Triangle calculator VC

Please enter the coordinates of the three vertices


Acute isosceles triangle.

Sides: a = 57.91992201087   b = 22.186   c = 57.91992201087

Area: T = 630.6043771
Perimeter: p = 138.0244440217
Semiperimeter: s = 69.01222201087

Angle ∠ A = α = 78.95881836183° = 78°57'29″ = 1.37880802755 rad
Angle ∠ B = β = 22.08436327634° = 22°5'1″ = 0.38554321025 rad
Angle ∠ C = γ = 78.95881836183° = 78°57'29″ = 1.37880802755 rad

Height: ha = 21.77552852962
Height: hb = 56.847
Height: hc = 21.77552852962

Median: ma = 32.93658211147
Median: mb = 56.847
Median: mc = 32.93658211147

Inradius: r = 9.13875667962
Circumradius: R = 29.50658319524

Vertex coordinates: A[-11.093; 56.847] B[0; 0] C[11.093; 56.847]
Centroid: CG[0; 37.898]
Coordinates of the circumscribed circle: U[0; 0]
Coordinates of the inscribed circle: I[22.52215675941; 9.13875667962]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 101.0421816382° = 101°2'31″ = 1.37880802755 rad
∠ B' = β' = 157.9166367237° = 157°54'59″ = 0.38554321025 rad
∠ C' = γ' = 101.0421816382° = 101°2'31″ = 1.37880802755 rad

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How did we calculate this triangle?

1. We compute side a from coordinates using the Pythagorean theorem

a = | beta gamma | = | beta - gamma | ; ; a**2 = ( beta _x- gamma _x)**2 + ( beta _y- gamma _y)**2 ; ; a = sqrt{ ( beta _x- gamma _x)**2 + ( beta _y- gamma _y)**2 } ; ; a = sqrt{ (0-11.093)**2 + (0-56.847)**2 } ; ; a = sqrt{ 3354.636 } = 57.92 ; ;

2. We compute side b from coordinates using the Pythagorean theorem

b = | alpha gamma | = | alpha - gamma | ; ; b**2 = ( alpha _x- gamma _x)**2 + ( alpha _y- gamma _y)**2 ; ; b = sqrt{ ( alpha _x- gamma _x)**2 + ( alpha _y- gamma _y)**2 } ; ; b = sqrt{ (-11.093-11.093)**2 + (56.847-56.847)**2 } ; ; b = sqrt{ 492.219 } = 22.19 ; ;

3. We compute side c from coordinates using the Pythagorean theorem

c = | alpha beta | = | alpha - beta | ; ; c**2 = ( alpha _x- beta _x)**2 + ( alpha _y- beta _y)**2 ; ; c = sqrt{ ( alpha _x- beta _x)**2 + ( alpha _y- beta _y)**2 } ; ; c = sqrt{ (-11.093-0)**2 + (56.847-0)**2 } ; ; c = sqrt{ 3354.636 } = 57.92 ; ;


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 = 57.92 ; ; b = 22.19 ; ; c = 57.92 ; ;

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

p = a+b+c = 57.92+22.19+57.92 = 138.02 ; ;

5. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 138.02 }{ 2 } = 69.01 ; ;

6. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 69.01 * (69.01-57.92)(69.01-22.19)(69.01-57.92) } ; ; T = sqrt{ 397661.12 } = 630.6 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 630.6 }{ 57.92 } = 21.78 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 630.6 }{ 22.19 } = 56.85 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 630.6 }{ 57.92 } = 21.78 ; ;

8. 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{ 57.92**2-22.19**2-57.92**2 }{ 2 * 22.19 * 57.92 } ) = 78° 57'29" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22.19**2-57.92**2-57.92**2 }{ 2 * 57.92 * 57.92 } ) = 22° 5'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 57.92**2-57.92**2-22.19**2 }{ 2 * 22.19 * 57.92 } ) = 78° 57'29" ; ;

9. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 630.6 }{ 69.01 } = 9.14 ; ;

10. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 57.92 }{ 2 * sin 78° 57'29" } = 29.51 ; ;




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