Triangle calculator SSA

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Triangle has two solutions with side c=191.4211356237 and with side c=91.42113562373

#1 Acute scalene triangle.

Sides: a = 200   b = 150   c = 191.4211356237

Area: T = 13535.53439059
Perimeter: p = 541.4211356237
Semiperimeter: s = 270.7110678119

Angle ∠ A = α = 70.52987793655° = 70°31'44″ = 1.23109594173 rad
Angle ∠ B = β = 45° = 0.78553981634 rad
Angle ∠ C = γ = 64.47112206345° = 64°28'16″ = 1.12552350729 rad

Height: ha = 135.3555339059
Height: hb = 180.4743785412
Height: hc = 141.4211356237

Median: ma = 139.8976632597
Median: mb = 180.8210540348
Median: mc = 148.6255253891

Inradius: r = 50
Circumradius: R = 106.0666017178

Vertex coordinates: A[191.4211356237; 0] B[0; 0] C[141.4211356237; 141.4211356237]
Centroid: CG[110.9487570825; 47.14404520791]
Coordinates of the circumscribed circle: U[95.71106781187; 45.71106781187]
Coordinates of the inscribed circle: I[120.7110678119; 50]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 109.4711220634° = 109°28'16″ = 1.23109594173 rad
∠ B' = β' = 135° = 0.78553981634 rad
∠ C' = γ' = 115.5298779366° = 115°31'44″ = 1.12552350729 rad




How did we calculate this triangle?

1. Use Law of Cosines

a = 200 ; ; b = 150 ; ; beta = 45° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 150**2 = 200**2 + c**2 -2 * 200 * c * cos (45° ) ; ; ; ; c**2 -282.843c +17500 =0 ; ; p=1; q=-282.843; r=17500 ; ; D = q**2 - 4pr = 282.843**2 - 4 * 1 * 17500 = 10000 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 282.84 ± sqrt{ 10000 } }{ 2 } ; ; c_{1,2} = fraction{ 282.84 ± 100 }{ 2 } ; ; c_{1,2} = 141.42135624 ± 50 ; ;
c_{1} = 191.42135624 ; ; c_{2} = 91.42135624 ; ; ; ; (c -191.42135624) (c -91.42135624) = 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 = 150 ; ; c = 191.42 ; ;

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

p = a+b+c = 200+150+191.42 = 541.42 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 541.42 }{ 2 } = 270.71 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 270.71 * (270.71-200)(270.71-150)(270.71-191.42) } ; ; T = sqrt{ 183210678.12 } = 13535.53 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 13535.53 }{ 200 } = 135.36 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 13535.53 }{ 150 } = 180.47 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 13535.53 }{ 191.42 } = 141.42 ; ;

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**2-150**2-191.42**2 }{ 2 * 150 * 191.42 } ) = 70° 31'44" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 150**2-200**2-191.42**2 }{ 2 * 200 * 191.42 } ) = 45° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 191.42**2-200**2-150**2 }{ 2 * 150 * 200 } ) = 64° 28'16" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 13535.53 }{ 270.71 } = 50 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 200 }{ 2 * sin 70° 31'44" } = 106.07 ; ;





#2 Obtuse scalene triangle.

Sides: a = 200   b = 150   c = 91.42113562373

Area: T = 6464.466609407
Perimeter: p = 441.4211356237
Semiperimeter: s = 220.7110678119

Angle ∠ A = α = 109.4711220634° = 109°28'16″ = 1.91106332362 rad
Angle ∠ B = β = 45° = 0.78553981634 rad
Angle ∠ C = γ = 25.52987793655° = 25°31'44″ = 0.44655612539 rad

Height: ha = 64.64546609407
Height: hb = 86.19328812542
Height: hc = 141.4211356237

Median: ma = 73.68112879104
Median: mb = 136.2132819471
Median: mc = 170.7654556937

Inradius: r = 29.28993218813
Circumradius: R = 106.0666017178

Vertex coordinates: A[91.42113562373; 0] B[0; 0] C[141.4211356237; 141.4211356237]
Centroid: CG[77.61442374915; 47.14404520791]
Coordinates of the circumscribed circle: U[45.71106781187; 95.71106781187]
Coordinates of the inscribed circle: I[70.71106781187; 29.28993218813]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 70.52987793655° = 70°31'44″ = 1.91106332362 rad
∠ B' = β' = 135° = 0.78553981634 rad
∠ C' = γ' = 154.4711220634° = 154°28'16″ = 0.44655612539 rad

Calculate another triangle

How did we calculate this triangle?

1. Use Law of Cosines

a = 200 ; ; b = 150 ; ; beta = 45° ; ; ; ; b**2 = a**2 + c**2 - 2ac cos beta ; ; 150**2 = 200**2 + c**2 -2 * 200 * c * cos (45° ) ; ; ; ; c**2 -282.843c +17500 =0 ; ; p=1; q=-282.843; r=17500 ; ; D = q**2 - 4pr = 282.843**2 - 4 * 1 * 17500 = 10000 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 282.84 ± sqrt{ 10000 } }{ 2 } ; ; c_{1,2} = fraction{ 282.84 ± 100 }{ 2 } ; ; c_{1,2} = 141.42135624 ± 50 ; ; : Nr. 1
c_{1} = 191.42135624 ; ; c_{2} = 91.42135624 ; ; ; ; (c -191.42135624) (c -91.42135624) = 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 = 150 ; ; c = 91.42 ; ;

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

p = a+b+c = 200+150+91.42 = 441.42 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 441.42 }{ 2 } = 220.71 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 220.71 * (220.71-200)(220.71-150)(220.71-91.42) } ; ; T = sqrt{ 41789321.88 } = 6464.47 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 6464.47 }{ 200 } = 64.64 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 6464.47 }{ 150 } = 86.19 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 6464.47 }{ 91.42 } = 141.42 ; ;

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**2-150**2-91.42**2 }{ 2 * 150 * 91.42 } ) = 109° 28'16" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 150**2-200**2-91.42**2 }{ 2 * 200 * 91.42 } ) = 45° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 91.42**2-200**2-150**2 }{ 2 * 150 * 200 } ) = 25° 31'44" ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 6464.47 }{ 220.71 } = 29.29 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 200 }{ 2 * sin 109° 28'16" } = 106.07 ; ;




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