4 27 28 triangle

Obtuse scalene triangle.

Sides: a = 4   b = 27   c = 28

Area: T = 53.11224985291
Perimeter: p = 59
Semiperimeter: s = 29.5

Angle ∠ A = α = 8.07773157229° = 8°4'38″ = 0.14109757541 rad
Angle ∠ B = β = 71.52105632681° = 71°31'14″ = 1.24882693119 rad
Angle ∠ C = γ = 100.4022121009° = 100°24'8″ = 1.75223475876 rad

Height: ha = 26.55662492645
Height: hb = 3.93442591503
Height: hc = 3.79437498949

Median: ma = 27.43217334487
Median: mb = 14.75663545634
Median: mc = 13.28553302556

Inradius: r = 1.8800423679
Circumradius: R = 14.23439377912

Vertex coordinates: A[28; 0] B[0; 0] C[1.26878571429; 3.79437498949]
Centroid: CG[9.7565952381; 1.26545832983]
Coordinates of the circumscribed circle: U[14; -2.57700165456]
Coordinates of the inscribed circle: I[2.5; 1.8800423679]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 171.9232684277° = 171°55'22″ = 0.14109757541 rad
∠ B' = β' = 108.4799436732° = 108°28'46″ = 1.24882693119 rad
∠ C' = γ' = 79.5987878991° = 79°35'52″ = 1.75223475876 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 = 4 ; ; b = 27 ; ; c = 28 ; ;

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

p = a+b+c = 4+27+28 = 59 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 59 }{ 2 } = 29.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29.5 * (29.5-4)(29.5-27)(29.5-28) } ; ; T = sqrt{ 2820.94 } = 53.11 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 53.11 }{ 4 } = 26.56 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 53.11 }{ 27 } = 3.93 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 53.11 }{ 28 } = 3.79 ; ;

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{ 4**2-27**2-28**2 }{ 2 * 27 * 28 } ) = 8° 4'38" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-4**2-28**2 }{ 2 * 4 * 28 } ) = 71° 31'14" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-4**2-27**2 }{ 2 * 27 * 4 } ) = 100° 24'8" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 53.11 }{ 29.5 } = 1.8 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 4 }{ 2 * sin 8° 4'38" } = 14.23 ; ;




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

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