11 29 29 triangle

Acute isosceles triangle.

Sides: a = 11 b = 29 c = 29

Area: T = 156.6055196274
Perimeter: p = 69
Semiperimeter: s = 34.5

Angle ∠ A = α = 21.86553222896° = 21°51'55″ = 0.3821621866 rad
Angle ∠ B = β = 79.06773388552° = 79°4'2″ = 1.38799853938 rad
Angle ∠ C = γ = 79.06773388552° = 79°4'2″ = 1.38799853938 rad

Height: h_a = 28.47436720498
Height: h_b = 10.88003583637
Height: h_c = 10.88003583637

Median: m_a = 28.47436720498
Median: m_b = 16.45444826719
Median: m_c = 16.45444826719

Inradius: r = 4.53992810514
Circumradius: R = 14.76880284884

Vertex coordinates: A[29; 0] B[0; 0] C[2.08662068966; 10.88003583637]
Centroid: CG[10.36220689655; 3.66001194546]
Coordinates of the circumscribed circle: U[14.5; 2.80108329892]
Coordinates of the inscribed circle: I[5.5; 4.53992810514]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 158.135467771° = 158°8'5″ = 0.3821621866 rad
∠ B' = β' = 100.9332661145° = 100°55'58″ = 1.38799853938 rad
∠ C' = γ' = 100.9332661145° = 100°55'58″ = 1.38799853938 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.

$a = 11 ; ; b = 29 ; ; c = 29 ; ;$

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

$p = a+b+c = 11+29+29 = 69 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 69 }{ 2 } = 34.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 34.5 * (34.5-11)(34.5-29)(34.5-29) } ; ; T = sqrt{ 24525.19 } = 156.61 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 156.61 }{ 11 } = 28.47 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 156.61 }{ 29 } = 10.8 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 156.61 }{ 29 } = 10.8 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 11**2-29**2-29**2 }{ 2 * 29 * 29 } ) = 21° 51'55" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-11**2-29**2 }{ 2 * 11 * 29 } ) = 79° 4'2" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-11**2-29**2 }{ 2 * 29 * 11 } ) = 79° 4'2" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 156.61 }{ 34.5 } = 4.54 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 21° 51'55" } = 14.77 ; ;$

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