15 23 24 triangle

Acute scalene triangle.

Sides: a = 15 b = 23 c = 24

Area: T = 166.6611333248
Perimeter: p = 62
Semiperimeter: s = 31

Angle ∠ A = α = 37.14658041715° = 37°8'45″ = 0.64883165861 rad
Angle ∠ B = β = 67.80438991432° = 67°48'14″ = 1.18334012857 rad
Angle ∠ C = γ = 75.05502966853° = 75°3'1″ = 1.31098747818 rad

Height: h_a = 22.22215110997
Height: h_b = 14.49222898477
Height: h_c = 13.88884444373

Median: m_a = 22.27766694099
Median: m_b = 16.37883393542
Median: m_c = 15.26443375225

Inradius: r = 5.37661720403
Circumradius: R = 12.42203974591

Vertex coordinates: A[24; 0] B[0; 0] C[5.66766666667; 13.88884444373]
Centroid: CG[9.88988888889; 4.62994814791]
Coordinates of the circumscribed circle: U[12; 3.20441025329]
Coordinates of the inscribed circle: I[8; 5.37661720403]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 142.8544195828° = 142°51'15″ = 0.64883165861 rad
∠ B' = β' = 112.1966100857° = 112°11'46″ = 1.18334012857 rad
∠ C' = γ' = 104.9549703315° = 104°56'59″ = 1.31098747818 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.

$a = 15 ; ; b = 23 ; ; c = 24 ; ;$

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

$p = a+b+c = 15+23+24 = 62 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 62 }{ 2 } = 31 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31 * (31-15)(31-23)(31-24) } ; ; T = sqrt{ 27776 } = 166.66 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 166.66 }{ 15 } = 22.22 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 166.66 }{ 23 } = 14.49 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 166.66 }{ 24 } = 13.89 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 15**2-23**2-24**2 }{ 2 * 23 * 24 } ) = 37° 8'45" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-15**2-24**2 }{ 2 * 15 * 24 } ) = 67° 48'14" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 24**2-15**2-23**2 }{ 2 * 23 * 15 } ) = 75° 3'1" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 166.66 }{ 31 } = 5.38 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 37° 8'45" } = 12.42 ; ;$

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