10 24 30 triangle

Obtuse scalene triangle.

Sides: a = 10   b = 24   c = 30

Area: T = 106.1321993291
Perimeter: p = 64
Semiperimeter: s = 32

Angle ∠ A = α = 17.14662099989° = 17°8'46″ = 0.29992578187 rad
Angle ∠ B = β = 45.03656507165° = 45°2'8″ = 0.78660203858 rad
Angle ∠ C = γ = 117.8188139285° = 117°49'5″ = 2.05663144491 rad

Height: ha = 21.22663986583
Height: hb = 8.84443327743
Height: hc = 7.07554662194

Median: ma = 26.70220598456
Median: mb = 18.86879622641
Median: mc = 10.63301458127

Inradius: r = 3.31766247904
Circumradius: R = 16.96600131325

Vertex coordinates: A[30; 0] B[0; 0] C[7.06766666667; 7.07554662194]
Centroid: CG[12.35655555556; 2.35884887398]
Coordinates of the circumscribed circle: U[15; -7.91546727952]
Coordinates of the inscribed circle: I[8; 3.31766247904]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 162.8543790001° = 162°51'14″ = 0.29992578187 rad
∠ B' = β' = 134.9644349284° = 134°57'52″ = 0.78660203858 rad
∠ C' = γ' = 62.18218607153° = 62°10'55″ = 2.05663144491 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 = 24 ; ; c = 30 ; ;

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

p = a+b+c = 10+24+30 = 64 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 64 }{ 2 } = 32 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 32 * (32-10)(32-24)(32-30) } ; ; T = sqrt{ 11264 } = 106.13 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 106.13 }{ 10 } = 21.23 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 106.13 }{ 24 } = 8.84 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 106.13 }{ 30 } = 7.08 ; ;

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**2-24**2-30**2 }{ 2 * 24 * 30 } ) = 17° 8'46" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-10**2-30**2 }{ 2 * 10 * 30 } ) = 45° 2'8" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-10**2-24**2 }{ 2 * 24 * 10 } ) = 117° 49'5" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 106.13 }{ 32 } = 3.32 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 17° 8'46" } = 16.96 ; ;




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

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