Triangle calculator SSA

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Triangle has two solutions with side c=193.2488396363 and with side c=54.95500032075

#1 Acute scalene triangle.

Sides: a = 162   b = 125   c = 193.2488396363

Area: T = 10061.63216567
Perimeter: p = 480.2488396363
Semiperimeter: s = 240.1244198181

Angle ∠ A = α = 56.41436175293° = 56°24'49″ = 0.98546033688 rad
Angle ∠ B = β = 40° = 0.69881317008 rad
Angle ∠ C = γ = 83.58663824707° = 83°35'11″ = 1.45988575839 rad

Height: ha = 124.2187674774
Height: hb = 160.9866106507
Height: hc = 104.1321592769

Median: ma = 141.1522298417
Median: mb = 166.9987668692
Median: mc = 107.6955238176

Inradius: r = 41.90217813818
Circumradius: R = 97.23327391788

Vertex coordinates: A[193.2488396363; 0] B[0; 0] C[124.0999199785; 104.1321592769]
Centroid: CG[105.7832532049; 34.71105309231]
Coordinates of the circumscribed circle: U[96.62441981815; 10.86113946614]
Coordinates of the inscribed circle: I[115.1244198182; 41.90217813818]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 123.5866382471° = 123°35'11″ = 0.98546033688 rad
∠ B' = β' = 140° = 0.69881317008 rad
∠ C' = γ' = 96.41436175293° = 96°24'49″ = 1.45988575839 rad




How did we calculate this triangle?

1. Use Law of Cosines

a = 162 ; ; b = 125 ; ; beta = 40° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 125**2 = 162**2 + c**2 -2 * 162 * c * cos (40° ) ; ; ; ; c**2 -248.198c +10619 =0 ; ; p=1; q=-248.198; r=10619 ; ; D = q**2 - 4pr = 248.198**2 - 4 * 1 * 10619 = 19126.4455494 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 248.2 ± sqrt{ 19126.45 } }{ 2 } ; ; c_{1,2} = 124.09919979 ± 69.1491965777 ; ; c_{1} = 193.248396368 ; ;
c_{2} = 54.9500032123 ; ; ; ; (c -193.248396368) (c -54.9500032123) = 0 ; ; ; ; c>0 ; ;


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 = 162 ; ; b = 125 ; ; c = 193.25 ; ;

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

p = a+b+c = 162+125+193.25 = 480.25 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 480.25 }{ 2 } = 240.12 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 240.12 * (240.12-162)(240.12-125)(240.12-193.25) } ; ; T = sqrt{ 101236431.59 } = 10061.63 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 10061.63 }{ 162 } = 124.22 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 10061.63 }{ 125 } = 160.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 10061.63 }{ 193.25 } = 104.13 ; ;

6. 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{ 162**2-125**2-193.25**2 }{ 2 * 125 * 193.25 } ) = 56° 24'49" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 125**2-162**2-193.25**2 }{ 2 * 162 * 193.25 } ) = 40° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 193.25**2-162**2-125**2 }{ 2 * 125 * 162 } ) = 83° 35'11" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 10061.63 }{ 240.12 } = 41.9 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 162 }{ 2 * sin 56° 24'49" } = 97.23 ; ;





#2 Obtuse scalene triangle.

Sides: a = 162   b = 125   c = 54.95500032075

Area: T = 2861.016567834
Perimeter: p = 341.9550003208
Semiperimeter: s = 170.9755001604

Angle ∠ A = α = 123.5866382471° = 123°35'11″ = 2.15769892847 rad
Angle ∠ B = β = 40° = 0.69881317008 rad
Angle ∠ C = γ = 16.41436175293° = 16°24'49″ = 0.28664716681 rad

Height: ha = 35.3211181214
Height: hb = 45.77662508534
Height: hc = 104.1321592769

Median: ma = 52.54876110423
Median: mb = 103.5643996766
Median: mc = 142.0555004441

Inradius: r = 16.73435321041
Circumradius: R = 97.23327391788

Vertex coordinates: A[54.95500032075; 0] B[0; 0] C[124.0999199785; 104.1321592769]
Centroid: CG[59.68330676643; 34.71105309231]
Coordinates of the circumscribed circle: U[27.47550016038; 93.27701981079]
Coordinates of the inscribed circle: I[45.97550016038; 16.73435321041]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 56.41436175293° = 56°24'49″ = 2.15769892847 rad
∠ B' = β' = 140° = 0.69881317008 rad
∠ C' = γ' = 163.5866382471° = 163°35'11″ = 0.28664716681 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 162 ; ; b = 125 ; ; beta = 40° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 125**2 = 162**2 + c**2 -2 * 162 * c * cos (40° ) ; ; ; ; c**2 -248.198c +10619 =0 ; ; p=1; q=-248.198; r=10619 ; ; D = q**2 - 4pr = 248.198**2 - 4 * 1 * 10619 = 19126.4455494 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 248.2 ± sqrt{ 19126.45 } }{ 2 } ; ; c_{1,2} = 124.09919979 ± 69.1491965777 ; ; c_{1} = 193.248396368 ; ; : Nr. 1
c_{2} = 54.9500032123 ; ; ; ; (c -193.248396368) (c -54.9500032123) = 0 ; ; ; ; c>0 ; ; : 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 = 162 ; ; b = 125 ; ; c = 54.95 ; ;

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

p = a+b+c = 162+125+54.95 = 341.95 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 341.95 }{ 2 } = 170.98 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 170.98 * (170.98-162)(170.98-125)(170.98-54.95) } ; ; T = sqrt{ 8185410.71 } = 2861.02 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2861.02 }{ 162 } = 35.32 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2861.02 }{ 125 } = 45.78 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2861.02 }{ 54.95 } = 104.13 ; ;

6. 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{ 162**2-125**2-54.95**2 }{ 2 * 125 * 54.95 } ) = 123° 35'11" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 125**2-162**2-54.95**2 }{ 2 * 162 * 54.95 } ) = 40° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 54.95**2-162**2-125**2 }{ 2 * 125 * 162 } ) = 16° 24'49" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 2861.02 }{ 170.98 } = 16.73 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 162 }{ 2 * sin 123° 35'11" } = 97.23 ; ;




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