17 17 19 triangle

Acute isosceles triangle.

Sides: a = 17 b = 17 c = 19

Area: T = 133.9329785709
Perimeter: p = 53
Semiperimeter: s = 26.5

Angle ∠ A = α = 56.02655240497° = 56°1'32″ = 0.97878298598 rad
Angle ∠ B = β = 56.02655240497° = 56°1'32″ = 0.97878298598 rad
Angle ∠ C = γ = 67.94989519006° = 67°56'56″ = 1.18659329339 rad

Height: h_a = 15.75664453775
Height: h_b = 15.75664453775
Height: h_c = 14.09878721799

Median: m_a = 15.89881130956
Median: m_b = 15.89881130956
Median: m_c = 14.09878721799

Inradius: r = 5.05439541777
Circumradius: R = 10.25497737358

Vertex coordinates: A[19; 0] B[0; 0] C[9.5; 14.09878721799]
Centroid: CG[9.5; 4.69992907266]
Coordinates of the circumscribed circle: U[9.5; 3.84880984441]
Coordinates of the inscribed circle: I[9.5; 5.05439541777]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 123.974447595° = 123°58'28″ = 0.97878298598 rad
∠ B' = β' = 123.974447595° = 123°58'28″ = 0.97878298598 rad
∠ C' = γ' = 112.0511048099° = 112°3'4″ = 1.18659329339 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.

$a = 17 ; ; b = 17 ; ; c = 19 ; ;$

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

$p = a+b+c = 17+17+19 = 53 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 53 }{ 2 } = 26.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 26.5 * (26.5-17)(26.5-17)(26.5-19) } ; ; T = sqrt{ 17937.19 } = 133.93 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 133.93 }{ 17 } = 15.76 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 133.93 }{ 17 } = 15.76 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 133.93 }{ 19 } = 14.1 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 17**2-17**2-19**2 }{ 2 * 17 * 19 } ) = 56° 1'32" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 17**2-17**2-19**2 }{ 2 * 17 * 19 } ) = 56° 1'32" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 19**2-17**2-17**2 }{ 2 * 17 * 17 } ) = 67° 56'56" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 133.93 }{ 26.5 } = 5.05 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 17 }{ 2 * sin 56° 1'32" } = 10.25 ; ;$

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