18 22 30 triangle

Obtuse scalene triangle.

Sides: a = 18 b = 22 c = 30

Area: T = 196.6659604393
Perimeter: p = 70
Semiperimeter: s = 35

Angle ∠ A = α = 36.58795428375° = 36°34'46″ = 0.63884334614 rad
Angle ∠ B = β = 46.75498273458° = 46°44'59″ = 0.81659384119 rad
Angle ∠ C = γ = 96.67106298167° = 96°40'14″ = 1.68772207803 rad

Height: h_a = 21.85110671548
Height: h_b = 17.87881458539
Height: h_c = 13.11106402929

Median: m_a = 24.71884141886
Median: m_b = 22.15985198062
Median: m_c = 13.37990881603

Inradius: r = 5.61988458398
Circumradius: R = 15.10222372346

Vertex coordinates: A[30; 0] B[0; 0] C[12.33333333333; 13.11106402929]
Centroid: CG[14.11111111111; 4.3770213431]
Coordinates of the circumscribed circle: U[15; -1.75443002848]
Coordinates of the inscribed circle: I[13; 5.61988458398]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 143.4220457162° = 143°25'14″ = 0.63884334614 rad
∠ B' = β' = 133.2550172654° = 133°15'1″ = 0.81659384119 rad
∠ C' = γ' = 83.32993701833° = 83°19'46″ = 1.68772207803 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.

$a = 18 ; ; b = 22 ; ; c = 30 ; ;$

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

$p = a+b+c = 18+22+30 = 70 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 70 }{ 2 } = 35 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 35 * (35-18)(35-22)(35-30) } ; ; T = sqrt{ 38675 } = 196.66 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 196.66 }{ 18 } = 21.85 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 196.66 }{ 22 } = 17.88 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 196.66 }{ 30 } = 13.11 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 18**2-22**2-30**2 }{ 2 * 22 * 30 } ) = 36° 34'46" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-18**2-30**2 }{ 2 * 18 * 30 } ) = 46° 44'59" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-18**2-22**2 }{ 2 * 22 * 18 } ) = 96° 40'14" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 196.66 }{ 35 } = 5.62 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 18 }{ 2 * sin 36° 34'46" } = 15.1 ; ;$

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