Triangle calculator VC

Please enter the coordinates of the three vertices


Acute scalene triangle.

Sides: a = 4.12331056256   b = 5.38551648071   c = 6.32545553203

Area: T = 11
Perimeter: p = 15.83328257531
Semiperimeter: s = 7.91664128765

Angle ∠ A = α = 40.23663583093° = 40°14'11″ = 0.70222569315 rad
Angle ∠ B = β = 57.52988077092° = 57°31'44″ = 1.00440671093 rad
Angle ∠ C = γ = 82.23548339816° = 82°14'5″ = 1.43552686128 rad

Height: ha = 5.33657837508
Height: hb = 4.08552974399
Height: hc = 3.47985054262

Median: ma = 5.5
Median: mb = 4.61097722286
Median: mc = 3.60655512755

Inradius: r = 1.3989518229
Circumradius: R = 3.19215435785

Vertex coordinates: A[5; -1] B[-1; 1] C[0; -3]
Centroid: CG[1.33333333333; -1]
Coordinates of the circumscribed circle: U[0; 0]
Coordinates of the inscribed circle: I[0.8844238873; 1.3989518229]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 139.7643641691° = 139°45'49″ = 0.70222569315 rad
∠ B' = β' = 122.4711192291° = 122°28'16″ = 1.00440671093 rad
∠ C' = γ' = 97.76551660184° = 97°45'55″ = 1.43552686128 rad

Calculate another triangle




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{ (-1-0)**2 + (1-(-3))**2 } ; ; a = sqrt{ 17 } = 4.12 ; ;

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{ (5-0)**2 + (-1-(-3))**2 } ; ; b = sqrt{ 29 } = 5.39 ; ;

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{ (5-(-1))**2 + (-1-1)**2 } ; ; c = sqrt{ 40 } = 6.32 ; ;


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 = 4.12 ; ; b = 5.39 ; ; c = 6.32 ; ;

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

p = a+b+c = 4.12+5.39+6.32 = 15.83 ; ;

5. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 15.83 }{ 2 } = 7.92 ; ;

6. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 7.92 * (7.92-4.12)(7.92-5.39)(7.92-6.32) } ; ; T = sqrt{ 121 } = 11 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 11 }{ 4.12 } = 5.34 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 11 }{ 5.39 } = 4.09 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 11 }{ 6.32 } = 3.48 ; ;

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{ 4.12**2-5.39**2-6.32**2 }{ 2 * 5.39 * 6.32 } ) = 40° 14'11" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 5.39**2-4.12**2-6.32**2 }{ 2 * 4.12 * 6.32 } ) = 57° 31'44" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 6.32**2-4.12**2-5.39**2 }{ 2 * 5.39 * 4.12 } ) = 82° 14'5" ; ;

9. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 11 }{ 7.92 } = 1.39 ; ;

10. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 4.12 }{ 2 * sin 40° 14'11" } = 3.19 ; ;




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

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