1.76 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 23.8 }{ 28 } = 1.7 ; ;

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{ 2**2-27**2-28**2 }{ 2 * 27 * 28 } ) = 3° 36'36" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-2**2-28**2 }{ 2 * 2 * 28 } ) = 58° 12'41" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-2**2-27**2 }{ 2 * 27 * 2 } ) = 118° 10'43" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 23.8 }{ 28.5 } = 0.84 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2 }{ 2 * sin 3° 36'36" } = 15.88 ; ;




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

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