22 29 30 triangle

Acute scalene triangle.

Sides: a = 22   b = 29   c = 30

Area: T = 300.786553406
Perimeter: p = 81
Semiperimeter: s = 40.5

Angle ∠ A = α = 43.74658710599° = 43°44'45″ = 0.76435094841 rad
Angle ∠ B = β = 65.70994748682° = 65°42'34″ = 1.14768466862 rad
Angle ∠ C = γ = 70.54546540718° = 70°32'41″ = 1.23112364832 rad

Height: ha = 27.344413946
Height: hb = 20.74438299352
Height: hc = 20.05223689374

Median: ma = 27.3776997644
Median: mb = 21.94988040676
Median: mc = 20.91765006634

Inradius: r = 7.42768033101
Circumradius: R = 15.90883448443

Vertex coordinates: A[30; 0] B[0; 0] C[9.05; 20.05223689374]
Centroid: CG[13.01766666667; 6.68441229791]
Coordinates of the circumscribed circle: U[15; 5.29986258298]
Coordinates of the inscribed circle: I[11.5; 7.42768033101]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 136.254412894° = 136°15'15″ = 0.76435094841 rad
∠ B' = β' = 114.2910525132° = 114°17'26″ = 1.14768466862 rad
∠ C' = γ' = 109.4555345928° = 109°27'19″ = 1.23112364832 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 = 22 ; ; b = 29 ; ; c = 30 ; ;

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

p = a+b+c = 22+29+30 = 81 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 81 }{ 2 } = 40.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 40.5 * (40.5-22)(40.5-29)(40.5-30) } ; ; T = sqrt{ 90471.94 } = 300.79 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 300.79 }{ 22 } = 27.34 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 300.79 }{ 29 } = 20.74 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 300.79 }{ 30 } = 20.05 ; ;

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{ 22**2-29**2-30**2 }{ 2 * 29 * 30 } ) = 43° 44'45" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-22**2-30**2 }{ 2 * 22 * 30 } ) = 65° 42'34" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-22**2-29**2 }{ 2 * 29 * 22 } ) = 70° 32'41" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 300.79 }{ 40.5 } = 7.43 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 22 }{ 2 * sin 43° 44'45" } = 15.91 ; ;




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

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