20 27 30 triangle

Acute scalene triangle.

Sides: a = 20   b = 27   c = 30

Area: T = 263.8610640301
Perimeter: p = 77
Semiperimeter: s = 38.5

Angle ∠ A = α = 40.65553767418° = 40°39'19″ = 0.71095701828 rad
Angle ∠ B = β = 61.58663778464° = 61°35'11″ = 1.07548850678 rad
Angle ∠ C = γ = 77.75882454118° = 77°45'30″ = 1.3577137403 rad

Height: ha = 26.38660640301
Height: hb = 19.54552326149
Height: hc = 17.59107093534

Median: ma = 26.73301328092
Median: mb = 21.62875287539
Median: mc = 18.42655257727

Inradius: r = 6.85435231247
Circumradius: R = 15.3499011491

Vertex coordinates: A[30; 0] B[0; 0] C[9.51766666667; 17.59107093534]
Centroid: CG[13.17222222222; 5.86435697845]
Coordinates of the circumscribed circle: U[15; 3.2554558918]
Coordinates of the inscribed circle: I[11.5; 6.85435231247]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 139.3454623258° = 139°20'41″ = 0.71095701828 rad
∠ B' = β' = 118.4143622154° = 118°24'49″ = 1.07548850678 rad
∠ C' = γ' = 102.2421754588° = 102°14'30″ = 1.3577137403 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 = 20 ; ; b = 27 ; ; c = 30 ; ;

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

p = a+b+c = 20+27+30 = 77 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 77 }{ 2 } = 38.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 38.5 * (38.5-20)(38.5-27)(38.5-30) } ; ; T = sqrt{ 69622.44 } = 263.86 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 263.86 }{ 20 } = 26.39 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 263.86 }{ 27 } = 19.55 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 263.86 }{ 30 } = 17.59 ; ;

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{ 20**2-27**2-30**2 }{ 2 * 27 * 30 } ) = 40° 39'19" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-20**2-30**2 }{ 2 * 20 * 30 } ) = 61° 35'11" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-20**2-27**2 }{ 2 * 27 * 20 } ) = 77° 45'30" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 263.86 }{ 38.5 } = 6.85 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 40° 39'19" } = 15.35 ; ;




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

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