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=125; b=150; c=28.21333475737 and a=125; b=150; c=243.6798988537.

#1 Obtuse scalene triangle.

Sides: a = 125   b = 150   c = 28.21333475737

Area: T = 894.261069321
Perimeter: p = 303.2133347574
Semiperimeter: s = 151.6076673787

Angle ∠ A = α = 25° = 0.4366332313 rad
Angle ∠ B = β = 149.5266359102° = 149°31'35″ = 2.61097272848 rad
Angle ∠ C = γ = 5.47436408984° = 5°28'25″ = 0.09655330557 rad

Height: ha = 14.30881710914
Height: hb = 11.92334759095
Height: hc = 63.39327392611

Median: ma = 87.9877195038
Median: mb = 50.84877776373
Median: mc = 137.3444463866

Inradius: r = 5.89985575692
Circumradius: R = 147.8887598947

Vertex coordinates: A[28.21333475737; 0] B[0; 0] C[-107.7332820482; 63.39327392611]
Centroid: CG[-26.50664909694; 21.1310913087]
Coordinates of the circumscribed circle: U[14.10766737868; 147.2133259175]
Coordinates of the inscribed circle: I[1.60766737868; 5.89985575692]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155° = 0.4366332313 rad
∠ B' = β' = 30.47436408984° = 30°28'25″ = 2.61097272848 rad
∠ C' = γ' = 174.5266359102° = 174°31'35″ = 0.09655330557 rad




How did we calculate this triangle?

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

a = 125 ; ; b = 150 ; ; alpha = 25° ; ;

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

a**2 = b**2 + c**2 - 2b c cos alpha ; ; ; ; 125**2 = 150**2 + c**2 - 2 * 150 * c * cos(25° ) ; ; ; ; ; ; c**2 -271.892c +6875 =0 ; ; a=1; b=-271.892; c=6875 ; ; D = b**2 - 4ac = 271.892**2 - 4 * 1 * 6875 = 46425.4424359 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 271.89 ± sqrt{ 46425.44 } }{ 2 } ; ; c_{1,2} = 135.94616806 ± 107.732820482 ; ; c_{1} = 243.678988542 ; ; c_{2} = 28.2133475782 ; ;
 ; ; text{ Factored form: } ; ; (c -243.678988542) (c -28.2133475782) = 0 ; ; ; ; c > 0 ; ; ; ; c = 243.679 ; ;


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 = 125 ; ; b = 150 ; ; c = 28.21 ; ;

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

p = a+b+c = 125+150+28.21 = 303.21 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 303.21 }{ 2 } = 151.61 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 151.61 * (151.61-125)(151.61-150)(151.61-28.21) } ; ; T = sqrt{ 799702.19 } = 894.26 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 894.26 }{ 125 } = 14.31 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 894.26 }{ 150 } = 11.92 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 894.26 }{ 28.21 } = 63.39 ; ;

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{ 150**2+28.21**2-125**2 }{ 2 * 150 * 28.21 } ) = 25° ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 125**2+28.21**2-150**2 }{ 2 * 125 * 28.21 } ) = 149° 31'35" ; ; gamma = arccos( fraction{ a**2+b**2-c**2 }{ 2ab } ) = arccos( fraction{ 125**2+150**2-28.21**2 }{ 2 * 125 * 150 } ) = 5° 28'25" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 894.26 }{ 151.61 } = 5.9 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 125 }{ 2 * sin 25° } = 147.89 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 150**2+2 * 28.21**2 - 125**2 } }{ 2 } = 87.987 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 28.21**2+2 * 125**2 - 150**2 } }{ 2 } = 50.848 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 150**2+2 * 125**2 - 28.21**2 } }{ 2 } = 137.344 ; ;







#2 Obtuse scalene triangle.

Sides: a = 125   b = 150   c = 243.6798988537

Area: T = 7723.739929188
Perimeter: p = 518.6798988537
Semiperimeter: s = 259.3399494269

Angle ∠ A = α = 25° = 0.4366332313 rad
Angle ∠ B = β = 30.47436408984° = 30°28'25″ = 0.53218653687 rad
Angle ∠ C = γ = 124.5266359102° = 124°31'35″ = 2.17333949718 rad

Height: ha = 123.587982867
Height: hb = 102.9833190558
Height: hc = 63.39327392611

Median: ma = 192.4410834355
Median: mb = 178.5421941087
Median: mc = 64.943334174

Inradius: r = 29.78223488615
Circumradius: R = 147.8887598947

Vertex coordinates: A[243.6798988537; 0] B[0; 0] C[107.7332820482; 63.39327392611]
Centroid: CG[117.1377269673; 21.1310913087]
Coordinates of the circumscribed circle: U[121.8399494269; -83.82105199141]
Coordinates of the inscribed circle: I[109.3399494269; 29.78223488615]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155° = 0.4366332313 rad
∠ B' = β' = 149.5266359102° = 149°31'35″ = 0.53218653687 rad
∠ C' = γ' = 55.47436408984° = 55°28'25″ = 2.17333949718 rad

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How did we calculate this triangle?

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

a = 125 ; ; b = 150 ; ; alpha = 25° ; ; : Nr. 1

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

a**2 = b**2 + c**2 - 2b c cos alpha ; ; ; ; 125**2 = 150**2 + c**2 - 2 * 150 * c * cos(25° ) ; ; ; ; ; ; c**2 -271.892c +6875 =0 ; ; a=1; b=-271.892; c=6875 ; ; D = b**2 - 4ac = 271.892**2 - 4 * 1 * 6875 = 46425.4424359 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 271.89 ± sqrt{ 46425.44 } }{ 2 } ; ; c_{1,2} = 135.94616806 ± 107.732820482 ; ; c_{1} = 243.678988542 ; ; c_{2} = 28.2133475782 ; ; : Nr. 1
 ; ; text{ Factored form: } ; ; (c -243.678988542) (c -28.2133475782) = 0 ; ; ; ; c > 0 ; ; ; ; c = 243.679 ; ; : 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 = 125 ; ; b = 150 ; ; c = 243.68 ; ;

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

p = a+b+c = 125+150+243.68 = 518.68 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 518.68 }{ 2 } = 259.34 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 259.34 * (259.34-125)(259.34-150)(259.34-243.68) } ; ; T = sqrt{ 59656148.65 } = 7723.74 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 7723.74 }{ 125 } = 123.58 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 7723.74 }{ 150 } = 102.98 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 7723.74 }{ 243.68 } = 63.39 ; ;

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{ 150**2+243.68**2-125**2 }{ 2 * 150 * 243.68 } ) = 25° ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 125**2+243.68**2-150**2 }{ 2 * 125 * 243.68 } ) = 30° 28'25" ; ; gamma = arccos( fraction{ a**2+b**2-c**2 }{ 2ab } ) = arccos( fraction{ 125**2+150**2-243.68**2 }{ 2 * 125 * 150 } ) = 124° 31'35" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 7723.74 }{ 259.34 } = 29.78 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 125 }{ 2 * sin 25° } = 147.89 ; ; : Nr. 1

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 150**2+2 * 243.68**2 - 125**2 } }{ 2 } = 192.441 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 243.68**2+2 * 125**2 - 150**2 } }{ 2 } = 178.542 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 150**2+2 * 125**2 - 243.68**2 } }{ 2 } = 64.943 ; ;
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