16 26 29 triangle

Acute scalene triangle.

Sides: a = 16   b = 26   c = 29

Area: T = 206.7522116071
Perimeter: p = 71
Semiperimeter: s = 35.5

Angle ∠ A = α = 33.25882815588° = 33°15'30″ = 0.58804665168 rad
Angle ∠ B = β = 63.02110067637° = 63°1'16″ = 1.10999240659 rad
Angle ∠ C = γ = 83.72107116774° = 83°43'15″ = 1.46112020709 rad

Height: ha = 25.84440145089
Height: hb = 15.90440089286
Height: hc = 14.25987666256

Median: ma = 26.35333679062
Median: mb = 19.48107597388
Median: mc = 15.99221855917

Inradius: r = 5.82440032696
Circumradius: R = 14.58875169614

Vertex coordinates: A[29; 0] B[0; 0] C[7.25986206897; 14.25987666256]
Centroid: CG[12.08662068966; 4.75329222085]
Coordinates of the circumscribed circle: U[14.5; 1.59655096677]
Coordinates of the inscribed circle: I[9.5; 5.82440032696]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 146.7421718441° = 146°44'30″ = 0.58804665168 rad
∠ B' = β' = 116.9798993236° = 116°58'44″ = 1.10999240659 rad
∠ C' = γ' = 96.27992883226° = 96°16'45″ = 1.46112020709 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 = 26 ; ; c = 29 ; ;

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

p = a+b+c = 16+26+29 = 71 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 71 }{ 2 } = 35.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 35.5 * (35.5-16)(35.5-26)(35.5-29) } ; ; T = sqrt{ 42746.44 } = 206.75 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 206.75 }{ 16 } = 25.84 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 206.75 }{ 26 } = 15.9 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 206.75 }{ 29 } = 14.26 ; ;

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**2-26**2-29**2 }{ 2 * 26 * 29 } ) = 33° 15'30" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 26**2-16**2-29**2 }{ 2 * 16 * 29 } ) = 63° 1'16" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-16**2-26**2 }{ 2 * 26 * 16 } ) = 83° 43'15" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 206.75 }{ 35.5 } = 5.82 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 33° 15'30" } = 14.59 ; ;




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

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