9 25 29 triangle

Obtuse scalene triangle.

Sides: a = 9 b = 25 c = 29

Area: T = 107.3188160159
Perimeter: p = 63
Semiperimeter: s = 31.5

Angle ∠ A = α = 17.22105164629° = 17°13'14″ = 0.30105547112 rad
Angle ∠ B = β = 55.32218804133° = 55°19'19″ = 0.96655489616 rad
Angle ∠ C = γ = 107.4587603124° = 107°27'27″ = 1.87554889808 rad

Height: h_a = 23.84884800354
Height: h_b = 8.58554528128
Height: h_c = 7.40112524248

Median: m_a = 26.69773781484
Median: m_b = 17.45770902501
Median: m_c = 11.94878031453

Inradius: r = 3.40769257193
Circumradius: R = 15.22001301325

Vertex coordinates: A[29; 0] B[0; 0] C[5.12106896552; 7.40112524248]
Centroid: CG[11.37435632184; 2.46770841416]
Coordinates of the circumscribed circle: U[14.5; -4.56600390397]
Coordinates of the inscribed circle: I[6.5; 3.40769257193]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 162.7799483537° = 162°46'46″ = 0.30105547112 rad
∠ B' = β' = 124.6788119587° = 124°40'41″ = 0.96655489616 rad
∠ C' = γ' = 72.54223968763° = 72°32'33″ = 1.87554889808 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.

$a = 9 ; ; b = 25 ; ; c = 29 ; ;$

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

$p = a+b+c = 9+25+29 = 63 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 63 }{ 2 } = 31.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31.5 * (31.5-9)(31.5-25)(31.5-29) } ; ; T = sqrt{ 11517.19 } = 107.32 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 107.32 }{ 9 } = 23.85 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 107.32 }{ 25 } = 8.59 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 107.32 }{ 29 } = 7.4 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 9**2-25**2-29**2 }{ 2 * 25 * 29 } ) = 17° 13'14" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-9**2-29**2 }{ 2 * 9 * 29 } ) = 55° 19'19" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-9**2-25**2 }{ 2 * 25 * 9 } ) = 107° 27'27" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 107.32 }{ 31.5 } = 3.41 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 17° 13'14" } = 15.2 ; ;$

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