21 22 23 triangle

Acute scalene triangle.

Sides: a = 21   b = 22   c = 23

Area: T = 208.7110325571
Perimeter: p = 66
Semiperimeter: s = 33

Angle ∠ A = α = 55.58326112896° = 55°34'57″ = 0.97700995739 rad
Angle ∠ B = β = 59.79443222356° = 59°47'40″ = 1.0443607797 rad
Angle ∠ C = γ = 64.62330664748° = 64°37'23″ = 1.12878852827 rad

Height: ha = 19.87771738639
Height: hb = 18.9743665961
Height: hc = 18.14987239627

Median: ma = 19.90660292374
Median: mb = 19.07987840283
Median: mc = 18.17327818454

Inradius: r = 6.32545553203
Circumradius: R = 12.72881675822

Vertex coordinates: A[23; 0] B[0; 0] C[10.56552173913; 18.14987239627]
Centroid: CG[11.18884057971; 6.05495746542]
Coordinates of the circumscribed circle: U[11.5; 5.45549289638]
Coordinates of the inscribed circle: I[11; 6.32545553203]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 124.417738871° = 124°25'3″ = 0.97700995739 rad
∠ B' = β' = 120.2065677764° = 120°12'20″ = 1.0443607797 rad
∠ C' = γ' = 115.3776933525° = 115°22'37″ = 1.12878852827 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 = 21 ; ; b = 22 ; ; c = 23 ; ;

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

p = a+b+c = 21+22+23 = 66 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 66 }{ 2 } = 33 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33 * (33-21)(33-22)(33-23) } ; ; T = sqrt{ 43560 } = 208.71 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 208.71 }{ 21 } = 19.88 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 208.71 }{ 22 } = 18.97 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 208.71 }{ 23 } = 18.15 ; ;

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{ 21**2-22**2-23**2 }{ 2 * 22 * 23 } ) = 55° 34'57" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-21**2-23**2 }{ 2 * 21 * 23 } ) = 59° 47'40" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-21**2-22**2 }{ 2 * 22 * 21 } ) = 64° 37'23" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 208.71 }{ 33 } = 6.32 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 55° 34'57" } = 12.73 ; ;




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

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