Triangle calculator

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

You have entered side a, b and angle α.

Triangle has two solutions: a=115; b=250; c=181.166645455 and a=115; b=250; c=271.9877438969.

#1 Obtuse scalene triangle.

Sides: a = 115   b = 250   c = 181.166645455

Area: T = 9570.532151344
Perimeter: p = 546.166645455
Semiperimeter: s = 273.0833227275

Angle ∠ A = α = 25° = 0.4366332313 rad
Angle ∠ B = β = 113.2588106349° = 113°15'29″ = 1.97767268604 rad
Angle ∠ C = γ = 41.74218936505° = 41°44'31″ = 0.72985334802 rad

Height: ha = 166.4444026321
Height: hb = 76.56442521075
Height: hc = 105.6554565435

Median: ma = 210.6054824558
Median: mb = 86.01224533254
Median: mc = 172.2132598077

Inradius: r = 35.04662077402
Circumradius: R = 136.0576591031

Vertex coordinates: A[181.166645455; 0] B[0; 0] C[-45.41104922094; 105.6554565435]
Centroid: CG[45.25219874468; 35.21881884784]
Coordinates of the circumscribed circle: U[90.58332272749; 101.5198840121]
Coordinates of the inscribed circle: I[23.08332272749; 35.04662077402]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155° = 0.4366332313 rad
∠ B' = β' = 66.74218936505° = 66°44'31″ = 1.97767268604 rad
∠ C' = γ' = 138.2588106349° = 138°15'29″ = 0.72985334802 rad




How did we calculate this triangle?

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

a = 115 ; ; b = 250 ; ; alpha = 25° ; ;

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

a**2 = b**2 + c**2 - 2b c cos alpha ; ; ; ; 115**2 = 250**2 + c**2 - 2 * 250 * c * cos 25° ; ; ; ; ; ; c**2 -453.154c +49275 =0 ; ; p=1; q=-453.154; r=49275 ; ; D = q**2 - 4pr = 453.154**2 - 4 * 1 * 49275 = 8248.45121082 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 453.15 ± sqrt{ 8248.45 } }{ 2 } ; ; c_{1,2} = 226.57694676 ± 45.4104922094 ; ; c_{1} = 271.987438969 ; ; c_{2} = 181.166454551 ; ; ; ; text{ Factored form: } ; ; (c -271.987438969) (c -181.166454551) = 0 ; ; ; ; c > 0 ; ; ; ; c = 271.987 ; ;


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 = 115 ; ; b = 250 ; ; c = 181.17 ; ;

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

p = a+b+c = 115+250+181.17 = 546.17 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 546.17 }{ 2 } = 273.08 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 273.08 * (273.08-115)(273.08-250)(273.08-181.17) } ; ; T = sqrt{ 91595073.45 } = 9570.53 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 9570.53 }{ 115 } = 166.44 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 9570.53 }{ 250 } = 76.56 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 9570.53 }{ 181.17 } = 105.65 ; ;

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+181.17**2-115**2 }{ 2 * 250 * 181.17 } ) = 25° ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 115**2+181.17**2-250**2 }{ 2 * 115 * 181.17 } ) = 113° 15'29" ; ; gamma = 180° - alpha - beta = 180° - 25° - 113° 15'29" = 41° 44'31" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 9570.53 }{ 273.08 } = 35.05 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 115 }{ 2 * sin 25° } = 136.06 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 250**2+2 * 181.17**2 - 115**2 } }{ 2 } = 210.605 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 181.17**2+2 * 115**2 - 250**2 } }{ 2 } = 86.012 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 250**2+2 * 115**2 - 181.17**2 } }{ 2 } = 172.213 ; ;







#2 Acute scalene triangle.

Sides: a = 115   b = 250   c = 271.9877438969

Area: T = 14368.3577334
Perimeter: p = 636.9877438969
Semiperimeter: s = 318.4943719484

Angle ∠ A = α = 25° = 0.4366332313 rad
Angle ∠ B = β = 66.74218936505° = 66°44'31″ = 1.16548657932 rad
Angle ∠ C = γ = 88.25881063495° = 88°15'29″ = 1.54403945474 rad

Height: ha = 249.8844475374
Height: hb = 114.9476858672
Height: hc = 105.6554565435

Median: ma = 254.8188236157
Median: mb = 167.2610525763
Median: mc = 139.1769710285

Inradius: r = 45.1133471491
Circumradius: R = 136.0576591031

Vertex coordinates: A[271.9877438969; 0] B[0; 0] C[45.41104922094; 105.6554565435]
Centroid: CG[105.7999310393; 35.21881884784]
Coordinates of the circumscribed circle: U[135.9943719484; 4.13657253141]
Coordinates of the inscribed circle: I[68.49437194843; 45.1133471491]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155° = 0.4366332313 rad
∠ B' = β' = 113.2588106349° = 113°15'29″ = 1.16548657932 rad
∠ C' = γ' = 91.74218936505° = 91°44'31″ = 1.54403945474 rad

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How did we calculate this triangle?

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

a = 115 ; ; b = 250 ; ; alpha = 25° ; ; : Nr. 1

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

a**2 = b**2 + c**2 - 2b c cos alpha ; ; ; ; 115**2 = 250**2 + c**2 - 2 * 250 * c * cos 25° ; ; ; ; ; ; c**2 -453.154c +49275 =0 ; ; p=1; q=-453.154; r=49275 ; ; D = q**2 - 4pr = 453.154**2 - 4 * 1 * 49275 = 8248.45121082 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 453.15 ± sqrt{ 8248.45 } }{ 2 } ; ; c_{1,2} = 226.57694676 ± 45.4104922094 ; ; c_{1} = 271.987438969 ; ; c_{2} = 181.166454551 ; ; ; ; text{ Factored form: } ; ; (c -271.987438969) (c -181.166454551) = 0 ; ; ; ; c > 0 ; ; ; ; c = 271.987 ; ; : 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 = 115 ; ; b = 250 ; ; c = 271.99 ; ;

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

p = a+b+c = 115+250+271.99 = 636.99 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 636.99 }{ 2 } = 318.49 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 318.49 * (318.49-115)(318.49-250)(318.49-271.99) } ; ; T = sqrt{ 206449692.48 } = 14368.36 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 14368.36 }{ 115 } = 249.88 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 14368.36 }{ 250 } = 114.95 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 14368.36 }{ 271.99 } = 105.65 ; ;

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+271.99**2-115**2 }{ 2 * 250 * 271.99 } ) = 25° ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 115**2+271.99**2-250**2 }{ 2 * 115 * 271.99 } ) = 66° 44'31" ; ; gamma = 180° - alpha - beta = 180° - 25° - 66° 44'31" = 88° 15'29" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 14368.36 }{ 318.49 } = 45.11 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 115 }{ 2 * sin 25° } = 136.06 ; ; : Nr. 1

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 250**2+2 * 271.99**2 - 115**2 } }{ 2 } = 254.818 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 271.99**2+2 * 115**2 - 250**2 } }{ 2 } = 167.261 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 250**2+2 * 115**2 - 271.99**2 } }{ 2 } = 139.17 ; ;
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