17 19 27 triangle

Obtuse scalene triangle.

Sides: a = 17   b = 19   c = 27

Area: T = 160.2887827049
Perimeter: p = 63
Semiperimeter: s = 31.5

Angle ∠ A = α = 38.67551276498° = 38°40'30″ = 0.67550083161 rad
Angle ∠ B = β = 44.30105298893° = 44°18'2″ = 0.77331901069 rad
Angle ∠ C = γ = 97.02443424609° = 97°1'28″ = 1.69333942305 rad

Height: ha = 18.85773914175
Height: hb = 16.87224028472
Height: hc = 11.8733172374

Median: ma = 21.74328149052
Median: mb = 20.46333819297
Median: mc = 11.94878031453

Inradius: r = 5.0898502446
Circumradius: R = 13.60220934349

Vertex coordinates: A[27; 0] B[0; 0] C[12.16766666667; 11.8733172374]
Centroid: CG[13.05655555556; 3.95877241247]
Coordinates of the circumscribed circle: U[13.5; -1.6633413903]
Coordinates of the inscribed circle: I[12.5; 5.0898502446]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 141.325487235° = 141°19'30″ = 0.67550083161 rad
∠ B' = β' = 135.6999470111° = 135°41'58″ = 0.77331901069 rad
∠ C' = γ' = 82.97656575391° = 82°58'32″ = 1.69333942305 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 = 17 ; ; b = 19 ; ; c = 27 ; ;

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

p = a+b+c = 17+19+27 = 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 * (31.5-17)(31.5-19)(31.5-27) } ; ; T = sqrt{ 25692.19 } = 160.29 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 160.29 }{ 17 } = 18.86 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 160.29 }{ 19 } = 16.87 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 160.29 }{ 27 } = 11.87 ; ;

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{ 17**2-19**2-27**2 }{ 2 * 19 * 27 } ) = 38° 40'30" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 19**2-17**2-27**2 }{ 2 * 17 * 27 } ) = 44° 18'2" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-17**2-19**2 }{ 2 * 19 * 17 } ) = 97° 1'28" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 160.29 }{ 31.5 } = 5.09 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 17 }{ 2 * sin 38° 40'30" } = 13.6 ; ;




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

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