20 20 23 triangle

Acute isosceles triangle.

Sides: a = 20 b = 20 c = 23

Area: T = 188.1755283977
Perimeter: p = 63
Semiperimeter: s = 31.5

Angle ∠ A = α = 54.99003678046° = 54°54'1″ = 0.95881921787 rad
Angle ∠ B = β = 54.99003678046° = 54°54'1″ = 0.95881921787 rad
Angle ∠ C = γ = 70.19992643908° = 70°11'57″ = 1.22552082961 rad

Height: h_a = 18.81875283977
Height: h_b = 18.81875283977
Height: h_c = 16.3633068172

Median: m_a = 19.0921883092
Median: m_b = 19.0921883092
Median: m_c = 16.3633068172

Inradius: r = 5.9743818539
Circumradius: R = 12.2232646627

Vertex coordinates: A[23; 0] B[0; 0] C[11.5; 16.3633068172]
Centroid: CG[11.5; 5.45443560573]
Coordinates of the circumscribed circle: U[11.5; 4.14404215449]
Coordinates of the inscribed circle: I[11.5; 5.9743818539]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 125.1099632195° = 125°5'59″ = 0.95881921787 rad
∠ B' = β' = 125.1099632195° = 125°5'59″ = 0.95881921787 rad
∠ C' = γ' = 109.8010735609° = 109°48'3″ = 1.22552082961 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.

$a = 20 ; ; b = 20 ; ; c = 23 ; ;$

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

$p = a+b+c = 20+20+23 = 63 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 63 }{ 2 } = 31.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31.5 * (31.5-20)(31.5-20)(31.5-23) } ; ; T = sqrt{ 35409.94 } = 188.18 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 188.18 }{ 20 } = 18.82 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 188.18 }{ 20 } = 18.82 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 188.18 }{ 23 } = 16.36 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 20**2-20**2-23**2 }{ 2 * 20 * 23 } ) = 54° 54'1" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-20**2-23**2 }{ 2 * 20 * 23 } ) = 54° 54'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-20**2-20**2 }{ 2 * 20 * 20 } ) = 70° 11'57" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 188.18 }{ 31.5 } = 5.97 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 54° 54'1" } = 12.22 ; ;$

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