18 28 28 triangle

Acute isosceles triangle.

Sides: a = 18 b = 28 c = 28

Area: T = 238.6277324504
Perimeter: p = 74
Semiperimeter: s = 37

Angle ∠ A = α = 37.4998681694° = 37°29'55″ = 0.65444754607 rad
Angle ∠ B = β = 71.2510659153° = 71°15'2″ = 1.24435585964 rad
Angle ∠ C = γ = 71.2510659153° = 71°15'2″ = 1.24435585964 rad

Height: h_a = 26.51441471671
Height: h_b = 17.04548088932
Height: h_c = 17.04548088932

Median: m_a = 26.51441471671
Median: m_b = 18.92108879284
Median: m_c = 18.92108879284

Inradius: r = 6.44993871488
Circumradius: R = 14.78545600135

Vertex coordinates: A[28; 0] B[0; 0] C[5.78657142857; 17.04548088932]
Centroid: CG[11.26219047619; 5.68216029644]
Coordinates of the circumscribed circle: U[14; 4.75221800043]
Coordinates of the inscribed circle: I[9; 6.44993871488]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 142.5011318306° = 142°30'5″ = 0.65444754607 rad
∠ B' = β' = 108.7499340847° = 108°44'58″ = 1.24435585964 rad
∠ C' = γ' = 108.7499340847° = 108°44'58″ = 1.24435585964 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.

$a = 18 ; ; b = 28 ; ; c = 28 ; ;$

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

$p = a+b+c = 18+28+28 = 74 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 74 }{ 2 } = 37 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 37 * (37-18)(37-28)(37-28) } ; ; T = sqrt{ 56943 } = 238.63 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 238.63 }{ 18 } = 26.51 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 238.63 }{ 28 } = 17.04 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 238.63 }{ 28 } = 17.04 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 18**2-28**2-28**2 }{ 2 * 28 * 28 } ) = 37° 29'55" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 28**2-18**2-28**2 }{ 2 * 18 * 28 } ) = 71° 15'2" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-18**2-28**2 }{ 2 * 28 * 18 } ) = 71° 15'2" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 238.63 }{ 37 } = 6.45 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 18 }{ 2 * sin 37° 29'55" } = 14.78 ; ;$

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