16 20 21 triangle

Acute scalene triangle.

Sides: a = 16   b = 20   c = 21

Area: T = 150.7011484731
Perimeter: p = 57
Semiperimeter: s = 28.5

Angle ∠ A = α = 45.85988345181° = 45°51'32″ = 0.88003876535 rad
Angle ∠ B = β = 63.77107223958° = 63°46'15″ = 1.11330090722 rad
Angle ∠ C = γ = 70.37704430861° = 70°22'14″ = 1.22881959279 rad

Height: ha = 18.83876855913
Height: hb = 15.07701484731
Height: hc = 14.35325223553

Median: ma = 18.88112075885
Median: mb = 15.76438827704
Median: mc = 14.75663545634

Inradius: r = 5.28877713941
Circumradius: R = 11.14878662802

Vertex coordinates: A[21; 0] B[0; 0] C[7.07114285714; 14.35325223553]
Centroid: CG[9.35771428571; 4.78441741184]
Coordinates of the circumscribed circle: U[10.5; 3.74549863285]
Coordinates of the inscribed circle: I[8.5; 5.28877713941]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 134.1411165482° = 134°8'28″ = 0.88003876535 rad
∠ B' = β' = 116.2299277604° = 116°13'45″ = 1.11330090722 rad
∠ C' = γ' = 109.6329556914° = 109°37'46″ = 1.22881959279 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 = 20 ; ; c = 21 ; ;

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

p = a+b+c = 16+20+21 = 57 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 57 }{ 2 } = 28.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 28.5 * (28.5-16)(28.5-20)(28.5-21) } ; ; T = sqrt{ 22710.94 } = 150.7 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 150.7 }{ 16 } = 18.84 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 150.7 }{ 20 } = 15.07 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 150.7 }{ 21 } = 14.35 ; ;

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**2-20**2-21**2 }{ 2 * 20 * 21 } ) = 45° 51'32" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-16**2-21**2 }{ 2 * 16 * 21 } ) = 63° 46'15" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 21**2-16**2-20**2 }{ 2 * 20 * 16 } ) = 70° 22'14" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 150.7 }{ 28.5 } = 5.29 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 45° 51'32" } = 11.15 ; ;




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

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