17 17 29 triangle

Obtuse isosceles triangle.

Sides: a = 17   b = 17   c = 29

Area: T = 128.6754735282
Perimeter: p = 63
Semiperimeter: s = 31.5

Angle ∠ A = α = 31.46769762933° = 31°28'1″ = 0.5499202342 rad
Angle ∠ B = β = 31.46769762933° = 31°28'1″ = 0.5499202342 rad
Angle ∠ C = γ = 117.0666047413° = 117°3'58″ = 2.04331879697 rad

Height: ha = 15.13882041509
Height: hb = 15.13882041509
Height: hc = 8.87441196746

Median: ma = 22.19879728804
Median: mb = 22.19879728804
Median: mc = 8.87441196746

Inradius: r = 4.08549122312
Circumradius: R = 16.28333053078

Vertex coordinates: A[29; 0] B[0; 0] C[14.5; 8.87441196746]
Centroid: CG[14.5; 2.95880398915]
Coordinates of the circumscribed circle: U[14.5; -7.40991856331]
Coordinates of the inscribed circle: I[14.5; 4.08549122312]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 148.5333023707° = 148°31'59″ = 0.5499202342 rad
∠ B' = β' = 148.5333023707° = 148°31'59″ = 0.5499202342 rad
∠ C' = γ' = 62.93439525865° = 62°56'2″ = 2.04331879697 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 = 17 ; ; c = 29 ; ;

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

p = a+b+c = 17+17+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 * (31.5-17)(31.5-17)(31.5-29) } ; ; T = sqrt{ 16557.19 } = 128.67 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 128.67 }{ 17 } = 15.14 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 128.67 }{ 17 } = 15.14 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 128.67 }{ 29 } = 8.87 ; ;

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**2-17**2-29**2 }{ 2 * 17 * 29 } ) = 31° 28'1" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 17**2-17**2-29**2 }{ 2 * 17 * 29 } ) = 31° 28'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-17**2-17**2 }{ 2 * 17 * 17 } ) = 117° 3'58" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 128.67 }{ 31.5 } = 4.08 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 17 }{ 2 * sin 31° 28'1" } = 16.28 ; ;




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

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