23 27 27 triangle

Acute isosceles triangle.

Sides: a = 23   b = 27   c = 27

Area: T = 280.9277370507
Perimeter: p = 77
Semiperimeter: s = 38.5

Angle ∠ A = α = 50.41985711742° = 50°25'7″ = 0.88799700711 rad
Angle ∠ B = β = 64.79107144129° = 64°47'27″ = 1.13108112912 rad
Angle ∠ C = γ = 64.79107144129° = 64°47'27″ = 1.13108112912 rad

Height: ha = 24.42884670006
Height: hb = 20.80994348524
Height: hc = 20.80994348524

Median: ma = 24.42884670006
Median: mb = 21.13664613878
Median: mc = 21.13664613878

Inradius: r = 7.29768148184
Circumradius: R = 14.92111164168

Vertex coordinates: A[27; 0] B[0; 0] C[9.79662962963; 20.80994348524]
Centroid: CG[12.26554320988; 6.93664782841]
Coordinates of the circumscribed circle: U[13.5; 6.35552903257]
Coordinates of the inscribed circle: I[11.5; 7.29768148184]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 129.5811428826° = 129°34'53″ = 0.88799700711 rad
∠ B' = β' = 115.2099285587° = 115°12'33″ = 1.13108112912 rad
∠ C' = γ' = 115.2099285587° = 115°12'33″ = 1.13108112912 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 = 23 ; ; b = 27 ; ; c = 27 ; ;

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

p = a+b+c = 23+27+27 = 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 * (38.5-23)(38.5-27)(38.5-27) } ; ; T = sqrt{ 78920.19 } = 280.93 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 280.93 }{ 23 } = 24.43 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 280.93 }{ 27 } = 20.81 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 280.93 }{ 27 } = 20.81 ; ;

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{ 23**2-27**2-27**2 }{ 2 * 27 * 27 } ) = 50° 25'7" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-23**2-27**2 }{ 2 * 23 * 27 } ) = 64° 47'27" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-23**2-27**2 }{ 2 * 27 * 23 } ) = 64° 47'27" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 280.93 }{ 38.5 } = 7.3 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 23 }{ 2 * sin 50° 25'7" } = 14.92 ; ;




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