16 29 29 triangle

Acute isosceles triangle.

Sides: a = 16   b = 29   c = 29

Area: T = 222.9987757836
Perimeter: p = 74
Semiperimeter: s = 37

Angle ∠ A = α = 32.02767888479° = 32°1'36″ = 0.55989729142 rad
Angle ∠ B = β = 73.98766055761° = 73°59'12″ = 1.29113098697 rad
Angle ∠ C = γ = 73.98766055761° = 73°59'12″ = 1.29113098697 rad

Height: ha = 27.87547197295
Height: hb = 15.37991557128
Height: hc = 15.37991557128

Median: ma = 27.87547197295
Median: mb = 18.39215741577
Median: mc = 18.39215741577

Inradius: r = 6.0276966428
Circumradius: R = 15.08553534701

Vertex coordinates: A[29; 0] B[0; 0] C[4.41437931034; 15.37991557128]
Centroid: CG[11.13879310345; 5.12663852376]
Coordinates of the circumscribed circle: U[14.5; 4.16114768193]
Coordinates of the inscribed circle: I[8; 6.0276966428]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 147.9733211152° = 147°58'24″ = 0.55989729142 rad
∠ B' = β' = 106.0133394424° = 106°48″ = 1.29113098697 rad
∠ C' = γ' = 106.0133394424° = 106°48″ = 1.29113098697 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 = 16 ; ; b = 29 ; ; c = 29 ; ;

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

p = a+b+c = 16+29+29 = 74 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 74 }{ 2 } = 37 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 37 * (37-16)(37-29)(37-29) } ; ; T = sqrt{ 49728 } = 223 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 223 }{ 16 } = 27.87 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 223 }{ 29 } = 15.38 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 223 }{ 29 } = 15.38 ; ;

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{ 16**2-29**2-29**2 }{ 2 * 29 * 29 } ) = 32° 1'36" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-16**2-29**2 }{ 2 * 16 * 29 } ) = 73° 59'12" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-16**2-29**2 }{ 2 * 29 * 16 } ) = 73° 59'12" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 223 }{ 37 } = 6.03 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 32° 1'36" } = 15.09 ; ;




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

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