Triangle calculator SSA

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Triangle has two solutions with side c=2.88880901303 and with side c=1.66219979929

#1 Acute scalene triangle.

Sides: a = 3.4   b = 2.6   c = 2.88880901303

Area: T = 3.64986577009
Perimeter: p = 8.88880901303
Semiperimeter: s = 4.44440450651

Angle ∠ A = α = 76.36219847179° = 76°21'43″ = 1.33327680567 rad
Angle ∠ B = β = 48° = 0.8387758041 rad
Angle ∠ C = γ = 55.63880152821° = 55°38'17″ = 0.97110665559 rad

Height: ha = 2.14662692358
Height: hb = 2.80766597699
Height: hc = 2.52766924066

Median: ma = 2.15988266027
Median: mb = 2.87441141766
Median: mc = 2.66598371848

Inradius: r = 0.821102176
Circumradius: R = 1.74993225485

Vertex coordinates: A[2.88880901303; 0] B[0; 0] C[2.27550440616; 2.52766924066]
Centroid: CG[1.72110447306; 0.84222308022]
Coordinates of the circumscribed circle: U[1.44440450651; 0.98773516235]
Coordinates of the inscribed circle: I[1.84440450651; 0.821102176]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 103.6388015282° = 103°38'17″ = 1.33327680567 rad
∠ B' = β' = 132° = 0.8387758041 rad
∠ C' = γ' = 124.3621984718° = 124°21'43″ = 0.97110665559 rad




How did we calculate this triangle?

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 = 3.4 ; ; b = 2.6 ; ; c = 2.89 ; ;

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

p = a+b+c = 3.4+2.6+2.89 = 8.89 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 8.89 }{ 2 } = 4.44 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 4.44 * (4.44-3.4)(4.44-2.6)(4.44-2.89) } ; ; T = sqrt{ 13.31 } = 3.65 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 3.65 }{ 3.4 } = 2.15 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 3.65 }{ 2.6 } = 2.81 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 3.65 }{ 2.89 } = 2.53 ; ;

5. 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{ 3.4**2-2.6**2-2.89**2 }{ 2 * 2.6 * 2.89 } ) = 76° 21'43" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 2.6**2-3.4**2-2.89**2 }{ 2 * 3.4 * 2.89 } ) = 48° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 2.89**2-3.4**2-2.6**2 }{ 2 * 2.6 * 3.4 } ) = 55° 38'17" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 3.65 }{ 4.44 } = 0.82 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 3.4 }{ 2 * sin 76° 21'43" } = 1.75 ; ;





#2 Obtuse scalene triangle.

Sides: a = 3.4   b = 2.6   c = 1.66219979929

Area: T = 2.10996788543
Perimeter: p = 7.66219979929
Semiperimeter: s = 3.83109989965

Angle ∠ A = α = 103.6388015282° = 103°38'17″ = 1.80988245969 rad
Angle ∠ B = β = 48° = 0.8387758041 rad
Angle ∠ C = γ = 28.36219847179° = 28°21'43″ = 0.49550100157 rad

Height: ha = 1.23551052084
Height: hb = 1.61551375802
Height: hc = 2.52766924066

Median: ma = 1.36878883961
Median: mb = 2.33990422536
Median: mc = 2.91102303462

Inradius: r = 0.54880760648
Circumradius: R = 1.74993225485

Vertex coordinates: A[1.66219979929; 0] B[0; 0] C[2.27550440616; 2.52766924066]
Centroid: CG[1.31223473515; 0.84222308022]
Coordinates of the circumscribed circle: U[0.83109989965; 1.53993407831]
Coordinates of the inscribed circle: I[1.23109989965; 0.54880760648]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 76.36219847179° = 76°21'43″ = 1.80988245969 rad
∠ B' = β' = 132° = 0.8387758041 rad
∠ C' = γ' = 151.6388015282° = 151°38'17″ = 0.49550100157 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 3.4 ; ; b = 2.6 ; ; beta = 48° ; ; ; ; b**2 = a**2 + c**2 - 2bc cos( beta ) ; ; 2.6**2 = 3.4**2 + c**2 -2 * 2.6 * c * cos (48° ) ; ; ; ; c**2 -4.55c +4.8 =0 ; ; p=1; q=-4.55008812324; r=4.8 ; ; D = q**2 - 4pr = 4.55**2 - 4 * 1 * 4.8 = 1.50330192925 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 4.55 ± sqrt{ 1.5 } }{ 2 } ; ; c_{1,2} = 2.27504406162 ± 0.613046068671 ; ; c_{1} = 2.88809013029 ; ;
c_{2} = 1.66199799295 ; ; ; ; (c -2.88809013029) (c -1.66199799295) = 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 = 3.4 ; ; b = 2.6 ; ; c = 1.66 ; ;

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

p = a+b+c = 3.4+2.6+1.66 = 7.66 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 7.66 }{ 2 } = 3.83 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 3.83 * (3.83-3.4)(3.83-2.6)(3.83-1.66) } ; ; T = sqrt{ 4.41 } = 2.1 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2.1 }{ 3.4 } = 1.24 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2.1 }{ 2.6 } = 1.62 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2.1 }{ 1.66 } = 2.53 ; ;

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{ 3.4**2-2.6**2-1.66**2 }{ 2 * 2.6 * 1.66 } ) = 103° 38'17" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 2.6**2-3.4**2-1.66**2 }{ 2 * 3.4 * 1.66 } ) = 48° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 1.66**2-3.4**2-2.6**2 }{ 2 * 2.6 * 3.4 } ) = 28° 21'43" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 2.1 }{ 3.83 } = 0.55 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 3.4 }{ 2 * sin 103° 38'17" } = 1.75 ; ;




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