21 29 29 triangle

Acute isosceles triangle.

Sides: a = 21   b = 29   c = 29

Area: T = 283.8440073809
Perimeter: p = 79
Semiperimeter: s = 39.5

Angle ∠ A = α = 42.45546255238° = 42°27'17″ = 0.74109729981 rad
Angle ∠ B = β = 68.77326872381° = 68°46'22″ = 1.22003098277 rad
Angle ∠ C = γ = 68.77326872381° = 68°46'22″ = 1.22003098277 rad

Height: ha = 27.03223879818
Height: hb = 19.57551775041
Height: hc = 19.57551775041

Median: ma = 27.03223879818
Median: mb = 20.75545175805
Median: mc = 20.75545175805

Inradius: r = 7.18658246534
Circumradius: R = 15.55554145007

Vertex coordinates: A[29; 0] B[0; 0] C[7.60334482759; 19.57551775041]
Centroid: CG[12.20111494253; 6.5255059168]
Coordinates of the circumscribed circle: U[14.5; 5.63221328364]
Coordinates of the inscribed circle: I[10.5; 7.18658246534]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 137.5455374476° = 137°32'43″ = 0.74109729981 rad
∠ B' = β' = 111.2277312762° = 111°13'38″ = 1.22003098277 rad
∠ C' = γ' = 111.2277312762° = 111°13'38″ = 1.22003098277 rad

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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 = 29 ; ; c = 29 ; ;

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

p = a+b+c = 21+29+29 = 79 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 79 }{ 2 } = 39.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 39.5 * (39.5-21)(39.5-29)(39.5-29) } ; ; T = sqrt{ 80565.19 } = 283.84 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 283.84 }{ 21 } = 27.03 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 283.84 }{ 29 } = 19.58 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 283.84 }{ 29 } = 19.58 ; ;

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**2-29**2-29**2 }{ 2 * 29 * 29 } ) = 42° 27'17" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-21**2-29**2 }{ 2 * 21 * 29 } ) = 68° 46'22" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-21**2-29**2 }{ 2 * 29 * 21 } ) = 68° 46'22" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 283.84 }{ 39.5 } = 7.19 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 42° 27'17" } = 15.56 ; ;




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