21 27 29 triangle

Acute scalene triangle.

Sides: a = 21   b = 27   c = 29

Area: T = 271.306644574
Perimeter: p = 77
Semiperimeter: s = 38.5

Angle ∠ A = α = 43.86774365388° = 43°52'3″ = 0.7665631202 rad
Angle ∠ B = β = 62.99879108049° = 62°59'52″ = 1.10995209654 rad
Angle ∠ C = γ = 73.13546526563° = 73°8'5″ = 1.27664404862 rad

Height: ha = 25.83987091181
Height: hb = 20.09767737585
Height: hc = 18.71107893614

Median: ma = 25.97659504157
Median: mb = 21.41884499906
Median: mc = 19.35884606826

Inradius: r = 7.04769206686
Circumradius: R = 15.15216857213

Vertex coordinates: A[29; 0] B[0; 0] C[9.53444827586; 18.71107893614]
Centroid: CG[12.84548275862; 6.23769297871]
Coordinates of the circumscribed circle: U[14.5; 4.39658594376]
Coordinates of the inscribed circle: I[11.5; 7.04769206686]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 136.1332563461° = 136°7'57″ = 0.7665631202 rad
∠ B' = β' = 117.0022089195° = 117°8″ = 1.10995209654 rad
∠ C' = γ' = 106.8655347344° = 106°51'55″ = 1.27664404862 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 = 27 ; ; c = 29 ; ;

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

p = a+b+c = 21+27+29 = 77 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 77 }{ 2 } = 38.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 38.5 * (38.5-21)(38.5-27)(38.5-29) } ; ; T = sqrt{ 73607.19 } = 271.31 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 271.31 }{ 21 } = 25.84 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 271.31 }{ 27 } = 20.1 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 271.31 }{ 29 } = 18.71 ; ;

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**2-27**2-29**2 }{ 2 * 27 * 29 } ) = 43° 52'3" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-21**2-29**2 }{ 2 * 21 * 29 } ) = 62° 59'52" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-21**2-27**2 }{ 2 * 27 * 21 } ) = 73° 8'5" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 271.31 }{ 38.5 } = 7.05 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 43° 52'3" } = 15.15 ; ;




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

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