6 15 17 triangle

Obtuse scalene triangle.

Sides: a = 6 b = 15 c = 17

Area: T = 44.45222215418
Perimeter: p = 38
Semiperimeter: s = 19

Angle ∠ A = α = 20.40444519895° = 20°24'16″ = 0.35661248693 rad
Angle ∠ B = β = 60.64765299452° = 60°38'48″ = 1.05884816275 rad
Angle ∠ C = γ = 98.94990180653° = 98°56'56″ = 1.72769861569 rad

Height: h_a = 14.81774071806
Height: h_b = 5.92769628722
Height: h_c = 5.23296731226

Median: m_a = 15.7488015748
Median: m_b = 10.3087764064
Median: m_c = 7.63221687612

Inradius: r = 2.34395906075
Circumradius: R = 8.60547443015

Vertex coordinates: A[17; 0] B[0; 0] C[2.94111764706; 5.23296731226]
Centroid: CG[6.64770588235; 1.74332243742]
Coordinates of the circumscribed circle: U[8.5; -1.33985157802]
Coordinates of the inscribed circle: I[4; 2.34395906075]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 159.596554801° = 159°35'44″ = 0.35661248693 rad
∠ B' = β' = 119.3533470055° = 119°21'12″ = 1.05884816275 rad
∠ C' = γ' = 81.05109819347° = 81°3'4″ = 1.72769861569 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.

$a = 6 ; ; b = 15 ; ; c = 17 ; ;$

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

$p = a+b+c = 6+15+17 = 38 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 38 }{ 2 } = 19 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 19 * (19-6)(19-15)(19-17) } ; ; T = sqrt{ 1976 } = 44.45 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 44.45 }{ 6 } = 14.82 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 44.45 }{ 15 } = 5.93 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 44.45 }{ 17 } = 5.23 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 6**2-15**2-17**2 }{ 2 * 15 * 17 } ) = 20° 24'16" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-6**2-17**2 }{ 2 * 6 * 17 } ) = 60° 38'48" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 17**2-6**2-15**2 }{ 2 * 15 * 6 } ) = 98° 56'56" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 44.45 }{ 19 } = 2.34 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 20° 24'16" } = 8.6 ; ;$

Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.