21 22 30 triangle

Acute scalene triangle.

Sides: a = 21   b = 22   c = 30

Area: T = 230.9155433655
Perimeter: p = 73
Semiperimeter: s = 36.5

Angle ∠ A = α = 44.40664477069° = 44°24'23″ = 0.77550387216 rad
Angle ∠ B = β = 47.14439519766° = 47°8'38″ = 0.82328171844 rad
Angle ∠ C = γ = 88.45496003164° = 88°26'59″ = 1.54437367476 rad

Height: ha = 21.99219460624
Height: hb = 20.99223121504
Height: hc = 15.39443622437

Median: ma = 24.11994941904
Median: mb = 23.44114163395
Median: mc = 15.41110350074

Inradius: r = 6.32664502371
Circumradius: R = 15.00554933322

Vertex coordinates: A[30; 0] B[0; 0] C[14.28333333333; 15.39443622437]
Centroid: CG[14.76111111111; 5.13114540812]
Coordinates of the circumscribed circle: U[15; 0.4065992785]
Coordinates of the inscribed circle: I[14.5; 6.32664502371]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 135.5943552293° = 135°35'37″ = 0.77550387216 rad
∠ B' = β' = 132.8566048023° = 132°51'22″ = 0.82328171844 rad
∠ C' = γ' = 91.55503996836° = 91°33'1″ = 1.54437367476 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 = 30 ; ;

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

p = a+b+c = 21+22+30 = 73 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 73 }{ 2 } = 36.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 36.5 * (36.5-21)(36.5-22)(36.5-30) } ; ; T = sqrt{ 53321.94 } = 230.92 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 230.92 }{ 21 } = 21.99 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 230.92 }{ 22 } = 20.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 230.92 }{ 30 } = 15.39 ; ;

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**2-30**2 }{ 2 * 22 * 30 } ) = 44° 24'23" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-21**2-30**2 }{ 2 * 21 * 30 } ) = 47° 8'38" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-21**2-22**2 }{ 2 * 22 * 21 } ) = 88° 26'59" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 230.92 }{ 36.5 } = 6.33 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 44° 24'23" } = 15.01 ; ;




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

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