15 28 28 triangle

Acute isosceles triangle.

Sides: a = 15   b = 28   c = 28

Area: T = 202.3266314403
Perimeter: p = 71
Semiperimeter: s = 35.5

Angle ∠ A = α = 31.07435881363° = 31°4'25″ = 0.54223364234 rad
Angle ∠ B = β = 74.46332059318° = 74°27'48″ = 1.32996281151 rad
Angle ∠ C = γ = 74.46332059318° = 74°27'48″ = 1.32996281151 rad

Height: ha = 26.97768419204
Height: hb = 14.45218796002
Height: hc = 14.45218796002

Median: ma = 26.97768419204
Median: mb = 17.56441680703
Median: mc = 17.56441680703

Inradius: r = 5.69993328001
Circumradius: R = 14.53109818383

Vertex coordinates: A[28; 0] B[0; 0] C[4.01878571429; 14.45218796002]
Centroid: CG[10.67326190476; 4.81772932001]
Coordinates of the circumscribed circle: U[14; 3.89222272781]
Coordinates of the inscribed circle: I[7.5; 5.69993328001]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 148.9266411864° = 148°55'35″ = 0.54223364234 rad
∠ B' = β' = 105.5376794068° = 105°32'12″ = 1.32996281151 rad
∠ C' = γ' = 105.5376794068° = 105°32'12″ = 1.32996281151 rad

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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 = 28 ; ; c = 28 ; ;

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

p = a+b+c = 15+28+28 = 71 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 71 }{ 2 } = 35.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 35.5 * (35.5-15)(35.5-28)(35.5-28) } ; ; T = sqrt{ 40935.94 } = 202.33 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 202.33 }{ 15 } = 26.98 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 202.33 }{ 28 } = 14.45 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 202.33 }{ 28 } = 14.45 ; ;

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**2-28**2-28**2 }{ 2 * 28 * 28 } ) = 31° 4'25" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 28**2-15**2-28**2 }{ 2 * 15 * 28 } ) = 74° 27'48" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-15**2-28**2 }{ 2 * 28 * 15 } ) = 74° 27'48" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 202.33 }{ 35.5 } = 5.7 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 31° 4'25" } = 14.53 ; ;




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