Triangle calculator SSA

Triangle has two solutions with side c=302.5440188133 and with side c=25.12106295828

#1 Obtuse scalene triangle.

Sides: a = 200 b = 180 c = 302.5440188133

Area: T = 17352.99222962
Perimeter: p = 682.5440188133
Semiperimeter: s = 341.2770094066

Angle ∠ A = α = 39.59113123899° = 39°35'29″ = 0.69109987564 rad
Angle ∠ B = β = 35° = 0.61108652382 rad
Angle ∠ C = γ = 105.409868761° = 105°24'31″ = 1.8439728659 rad

Height: h_a = 173.5329922962
Height: h_b = 192.8111025513
Height: h_c = 114.715528727

Median: m_a = 227.9598949633
Median: m_b = 240.1365967147
Median: m_c = 115.401086066

Inradius: r = 50.84882653415
Circumradius: R = 156.9110211606

Vertex coordinates: A[302.5440188133; 0] B[0; 0] C[163.8330408858; 114.715528727]
Centroid: CG[155.4576865663; 38.23884290901]
Coordinates of the circumscribed circle: U[151.2770094066; -41.69114037585]
Coordinates of the inscribed circle: I[161.2770094066; 50.84882653415]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 140.409868761° = 140°24'31″ = 0.69109987564 rad
∠ B' = β' = 145° = 0.61108652382 rad
∠ C' = γ' = 74.59113123899° = 74°35'29″ = 1.8439728659 rad

How did we calculate this triangle?

1. Use Law of Cosines

$a = 200 ; ; b = 180 ; ; beta = 35° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 180**2 = 200**2 + c**2 -2 * 200 * c * cos (35° ) ; ; ; ; c**2 -327.661c +7600 =0 ; ; p=1; q=-327.661; r=7600 ; ; D = q**2 - 4pr = 327.661**2 - 4 * 1 * 7600 = 76961.6114661 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 327.66 ± sqrt{ 76961.61 } }{ 2 } ; ; c_{1,2} = 163.83040886 ± 138.709779275 ; ; c_{1} = 302.540188135 ; ;$
$c_{2} = 25.120629585 ; ; ; ; (c -302.540188135) (c -25.120629585) = 0 ; ; ; ; c>0 ; ;$

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.

$a = 200 ; ; b = 180 ; ; c = 302.54 ; ;$

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

$p = a+b+c = 200+180+302.54 = 682.54 ; ;$

3. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 682.54 }{ 2 } = 341.27 ; ;$

4. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 341.27 * (341.27-200)(341.27-180)(341.27-302.54) } ; ; T = sqrt{ 301126341.63 } = 17352.99 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 17352.99 }{ 200 } = 173.53 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 17352.99 }{ 180 } = 192.81 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 17352.99 }{ 302.54 } = 114.72 ; ;$

6. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 200**2-180**2-302.54**2 }{ 2 * 180 * 302.54 } ) = 39° 35'29" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 180**2-200**2-302.54**2 }{ 2 * 200 * 302.54 } ) = 35° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 302.54**2-200**2-180**2 }{ 2 * 180 * 200 } ) = 105° 24'31" ; ;$

7. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 17352.99 }{ 341.27 } = 50.85 ; ;$

8. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 200 }{ 2 * sin 39° 35'29" } = 156.91 ; ;$

#2 Obtuse scalene triangle.

Sides: a = 200 b = 180 c = 25.12106295828

Area: T = 1440.86601195
Perimeter: p = 405.1210629583
Semiperimeter: s = 202.5660314791

Angle ∠ A = α = 140.409868761° = 140°24'31″ = 2.45105938972 rad
Angle ∠ B = β = 35° = 0.61108652382 rad
Angle ∠ C = γ = 4.59113123899° = 4°35'29″ = 0.08801335182 rad

Height: h_a = 14.4098601195
Height: h_b = 16.01095568833
Height: h_c = 114.715528727

Median: m_a = 80.7198789729
Median: m_b = 110.5243857222
Median: m_c = 189.8487935181

Inradius: r = 7.11332399305
Circumradius: R = 156.9110211606

Vertex coordinates: A[25.12106295828; 0] B[0; 0] C[163.8330408858; 114.715528727]
Centroid: CG[62.98436794802; 38.23884290901]
Coordinates of the circumscribed circle: U[12.56603147914; 156.4076691029]
Coordinates of the inscribed circle: I[22.56603147914; 7.11332399305]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 39.59113123899° = 39°35'29″ = 2.45105938972 rad
∠ B' = β' = 145° = 0.61108652382 rad
∠ C' = γ' = 175.409868761° = 175°24'31″ = 0.08801335182 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

$a = 200 ; ; b = 180 ; ; beta = 35° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 180**2 = 200**2 + c**2 -2 * 200 * c * cos (35° ) ; ; ; ; c**2 -327.661c +7600 =0 ; ; p=1; q=-327.661; r=7600 ; ; D = q**2 - 4pr = 327.661**2 - 4 * 1 * 7600 = 76961.6114661 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 327.66 ± sqrt{ 76961.61 } }{ 2 } ; ; c_{1,2} = 163.83040886 ± 138.709779275 ; ; c_{1} = 302.540188135 ; ; : Nr. 1$
$c_{2} = 25.120629585 ; ; ; ; (c -302.540188135) (c -25.120629585) = 0 ; ; ; ; c>0 ; ; : Nr. 1$

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.

$a = 200 ; ; b = 180 ; ; c = 25.12 ; ;$

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

$p = a+b+c = 200+180+25.12 = 405.12 ; ;$

3. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 405.12 }{ 2 } = 202.56 ; ;$

4. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 202.56 * (202.56-200)(202.56-180)(202.56-25.12) } ; ; T = sqrt{ 2076077.88 } = 1440.86 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1440.86 }{ 200 } = 14.41 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1440.86 }{ 180 } = 16.01 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1440.86 }{ 25.12 } = 114.72 ; ;$

6. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 200**2-180**2-25.12**2 }{ 2 * 180 * 25.12 } ) = 140° 24'31" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 180**2-200**2-25.12**2 }{ 2 * 200 * 25.12 } ) = 35° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25.12**2-200**2-180**2 }{ 2 * 180 * 200 } ) = 4° 35'29" ; ;$

7. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 1440.86 }{ 202.56 } = 7.11 ; ;$

8. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 200 }{ 2 * sin 140° 24'31" } = 156.91 ; ;$

Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.