16 22 23 triangle

Acute scalene triangle.

Sides: a = 16   b = 22   c = 23

Area: T = 167.9099015541
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 41.58105251992° = 41°34'50″ = 0.72657170694 rad
Angle ∠ B = β = 65.86600155707° = 65°51'36″ = 1.14994741171 rad
Angle ∠ C = γ = 72.55994592301° = 72°33'34″ = 1.2666401467 rad

Height: ha = 20.98986269426
Height: hb = 15.26444559583
Height: hc = 14.60107839601

Median: ma = 21.0365683968
Median: mb = 16.47772570533
Median: mc = 15.41991439451

Inradius: r = 5.50552136243
Circumradius: R = 12.05441472623

Vertex coordinates: A[23; 0] B[0; 0] C[6.54334782609; 14.60107839601]
Centroid: CG[9.8487826087; 4.86769279867]
Coordinates of the circumscribed circle: U[11.5; 3.6132819705]
Coordinates of the inscribed circle: I[8.5; 5.50552136243]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 138.4199474801° = 138°25'10″ = 0.72657170694 rad
∠ B' = β' = 114.1439984429° = 114°8'24″ = 1.14994741171 rad
∠ C' = γ' = 107.441054077° = 107°26'26″ = 1.2666401467 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 = 16 ; ; b = 22 ; ; c = 23 ; ;

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

p = a+b+c = 16+22+23 = 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 * (30.5-16)(30.5-22)(30.5-23) } ; ; T = sqrt{ 28193.44 } = 167.91 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 167.91 }{ 16 } = 20.99 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 167.91 }{ 22 } = 15.26 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 167.91 }{ 23 } = 14.6 ; ;

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{ 16**2-22**2-23**2 }{ 2 * 22 * 23 } ) = 41° 34'50" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-16**2-23**2 }{ 2 * 16 * 23 } ) = 65° 51'36" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-16**2-22**2 }{ 2 * 22 * 16 } ) = 72° 33'34" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 167.91 }{ 30.5 } = 5.51 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 41° 34'50" } = 12.05 ; ;




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

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