Triangle calculator

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

You have entered side a, c and angle α.

Triangle has two solutions: a=120; b=32.92325682786; c=141 and a=120; b=166.4821544016; c=141.

#1 Obtuse scalene triangle.

Sides: a = 120   b = 32.92325682786   c = 141

Area: T = 1641.224387552
Perimeter: p = 293.9232568279
Semiperimeter: s = 146.9611284139

Angle ∠ A = α = 45° = 0.78553981634 rad
Angle ∠ B = β = 11.18662009828° = 11°11'10″ = 0.19552360379 rad
Angle ∠ C = γ = 123.8143799017° = 123°48'50″ = 2.16109584523 rad

Height: ha = 27.35437312586
Height: hb = 99.70220561473
Height: hc = 23.28797712839

Median: ma = 82.96105192307
Median: mb = 129.8832739902
Median: mc = 52.64769158739

Inradius: r = 11.16877295495
Circumradius: R = 84.85328137424

Vertex coordinates: A[141; 0] B[0; 0] C[117.7220228716; 23.28797712839]
Centroid: CG[86.24400762387; 7.76599237613]
Coordinates of the circumscribed circle: U[70.5; -47.22202287161]
Coordinates of the inscribed circle: I[114.0398715861; 11.16877295495]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 135° = 0.78553981634 rad
∠ B' = β' = 168.8143799017° = 168°48'50″ = 0.19552360379 rad
∠ C' = γ' = 56.18662009828° = 56°11'10″ = 2.16109584523 rad




How did we calculate this triangle?

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

a = 120 ; ; c = 141 ; ; alpha = 45° ; ;

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

a**2 = c**2 + b**2 - 2c b cos alpha ; ; ; ; 120**2 = 141**2 + b**2 - 2 * 141 * b * cos 45° ; ; ; ; ; ; b**2 -199.404b +5481 =0 ; ; p=1; q=-199.404; r=5481 ; ; D = q**2 - 4pr = 199.404**2 - 4 * 1 * 5481 = 17838 ; ; D>0 ; ; ; ; b_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 199.4 ± sqrt{ 17838 } }{ 2 } ; ; b_{1,2} = 99.70205615 ± 66.7794878687 ; ; b_{1} = 166.481544019 ; ; b_{2} = 32.9225682813 ; ; ; ; text{ Factored form: } ; ;
(b -166.481544019) (b -32.9225682813) = 0 ; ; ; ; b > 0 ; ; ; ; b = 166.482 ; ;
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 = 120 ; ; b = 32.92 ; ; c = 141 ; ;

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

p = a+b+c = 120+32.92+141 = 293.92 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 293.92 }{ 2 } = 146.96 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 146.96 * (146.96-120)(146.96-32.92)(146.96-141) } ; ; T = sqrt{ 2693615.81 } = 1641.22 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1641.22 }{ 120 } = 27.35 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1641.22 }{ 32.92 } = 99.7 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1641.22 }{ 141 } = 23.28 ; ;

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{ 32.92**2+141**2-120**2 }{ 2 * 32.92 * 141 } ) = 45° ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 120**2+141**2-32.92**2 }{ 2 * 120 * 141 } ) = 11° 11'10" ; ; gamma = 180° - alpha - beta = 180° - 45° - 11° 11'10" = 123° 48'50" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1641.22 }{ 146.96 } = 11.17 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 120 }{ 2 * sin 45° } = 84.85 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 32.92**2+2 * 141**2 - 120**2 } }{ 2 } = 82.961 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 141**2+2 * 120**2 - 32.92**2 } }{ 2 } = 129.883 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 32.92**2+2 * 120**2 - 141**2 } }{ 2 } = 52.647 ; ;







#2 Acute scalene triangle.

Sides: a = 120   b = 166.4821544016   c = 141

Area: T = 8299.276612448
Perimeter: p = 427.4821544016
Semiperimeter: s = 213.7410772008

Angle ∠ A = α = 45° = 0.78553981634 rad
Angle ∠ B = β = 78.81437990172° = 78°48'50″ = 1.37655602889 rad
Angle ∠ C = γ = 56.18662009828° = 56°11'10″ = 0.98106342013 rad

Height: ha = 138.3211268741
Height: hb = 99.70220561473
Height: hc = 117.7220228716

Median: ma = 142.1221610774
Median: mb = 101.0521837566
Median: mc = 126.8387700424

Inradius: r = 38.8298699113
Circumradius: R = 84.85328137424

Vertex coordinates: A[141; 0] B[0; 0] C[23.28797712839; 117.7220228716]
Centroid: CG[54.76599237613; 39.24400762387]
Coordinates of the circumscribed circle: U[70.5; 47.22202287161]
Coordinates of the inscribed circle: I[47.2599227992; 38.8298699113]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 135° = 0.78553981634 rad
∠ B' = β' = 101.1866200983° = 101°11'10″ = 1.37655602889 rad
∠ C' = γ' = 123.8143799017° = 123°48'50″ = 0.98106342013 rad

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How did we calculate this triangle?

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

a = 120 ; ; c = 141 ; ; alpha = 45° ; ; : Nr. 1

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

a**2 = c**2 + b**2 - 2c b cos alpha ; ; ; ; 120**2 = 141**2 + b**2 - 2 * 141 * b * cos 45° ; ; ; ; ; ; b**2 -199.404b +5481 =0 ; ; p=1; q=-199.404; r=5481 ; ; D = q**2 - 4pr = 199.404**2 - 4 * 1 * 5481 = 17838 ; ; D>0 ; ; ; ; b_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 199.4 ± sqrt{ 17838 } }{ 2 } ; ; b_{1,2} = 99.70205615 ± 66.7794878687 ; ; b_{1} = 166.481544019 ; ; b_{2} = 32.9225682813 ; ; ; ; text{ Factored form: } ; ; : Nr. 1
(b -166.481544019) (b -32.9225682813) = 0 ; ; ; ; b > 0 ; ; ; ; b = 166.482 ; ; : 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 = 120 ; ; b = 166.48 ; ; c = 141 ; ;

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

p = a+b+c = 120+166.48+141 = 427.48 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 427.48 }{ 2 } = 213.74 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 213.74 * (213.74-120)(213.74-166.48)(213.74-141) } ; ; T = sqrt{ 68877984.19 } = 8299.28 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 8299.28 }{ 120 } = 138.32 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 8299.28 }{ 166.48 } = 99.7 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 8299.28 }{ 141 } = 117.72 ; ;

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{ 166.48**2+141**2-120**2 }{ 2 * 166.48 * 141 } ) = 45° ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 120**2+141**2-166.48**2 }{ 2 * 120 * 141 } ) = 78° 48'50" ; ; gamma = 180° - alpha - beta = 180° - 45° - 78° 48'50" = 56° 11'10" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 8299.28 }{ 213.74 } = 38.83 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 120 }{ 2 * sin 45° } = 84.85 ; ; : Nr. 1

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 166.48**2+2 * 141**2 - 120**2 } }{ 2 } = 142.122 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 141**2+2 * 120**2 - 166.48**2 } }{ 2 } = 101.052 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 166.48**2+2 * 120**2 - 141**2 } }{ 2 } = 126.838 ; ;
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