10 23 28 triangle

Obtuse scalene triangle.

Sides: a = 10   b = 23   c = 28

Area: T = 108.2754823943
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 19.64990068155° = 19°38'56″ = 0.34329398637 rad
Angle ∠ B = β = 50.65994361643° = 50°39'34″ = 0.88441739583 rad
Angle ∠ C = γ = 109.692155702° = 109°41'30″ = 1.91444788316 rad

Height: ha = 21.65549647887
Height: hb = 9.4155202082
Height: hc = 7.7343915996

Median: ma = 25.13296637463
Median: mb = 17.65997159068
Median: mc = 10.88657705285

Inradius: r = 3.55499942277
Circumradius: R = 14.87695693178

Vertex coordinates: A[28; 0] B[0; 0] C[6.33992857143; 7.7343915996]
Centroid: CG[11.44664285714; 2.57879719987]
Coordinates of the circumscribed circle: U[14; -5.01103983571]
Coordinates of the inscribed circle: I[7.5; 3.55499942277]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 160.3510993185° = 160°21'4″ = 0.34329398637 rad
∠ B' = β' = 129.3410563836° = 129°20'26″ = 0.88441739583 rad
∠ C' = γ' = 70.30884429798° = 70°18'30″ = 1.91444788316 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 = 23 ; ; c = 28 ; ;

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

p = a+b+c = 10+23+28 = 61 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 61 }{ 2 } = 30.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30.5 * (30.5-10)(30.5-23)(30.5-28) } ; ; T = sqrt{ 11723.44 } = 108.27 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 108.27 }{ 10 } = 21.65 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 108.27 }{ 23 } = 9.42 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 108.27 }{ 28 } = 7.73 ; ;

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**2-23**2-28**2 }{ 2 * 23 * 28 } ) = 19° 38'56" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-10**2-28**2 }{ 2 * 10 * 28 } ) = 50° 39'34" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-10**2-23**2 }{ 2 * 23 * 10 } ) = 109° 41'30" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 108.27 }{ 30.5 } = 3.55 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 19° 38'56" } = 14.87 ; ;




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

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