3 25 27 triangle

Obtuse scalene triangle.

Sides: a = 3   b = 25   c = 27

Area: T = 29.02204669156
Perimeter: p = 55
Semiperimeter: s = 27.5

Angle ∠ A = α = 4.93327588085° = 4°55'58″ = 0.08660928824 rad
Angle ∠ B = β = 45.77107609523° = 45°46'15″ = 0.79988504798 rad
Angle ∠ C = γ = 129.2966480239° = 129°17'47″ = 2.25766492914 rad

Height: ha = 19.34769779437
Height: hb = 2.32216373532
Height: hc = 2.1549664216

Median: ma = 25.97659504157
Median: mb = 14.58659521458
Median: mc = 11.60881867662

Inradius: r = 1.0555289706
Circumradius: R = 17.44545849363

Vertex coordinates: A[27; 0] B[0; 0] C[2.09325925926; 2.1549664216]
Centroid: CG[9.69875308642; 0.71765547387]
Coordinates of the circumscribed circle: U[13.5; -11.04882371263]
Coordinates of the inscribed circle: I[2.5; 1.0555289706]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 175.0677241192° = 175°4'2″ = 0.08660928824 rad
∠ B' = β' = 134.2299239048° = 134°13'45″ = 0.79988504798 rad
∠ C' = γ' = 50.70435197608° = 50°42'13″ = 2.25766492914 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 = 3 ; ; b = 25 ; ; c = 27 ; ;

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

p = a+b+c = 3+25+27 = 55 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 55 }{ 2 } = 27.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 27.5 * (27.5-3)(27.5-25)(27.5-27) } ; ; T = sqrt{ 842.19 } = 29.02 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 29.02 }{ 3 } = 19.35 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 29.02 }{ 25 } = 2.32 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 29.02 }{ 27 } = 2.15 ; ;

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{ 3**2-25**2-27**2 }{ 2 * 25 * 27 } ) = 4° 55'58" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-3**2-27**2 }{ 2 * 3 * 27 } ) = 45° 46'15" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-3**2-25**2 }{ 2 * 25 * 3 } ) = 129° 17'47" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 29.02 }{ 27.5 } = 1.06 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 3 }{ 2 * sin 4° 55'58" } = 17.44 ; ;




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

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