35 60 90 triangle

Obtuse scalene triangle.

Sides: a = 35 b = 60 c = 90

Area: T = 657.387997954
Perimeter: p = 185
Semiperimeter: s = 92.5

Angle ∠ A = α = 14.09216737232° = 14°5'30″ = 0.24659461036 rad
Angle ∠ B = β = 24.67695454829° = 24°40'10″ = 0.43105647936 rad
Angle ∠ C = γ = 141.2398780794° = 141°14'20″ = 2.46550817564 rad

Height: h_a = 37.56545702594
Height: h_b = 21.91326659847
Height: h_c = 14.60884439898

Median: m_a = 74.45663630592
Median: m_b = 61.33992207319
Median: m_c = 19.6855019685

Inradius: r = 7.10768105896
Circumradius: R = 71.87662382041

Vertex coordinates: A[90; 0] B[0; 0] C[31.80655555556; 14.60884439898]
Centroid: CG[40.60218518519; 4.86994813299]
Coordinates of the circumscribed circle: U[45; -56.04663524091]
Coordinates of the inscribed circle: I[32.5; 7.10768105896]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 165.9088326277° = 165°54'30″ = 0.24659461036 rad
∠ B' = β' = 155.3330454517° = 155°19'50″ = 0.43105647936 rad
∠ C' = γ' = 38.76112192061° = 38°45'40″ = 2.46550817564 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.

$a = 35 ; ; b = 60 ; ; c = 90 ; ;$

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

$p = a+b+c = 35+60+90 = 185 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 185 }{ 2 } = 92.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 92.5 * (92.5-35)(92.5-60)(92.5-90) } ; ; T = sqrt{ 432148.44 } = 657.38 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 657.38 }{ 35 } = 37.56 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 657.38 }{ 60 } = 21.91 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 657.38 }{ 90 } = 14.61 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 35**2-60**2-90**2 }{ 2 * 60 * 90 } ) = 14° 5'30" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 60**2-35**2-90**2 }{ 2 * 35 * 90 } ) = 24° 40'10" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 90**2-35**2-60**2 }{ 2 * 60 * 35 } ) = 141° 14'20" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 657.38 }{ 92.5 } = 7.11 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 35 }{ 2 * sin 14° 5'30" } = 71.88 ; ;$

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