Triangle calculator

Please enter what you know about the triangle:
Symbols definition of ABC triangle

You have entered side a, b and angle β.

Triangle has two solutions: a=800; b=250; c=560.0187787443 and a=800; b=250; c=1031.217724515.

#1 Obtuse scalene triangle.

Sides: a = 800   b = 250   c = 560.0187787443

Area: T = 23415.11994896
Perimeter: p = 1610.018778744
Semiperimeter: s = 805.0098893722

Angle ∠ A = α = 160.4588406089° = 160°27'30″ = 2.80105274988 rad
Angle ∠ B = β = 6° = 0.10547197551 rad
Angle ∠ C = γ = 13.54215939107° = 13°32'30″ = 0.23663453997 rad

Height: ha = 58.5387798724
Height: hb = 187.3210955917
Height: hc = 83.62327706141

Median: ma = 167.5111077623
Median: mb = 679.1066001392
Median: mc = 522.3465689593

Inradius: r = 29.08767836023
Circumradius: R = 1195.847652919

Vertex coordinates: A[560.0187787443; 0] B[0; 0] C[795.6187516295; 83.62327706141]
Centroid: CG[451.8788434579; 27.87442568714]
Coordinates of the circumscribed circle: U[280.0098893722; 1162.602222811]
Coordinates of the inscribed circle: I[555.0098893722; 29.08767836023]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 19.54215939107° = 19°32'30″ = 2.80105274988 rad
∠ B' = β' = 174° = 0.10547197551 rad
∠ C' = γ' = 166.4588406089° = 166°27'30″ = 0.23663453997 rad




How did we calculate this triangle?

1. Input data entered: side a, b and angle β.

a = 800 ; ; b = 250 ; ; beta = 6° ; ;

2. From angle β, side a and side b we calculate side c - by using the law of cosines and quadratic equation:

b**2 = a**2 + c**2 - 2a c cos beta ; ; ; ; 250**2 = 800**2 + c**2 - 2 * 800 * c * cos 6° ; ; ; ; ; ; c**2 -1591.235c +577500 =0 ; ; p=1; q=-1591.235; r=577500 ; ; D = q**2 - 4pr = 1591.235**2 - 4 * 1 * 577500 = 222028.928939 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 1591.24 ± sqrt{ 222028.93 } }{ 2 } ; ; c_{1,2} = 795.61751629 ± 235.599728851 ; ; c_{1} = 1031.21724514 ; ; c_{2} = 560.017787439 ; ; ; ; text{ Factored form: } ; ; (c -1031.21724514) (c -560.017787439) = 0 ; ; ; ; c > 0 ; ; ; ; c = 1031.217 ; ;


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 = 800 ; ; b = 250 ; ; c = 560.02 ; ;

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

p = a+b+c = 800+250+560.02 = 1610.02 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1610.02 }{ 2 } = 805.01 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 805.01 * (805.01-800)(805.01-250)(805.01-560.02) } ; ; T = sqrt{ 548267820.71 } = 23415.12 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 23415.12 }{ 800 } = 58.54 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 23415.12 }{ 250 } = 187.32 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 23415.12 }{ 560.02 } = 83.62 ; ;

7. 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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 250**2+560.02**2-800**2 }{ 2 * 250 * 560.02 } ) = 160° 27'30" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 800**2+560.02**2-250**2 }{ 2 * 800 * 560.02 } ) = 6° ; ; gamma = 180° - alpha - beta = 180° - 160° 27'30" - 6° = 13° 32'30" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 23415.12 }{ 805.01 } = 29.09 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 800 }{ 2 * sin 160° 27'30" } = 1195.85 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 250**2+2 * 560.02**2 - 800**2 } }{ 2 } = 167.511 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 560.02**2+2 * 800**2 - 250**2 } }{ 2 } = 679.106 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 250**2+2 * 800**2 - 560.02**2 } }{ 2 } = 522.346 ; ;







#2 Obtuse scalene triangle.

Sides: a = 800   b = 250   c = 1031.217724515

Area: T = 43116.62215721
Perimeter: p = 2081.217724515
Semiperimeter: s = 1040.609862257

Angle ∠ A = α = 19.54215939107° = 19°32'30″ = 0.34110651548 rad
Angle ∠ B = β = 6° = 0.10547197551 rad
Angle ∠ C = γ = 154.4588406089° = 154°27'30″ = 2.69658077436 rad

Height: ha = 107.792155393
Height: hb = 344.9332972577
Height: hc = 83.62327706141

Median: ma = 634.7876974774
Median: mb = 914.3743831287
Median: mc = 292.2298931368

Inradius: r = 41.43440421911
Circumradius: R = 1195.847652919

Vertex coordinates: A[1031.217724515; 0] B[0; 0] C[795.6187516295; 83.62327706141]
Centroid: CG[608.945492048; 27.87442568714]
Coordinates of the circumscribed circle: U[515.6098622573; -1078.97994575]
Coordinates of the inscribed circle: I[790.6098622573; 41.43440421911]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 160.4588406089° = 160°27'30″ = 0.34110651548 rad
∠ B' = β' = 174° = 0.10547197551 rad
∠ C' = γ' = 25.54215939107° = 25°32'30″ = 2.69658077436 rad

Calculate another triangle

How did we calculate this triangle?

1. Input data entered: side a, b and angle β.

a = 800 ; ; b = 250 ; ; beta = 6° ; ; : Nr. 1

2. From angle β, side a and side b we calculate side c - by using the law of cosines and quadratic equation:

b**2 = a**2 + c**2 - 2a c cos beta ; ; ; ; 250**2 = 800**2 + c**2 - 2 * 800 * c * cos 6° ; ; ; ; ; ; c**2 -1591.235c +577500 =0 ; ; p=1; q=-1591.235; r=577500 ; ; D = q**2 - 4pr = 1591.235**2 - 4 * 1 * 577500 = 222028.928939 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 1591.24 ± sqrt{ 222028.93 } }{ 2 } ; ; c_{1,2} = 795.61751629 ± 235.599728851 ; ; c_{1} = 1031.21724514 ; ; c_{2} = 560.017787439 ; ; ; ; text{ Factored form: } ; ; (c -1031.21724514) (c -560.017787439) = 0 ; ; ; ; c > 0 ; ; ; ; c = 1031.217 ; ; : Nr. 1


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 = 800 ; ; b = 250 ; ; c = 1031.22 ; ;

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

p = a+b+c = 800+250+1031.22 = 2081.22 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 2081.22 }{ 2 } = 1040.61 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 1040.61 * (1040.61-800)(1040.61-250)(1040.61-1031.22) } ; ; T = sqrt{ 1859043055.79 } = 43116.62 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 43116.62 }{ 800 } = 107.79 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 43116.62 }{ 250 } = 344.93 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 43116.62 }{ 1031.22 } = 83.62 ; ;

7. 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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 250**2+1031.22**2-800**2 }{ 2 * 250 * 1031.22 } ) = 19° 32'30" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 800**2+1031.22**2-250**2 }{ 2 * 800 * 1031.22 } ) = 6° ; ; gamma = 180° - alpha - beta = 180° - 19° 32'30" - 6° = 154° 27'30" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 43116.62 }{ 1040.61 } = 41.43 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 800 }{ 2 * sin 19° 32'30" } = 1195.85 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 250**2+2 * 1031.22**2 - 800**2 } }{ 2 } = 634.787 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 1031.22**2+2 * 800**2 - 250**2 } }{ 2 } = 914.374 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 250**2+2 * 800**2 - 1031.22**2 } }{ 2 } = 292.229 ; ;
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