8 22 23 triangle

Acute scalene triangle.

Sides: a = 8   b = 22   c = 23

Area: T = 87.87217104647
Perimeter: p = 53
Semiperimeter: s = 26.5

Angle ∠ A = α = 20.32334214225° = 20°19'24″ = 0.35547106191 rad
Angle ∠ B = β = 72.77107522394° = 72°46'15″ = 1.27700892257 rad
Angle ∠ C = γ = 86.9065826338° = 86°54'21″ = 1.51767928088 rad

Height: ha = 21.96879276162
Height: hb = 7.9888337315
Height: hc = 7.64110183013

Median: ma = 22.14772345904
Median: mb = 13.24876412995
Median: mc = 11.90658808998

Inradius: r = 3.31659136024
Circumradius: R = 11.51767895862

Vertex coordinates: A[23; 0] B[0; 0] C[2.37695652174; 7.64110183013]
Centroid: CG[8.45765217391; 2.54770061004]
Coordinates of the circumscribed circle: U[11.5; 0.62216448924]
Coordinates of the inscribed circle: I[4.5; 3.31659136024]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 159.6776578577° = 159°40'36″ = 0.35547106191 rad
∠ B' = β' = 107.2299247761° = 107°13'45″ = 1.27700892257 rad
∠ C' = γ' = 93.0944173662° = 93°5'39″ = 1.51767928088 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 = 8 ; ; b = 22 ; ; c = 23 ; ;

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

p = a+b+c = 8+22+23 = 53 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 53 }{ 2 } = 26.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 26.5 * (26.5-8)(26.5-22)(26.5-23) } ; ; T = sqrt{ 7721.44 } = 87.87 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 87.87 }{ 8 } = 21.97 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 87.87 }{ 22 } = 7.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 87.87 }{ 23 } = 7.64 ; ;

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{ 8**2-22**2-23**2 }{ 2 * 22 * 23 } ) = 20° 19'24" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-8**2-23**2 }{ 2 * 8 * 23 } ) = 72° 46'15" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-8**2-22**2 }{ 2 * 22 * 8 } ) = 86° 54'21" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 87.87 }{ 26.5 } = 3.32 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 8 }{ 2 * sin 20° 19'24" } = 11.52 ; ;




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

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