10 29 30 triangle

Acute scalene triangle.

Sides: a = 10   b = 29   c = 30

Area: T = 144.6377261797
Perimeter: p = 69
Semiperimeter: s = 34.5

Angle ∠ A = α = 19.42105524779° = 19°25'14″ = 0.33989525833 rad
Angle ∠ B = β = 74.63330472022° = 74°37'59″ = 1.30325924045 rad
Angle ∠ C = γ = 85.94664003198° = 85°56'47″ = 1.55000476658 rad

Height: ha = 28.92774523593
Height: hb = 9.97549835722
Height: hc = 9.64224841198

Median: ma = 29.07774826971
Median: mb = 17.02220445305
Median: mc = 15.6688439616

Inradius: r = 4.19223843999
Circumradius: R = 15.03876187504

Vertex coordinates: A[30; 0] B[0; 0] C[2.65; 9.64224841198]
Centroid: CG[10.88333333333; 3.21441613733]
Coordinates of the circumscribed circle: U[15; 1.06330040841]
Coordinates of the inscribed circle: I[5.5; 4.19223843999]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 160.5799447522° = 160°34'46″ = 0.33989525833 rad
∠ B' = β' = 105.3676952798° = 105°22'1″ = 1.30325924045 rad
∠ C' = γ' = 94.05435996802° = 94°3'13″ = 1.55000476658 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 = 10 ; ; b = 29 ; ; c = 30 ; ;

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

p = a+b+c = 10+29+30 = 69 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 69 }{ 2 } = 34.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 34.5 * (34.5-10)(34.5-29)(34.5-30) } ; ; T = sqrt{ 20919.94 } = 144.64 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 144.64 }{ 10 } = 28.93 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 144.64 }{ 29 } = 9.97 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 144.64 }{ 30 } = 9.64 ; ;

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{ 10**2-29**2-30**2 }{ 2 * 29 * 30 } ) = 19° 25'14" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-10**2-30**2 }{ 2 * 10 * 30 } ) = 74° 37'59" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-10**2-29**2 }{ 2 * 29 * 10 } ) = 85° 56'47" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 144.64 }{ 34.5 } = 4.19 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 19° 25'14" } = 15.04 ; ;




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

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