15 23 28 triangle

Obtuse scalene triangle.

Sides: a = 15 b = 23 c = 28

Area: T = 172.3376879396
Perimeter: p = 66
Semiperimeter: s = 33

Angle ∠ A = α = 32.35879982101° = 32°21'29″ = 0.56547536081 rad
Angle ∠ B = β = 55.1550095421° = 55°9' = 0.96325507479 rad
Angle ∠ C = γ = 92.4921906369° = 92°29'31″ = 1.61442882976 rad

Height: h_a = 22.97882505862
Height: h_b = 14.98658155997
Height: h_c = 12.31097770997

Median: m_a = 24.5
Median: m_b = 19.29437813816
Median: m_c = 13.45436240471

Inradius: r = 5.22223296787
Circumradius: R = 14.01332513044

Vertex coordinates: A[28; 0] B[0; 0] C[8.57114285714; 12.31097770997]
Centroid: CG[12.19904761905; 4.10332590332]
Coordinates of the circumscribed circle: U[14; -0.60992717958]
Coordinates of the inscribed circle: I[10; 5.22223296787]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 147.642200179° = 147°38'31″ = 0.56547536081 rad
∠ B' = β' = 124.8549904579° = 124°51' = 0.96325507479 rad
∠ C' = γ' = 87.5088093631° = 87°30'29″ = 1.61442882976 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.

$a = 15 ; ; b = 23 ; ; c = 28 ; ;$

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

$p = a+b+c = 15+23+28 = 66 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 66 }{ 2 } = 33 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33 * (33-15)(33-23)(33-28) } ; ; T = sqrt{ 29700 } = 172.34 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 172.34 }{ 15 } = 22.98 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 172.34 }{ 23 } = 14.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 172.34 }{ 28 } = 12.31 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 15**2-23**2-28**2 }{ 2 * 23 * 28 } ) = 32° 21'29" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-15**2-28**2 }{ 2 * 15 * 28 } ) = 55° 9' ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-15**2-23**2 }{ 2 * 23 * 15 } ) = 92° 29'31" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 172.34 }{ 33 } = 5.22 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 32° 21'29" } = 14.01 ; ;$

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

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