15 23 25 triangle

Acute scalene triangle.

Sides: a = 15 b = 23 c = 25

Area: T = 169.4598512622
Perimeter: p = 63
Semiperimeter: s = 31.5

Angle ∠ A = α = 36.11659252263° = 36°6'57″ = 0.63303418076 rad
Angle ∠ B = β = 64.65992980761° = 64°39'33″ = 1.12985176435 rad
Angle ∠ C = γ = 79.22547766976° = 79°13'29″ = 1.38327332025 rad

Height: h_a = 22.59444683496
Height: h_b = 14.73655228367
Height: h_c = 13.55766810097

Median: m_a = 22.82199474145
Median: m_b = 17.11099386323
Median: m_c = 14.85876579581

Inradius: r = 5.38796353213
Circumradius: R = 12.72443533927

Vertex coordinates: A[25; 0] B[0; 0] C[6.42; 13.55766810097]
Centroid: CG[10.47333333333; 4.51988936699]
Coordinates of the circumscribed circle: U[12.5; 2.37989008517]
Coordinates of the inscribed circle: I[8.5; 5.38796353213]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 143.8844074774° = 143°53'3″ = 0.63303418076 rad
∠ B' = β' = 115.3410701924° = 115°20'27″ = 1.12985176435 rad
∠ C' = γ' = 100.7755223302° = 100°46'31″ = 1.38327332025 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.

$a = 15 ; ; b = 23 ; ; c = 25 ; ;$

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

$p = a+b+c = 15+23+25 = 63 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 63 }{ 2 } = 31.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31.5 * (31.5-15)(31.5-23)(31.5-25) } ; ; T = sqrt{ 28716.19 } = 169.46 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 169.46 }{ 15 } = 22.59 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 169.46 }{ 23 } = 14.74 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 169.46 }{ 25 } = 13.56 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 15**2-23**2-25**2 }{ 2 * 23 * 25 } ) = 36° 6'57" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-15**2-25**2 }{ 2 * 15 * 25 } ) = 64° 39'33" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-15**2-23**2 }{ 2 * 23 * 15 } ) = 79° 13'29" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 169.46 }{ 31.5 } = 5.38 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 36° 6'57" } = 12.72 ; ;$

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

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