26 29 30 triangle

Acute scalene triangle.

Sides: a = 26   b = 29   c = 30

Area: T = 3443.999909157
Perimeter: p = 85
Semiperimeter: s = 42.5

Angle ∠ A = α = 52.26107465423° = 52°15'39″ = 0.91221220967 rad
Angle ∠ B = β = 61.89107787174° = 61°53'27″ = 1.08801978652 rad
Angle ∠ C = γ = 65.84884747403° = 65°50'54″ = 1.14992726916 rad

Height: ha = 26.46215314736
Height: hb = 23.7244131666
Height: hc = 22.93333272771

Median: ma = 26.48658452763
Median: mb = 24.03664306834
Median: mc = 23.09876189249

Inradius: r = 8.09441155096
Circumradius: R = 16.43989578295

Vertex coordinates: A[30; 0] B[0; 0] C[12.25; 22.93333272771]
Centroid: CG[14.08333333333; 7.64444424257]
Coordinates of the circumscribed circle: U[15; 6.72660192181]
Coordinates of the inscribed circle: I[13.5; 8.09441155096]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 127.7399253458° = 127°44'21″ = 0.91221220967 rad
∠ B' = β' = 118.1099221283° = 118°6'33″ = 1.08801978652 rad
∠ C' = γ' = 114.152152526° = 114°9'6″ = 1.14992726916 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 = 26 ; ; b = 29 ; ; c = 30 ; ;

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

p = a+b+c = 26+29+30 = 85 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 85 }{ 2 } = 42.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 42.5 * (42.5-26)(42.5-29)(42.5-30) } ; ; T = sqrt{ 118335.94 } = 344 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 344 }{ 26 } = 26.46 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 344 }{ 29 } = 23.72 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 344 }{ 30 } = 22.93 ; ;

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{ 26**2-29**2-30**2 }{ 2 * 29 * 30 } ) = 52° 15'39" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-26**2-30**2 }{ 2 * 26 * 30 } ) = 61° 53'27" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-26**2-29**2 }{ 2 * 29 * 26 } ) = 65° 50'54" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 344 }{ 42.5 } = 8.09 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 26 }{ 2 * sin 52° 15'39" } = 16.44 ; ;




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

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