15 24 30 triangle

Obtuse scalene triangle.

Sides: a = 15   b = 24   c = 30

Area: T = 178.2990318021
Perimeter: p = 69
Semiperimeter: s = 34.5

Angle ∠ A = α = 29.68662952314° = 29°41'11″ = 0.51881235945 rad
Angle ∠ B = β = 52.41104970351° = 52°24'38″ = 0.91547357359 rad
Angle ∠ C = γ = 97.90332077335° = 97°54'12″ = 1.70987333232 rad

Height: ha = 23.77220424028
Height: hb = 14.85875265017
Height: hc = 11.88660212014

Median: ma = 26.11103427783
Median: mb = 20.45772725455
Median: mc = 13.24876412995

Inradius: r = 5.1687835305
Circumradius: R = 15.14438397215

Vertex coordinates: A[30; 0] B[0; 0] C[9.15; 11.88660212014]
Centroid: CG[13.05; 3.96220070671]
Coordinates of the circumscribed circle: U[15; -2.08222779617]
Coordinates of the inscribed circle: I[10.5; 5.1687835305]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 150.3143704769° = 150°18'49″ = 0.51881235945 rad
∠ B' = β' = 127.5989502965° = 127°35'22″ = 0.91547357359 rad
∠ C' = γ' = 82.09767922665° = 82°5'48″ = 1.70987333232 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 = 15 ; ; b = 24 ; ; c = 30 ; ;

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

p = a+b+c = 15+24+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 * (34.5-15)(34.5-24)(34.5-30) } ; ; T = sqrt{ 31787.44 } = 178.29 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 178.29 }{ 15 } = 23.77 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 178.29 }{ 24 } = 14.86 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 178.29 }{ 30 } = 11.89 ; ;

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{ 15**2-24**2-30**2 }{ 2 * 24 * 30 } ) = 29° 41'11" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-15**2-30**2 }{ 2 * 15 * 30 } ) = 52° 24'38" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-15**2-24**2 }{ 2 * 24 * 15 } ) = 97° 54'12" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 178.29 }{ 34.5 } = 5.17 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 29° 41'11" } = 15.14 ; ;




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

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