Triangle calculator SSA

Please enter two sides and a non-included angle
°


Triangle has two solutions with side c=80.92987410057 and with side c=12.35765495716

#1 Obtuse scalene triangle.

Sides: a = 55   b = 45   c = 80.92987410057

Area: T = 1179.357671945
Perimeter: p = 180.9298741006
Semiperimeter: s = 90.46443705028

Angle ∠ A = α = 40.36768437134° = 40°22'1″ = 0.70545343314 rad
Angle ∠ B = β = 32° = 0.55985053606 rad
Angle ∠ C = γ = 107.6333156287° = 107°37'59″ = 1.87985529615 rad

Height: ha = 42.8865698889
Height: hb = 52.41658541977
Height: hc = 29.14655595328

Median: ma = 59.42220544948
Median: mb = 65.42992026574
Median: mc = 29.79331992208

Inradius: r = 13.03766984581
Circumradius: R = 42.4599298083

Vertex coordinates: A[80.92987410057; 0] B[0; 0] C[46.64326452886; 29.14655595328]
Centroid: CG[42.52437954314; 9.71551865109]
Coordinates of the circumscribed circle: U[40.46443705028; -12.86218316546]
Coordinates of the inscribed circle: I[45.46443705028; 13.03766984581]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 139.6333156287° = 139°37'59″ = 0.70545343314 rad
∠ B' = β' = 148° = 0.55985053606 rad
∠ C' = γ' = 72.36768437134° = 72°22'1″ = 1.87985529615 rad




How did we calculate this triangle?

1. Use Law of Cosines

a = 55 ; ; b = 45 ; ; beta = 32° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 45**2 = 55**2 + c**2 -2 * 55 * c * cos (32° ) ; ; ; ; c**2 -93.285c +1000 =0 ; ; p=1; q=-93.285; r=1000 ; ; D = q**2 - 4pr = 93.285**2 - 4 * 1 * 1000 = 4702.14543807 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 93.29 ± sqrt{ 4702.15 } }{ 2 } ; ; c_{1,2} = 46.64264529 ± 34.286095717 ; ; c_{1} = 80.928741007 ; ;
c_{2} = 12.356549573 ; ; ; ; (c -80.928741007) (c -12.356549573) = 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 = 55 ; ; b = 45 ; ; c = 80.93 ; ;

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

p = a+b+c = 55+45+80.93 = 180.93 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 180.93 }{ 2 } = 90.46 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 90.46 * (90.46-55)(90.46-45)(90.46-80.93) } ; ; T = sqrt{ 1390882.27 } = 1179.36 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1179.36 }{ 55 } = 42.89 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1179.36 }{ 45 } = 52.42 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1179.36 }{ 80.93 } = 29.15 ; ;

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{ 55**2-45**2-80.93**2 }{ 2 * 45 * 80.93 } ) = 40° 22'1" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 45**2-55**2-80.93**2 }{ 2 * 55 * 80.93 } ) = 32° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 80.93**2-55**2-45**2 }{ 2 * 45 * 55 } ) = 107° 37'59" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1179.36 }{ 90.46 } = 13.04 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 55 }{ 2 * sin 40° 22'1" } = 42.46 ; ;





#2 Obtuse scalene triangle.

Sides: a = 55   b = 45   c = 12.35765495716

Area: T = 180.0699275579
Perimeter: p = 112.3576549572
Semiperimeter: s = 56.17882747858

Angle ∠ A = α = 139.6333156287° = 139°37'59″ = 2.43770583222 rad
Angle ∠ B = β = 32° = 0.55985053606 rad
Angle ∠ C = γ = 8.36768437134° = 8°22'1″ = 0.14660289708 rad

Height: ha = 6.54879736574
Height: hb = 8.00330789146
Height: hc = 29.14655595328

Median: ma = 18.23771093833
Median: mb = 32.90327682522
Median: mc = 49.86881152709

Inradius: r = 3.20553187156
Circumradius: R = 42.4599298083

Vertex coordinates: A[12.35765495716; 0] B[0; 0] C[46.64326452886; 29.14655595328]
Centroid: CG[19.66663982867; 9.71551865109]
Coordinates of the circumscribed circle: U[6.17882747858; 42.00773911874]
Coordinates of the inscribed circle: I[11.17882747858; 3.20553187156]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 40.36768437134° = 40°22'1″ = 2.43770583222 rad
∠ B' = β' = 148° = 0.55985053606 rad
∠ C' = γ' = 171.6333156287° = 171°37'59″ = 0.14660289708 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 55 ; ; b = 45 ; ; beta = 32° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 45**2 = 55**2 + c**2 -2 * 55 * c * cos (32° ) ; ; ; ; c**2 -93.285c +1000 =0 ; ; p=1; q=-93.285; r=1000 ; ; D = q**2 - 4pr = 93.285**2 - 4 * 1 * 1000 = 4702.14543807 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 93.29 ± sqrt{ 4702.15 } }{ 2 } ; ; c_{1,2} = 46.64264529 ± 34.286095717 ; ; c_{1} = 80.928741007 ; ; : Nr. 1
c_{2} = 12.356549573 ; ; ; ; (c -80.928741007) (c -12.356549573) = 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 = 55 ; ; b = 45 ; ; c = 12.36 ; ;

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

p = a+b+c = 55+45+12.36 = 112.36 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 112.36 }{ 2 } = 56.18 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 56.18 * (56.18-55)(56.18-45)(56.18-12.36) } ; ; T = sqrt{ 32424.94 } = 180.07 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 180.07 }{ 55 } = 6.55 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 180.07 }{ 45 } = 8 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 180.07 }{ 12.36 } = 29.15 ; ;

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{ 55**2-45**2-12.36**2 }{ 2 * 45 * 12.36 } ) = 139° 37'59" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 45**2-55**2-12.36**2 }{ 2 * 55 * 12.36 } ) = 32° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 12.36**2-55**2-45**2 }{ 2 * 45 * 55 } ) = 8° 22'1" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 180.07 }{ 56.18 } = 3.21 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 55 }{ 2 * sin 139° 37'59" } = 42.46 ; ;




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

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