5 26 28 triangle

Obtuse scalene triangle.

Sides: a = 5 b = 26 c = 28

Area: T = 61.59990056738
Perimeter: p = 59
Semiperimeter: s = 29.5

Angle ∠ A = α = 9.74329385734° = 9°44'35″ = 0.17700463569 rad
Angle ∠ B = β = 61.64106499718° = 61°38'26″ = 1.07658322951 rad
Angle ∠ C = γ = 108.6166411455° = 108°36'59″ = 1.89657140016 rad

Height: h_a = 24.64396022695
Height: h_b = 4.73883850518
Height: h_c = 4.43999289767

Median: m_a = 26.90326021046
Median: m_b = 15.34660092532
Median: m_c = 12.43298028947

Inradius: r = 2.08881018872
Circumradius: R = 14.77329657329

Vertex coordinates: A[28; 0] B[0; 0] C[2.375; 4.43999289767]
Centroid: CG[10.125; 1.46766429922]
Coordinates of the circumscribed circle: U[14; -4.71659852147]
Coordinates of the inscribed circle: I[3.5; 2.08881018872]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 170.2577061427° = 170°15'25″ = 0.17700463569 rad
∠ B' = β' = 118.3599350028° = 118°21'34″ = 1.07658322951 rad
∠ C' = γ' = 71.38435885453° = 71°23'1″ = 1.89657140016 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.

$a = 5 ; ; b = 26 ; ; c = 28 ; ;$

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

$p = a+b+c = 5+26+28 = 59 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 59 }{ 2 } = 29.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29.5 * (29.5-5)(29.5-26)(29.5-28) } ; ; T = sqrt{ 3794.44 } = 61.6 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 61.6 }{ 5 } = 24.64 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 61.6 }{ 26 } = 4.74 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 61.6 }{ 28 } = 4.4 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 5**2-26**2-28**2 }{ 2 * 26 * 28 } ) = 9° 44'35" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 26**2-5**2-28**2 }{ 2 * 5 * 28 } ) = 61° 38'26" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-5**2-26**2 }{ 2 * 26 * 5 } ) = 108° 36'59" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 61.6 }{ 29.5 } = 2.09 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 5 }{ 2 * sin 9° 44'35" } = 14.77 ; ;$

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