15 17 22 triangle

Acute scalene triangle.

Sides: a = 15 b = 17 c = 22

Area: T = 127.2799220614
Perimeter: p = 54
Semiperimeter: s = 27

Angle ∠ A = α = 42.89334830421° = 42°53'37″ = 0.74986325067 rad
Angle ∠ B = β = 50.47988036414° = 50°28'44″ = 0.8811021326 rad
Angle ∠ C = γ = 86.62877133166° = 86°37'40″ = 1.51219388208 rad

Height: h_a = 16.97105627485
Height: h_b = 14.97440259545
Height: h_c = 11.57108382376

Median: m_a = 18.17327818454
Median: m_b = 16.88002976164
Median: m_c = 11.66219037897

Inradius: r = 4.71440452079
Circumradius: R = 11.01990806735

Vertex coordinates: A[22; 0] B[0; 0] C[9.54554545455; 11.57108382376]
Centroid: CG[10.51551515152; 3.85769460792]
Coordinates of the circumscribed circle: U[11; 0.64881812161]
Coordinates of the inscribed circle: I[10; 4.71440452079]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 137.1076516958° = 137°6'23″ = 0.74986325067 rad
∠ B' = β' = 129.5211196359° = 129°31'16″ = 0.8811021326 rad
∠ C' = γ' = 93.37222866834° = 93°22'20″ = 1.51219388208 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.

$a = 15 ; ; b = 17 ; ; c = 22 ; ;$

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

$p = a+b+c = 15+17+22 = 54 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 54 }{ 2 } = 27 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 27 * (27-15)(27-17)(27-22) } ; ; T = sqrt{ 16200 } = 127.28 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 127.28 }{ 15 } = 16.97 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 127.28 }{ 17 } = 14.97 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 127.28 }{ 22 } = 11.57 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 15**2-17**2-22**2 }{ 2 * 17 * 22 } ) = 42° 53'37" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 17**2-15**2-22**2 }{ 2 * 15 * 22 } ) = 50° 28'44" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-15**2-17**2 }{ 2 * 17 * 15 } ) = 86° 37'40" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 127.28 }{ 27 } = 4.71 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 42° 53'37" } = 11.02 ; ;$

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