10 19 20 triangle

Acute scalene triangle.

Sides: a = 10   b = 19   c = 20

Area: T = 93.76879982723
Perimeter: p = 49
Semiperimeter: s = 24.5

Angle ∠ A = α = 29.5721926754° = 29°34'19″ = 0.5166127488 rad
Angle ∠ B = β = 69.66655200278° = 69°39'56″ = 1.21658926996 rad
Angle ∠ C = γ = 80.76325532182° = 80°45'45″ = 1.4109572466 rad

Height: ha = 18.75435996545
Height: hb = 9.87703156076
Height: hc = 9.37767998272

Median: ma = 18.8554707635
Median: mb = 12.63992246598
Median: mc = 11.42436596588

Inradius: r = 3.82772652356
Circumradius: R = 10.13113882935

Vertex coordinates: A[20; 0] B[0; 0] C[3.475; 9.37767998272]
Centroid: CG[7.825; 3.12655999424]
Coordinates of the circumscribed circle: U[10; 1.62663544366]
Coordinates of the inscribed circle: I[5.5; 3.82772652356]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 150.4288073246° = 150°25'41″ = 0.5166127488 rad
∠ B' = β' = 110.3344479972° = 110°20'4″ = 1.21658926996 rad
∠ C' = γ' = 99.23774467818° = 99°14'15″ = 1.4109572466 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 = 10 ; ; b = 19 ; ; c = 20 ; ;

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

p = a+b+c = 10+19+20 = 49 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 49 }{ 2 } = 24.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 24.5 * (24.5-10)(24.5-19)(24.5-20) } ; ; T = sqrt{ 8792.44 } = 93.77 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 93.77 }{ 10 } = 18.75 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 93.77 }{ 19 } = 9.87 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 93.77 }{ 20 } = 9.38 ; ;

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{ 10**2-19**2-20**2 }{ 2 * 19 * 20 } ) = 29° 34'19" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 19**2-10**2-20**2 }{ 2 * 10 * 20 } ) = 69° 39'56" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20**2-10**2-19**2 }{ 2 * 19 * 10 } ) = 80° 45'45" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 93.77 }{ 24.5 } = 3.83 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 29° 34'19" } = 10.13 ; ;




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

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