Triangle calculator

Please enter what you know about the triangle:
Symbols definition of ABC triangle

You have entered side b, c and angle β.

Triangle has two solutions: a=66.23224612987; b=121; c=164 and a=185.0330116044; b=121; c=164.

#1 Obtuse scalene triangle.

Sides: a = 66.23224612987   b = 121   c = 164

Area: T = 3491.019924951
Perimeter: p = 351.2322461299
Semiperimeter: s = 175.6166230649

Angle ∠ A = α = 20.66002863319° = 20°36'1″ = 0.36595428233 rad
Angle ∠ B = β = 40° = 0.69881317008 rad
Angle ∠ C = γ = 119.4399713668° = 119°23'59″ = 2.08439181294 rad

Height: ha = 105.4177167989
Height: hb = 57.70327975126
Height: hc = 42.57334054818

Median: ma = 140.2566248587
Median: mb = 109.4588300119
Median: mc = 52.81992149207

Inradius: r = 19.87986822642
Circumradius: R = 94.12112915251

Vertex coordinates: A[164; 0] B[0; 0] C[50.7377008932; 42.57334054818]
Centroid: CG[71.57990029773; 14.19111351606]
Coordinates of the circumscribed circle: U[82; -46.20440855157]
Coordinates of the inscribed circle: I[54.61662306494; 19.87986822642]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 159.4399713668° = 159°23'59″ = 0.36595428233 rad
∠ B' = β' = 140° = 0.69881317008 rad
∠ C' = γ' = 60.66002863319° = 60°36'1″ = 2.08439181294 rad




How did we calculate this triangle?

1. Input data entered: side b, c and angle β.

b = 121 ; ; c = 164 ; ; beta = 40° ; ;

2. From angle β, c and side b we calculate a - by using the law of cosines and quadratic equation:

b**2 = c**2 + a**2 - 2c a cos beta ; ; ; ; 121**2 = 164**2 + a**2 - 2 * 164 * a * cos(40° ) ; ; ; ; ; ; a**2 -251.263a +12255 =0 ; ; a=1; b=-251.263; c=12255 ; ; D = b**2 - 4ac = 251.263**2 - 4 * 1 * 12255 = 14112.8827731 ; ; D>0 ; ; ; ; a_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 251.26 ± sqrt{ 14112.88 } }{ 2 } ; ; a_{1,2} = 125.63128867 ± 59.3988273728 ; ; a_{1} = 185.030116043 ; ; a_{2} = 66.2324612972 ; ;
 ; ; (a -185.030116043) (a -66.2324612972) = 0 ; ; ; ; a > 0 ; ; ; ; a = 185.03 ; ;


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 = 66.23 ; ; b = 121 ; ; c = 164 ; ;

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

p = a+b+c = 66.23+121+164 = 351.23 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 351.23 }{ 2 } = 175.62 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 175.62 * (175.62-66.23)(175.62-121)(175.62-164) } ; ; T = sqrt{ 12187215.4 } = 3491.02 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 3491.02 }{ 66.23 } = 105.42 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 3491.02 }{ 121 } = 57.7 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 3491.02 }{ 164 } = 42.57 ; ;

7. 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{ 66.23**2-121**2-164**2 }{ 2 * 121 * 164 } ) = 20° 36'1" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 121**2-66.23**2-164**2 }{ 2 * 66.23 * 164 } ) = 40° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 164**2-66.23**2-121**2 }{ 2 * 121 * 66.23 } ) = 119° 23'59" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 3491.02 }{ 175.62 } = 19.88 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 66.23 }{ 2 * sin 20° 36'1" } = 94.12 ; ;





#2 Acute scalene triangle.

Sides: a = 185.0330116044   b = 121   c = 164

Area: T = 9752.6755413
Perimeter: p = 470.0330116044
Semiperimeter: s = 235.0155058022

Angle ∠ A = α = 79.43997136681° = 79°23'59″ = 1.38657864286 rad
Angle ∠ B = β = 40° = 0.69881317008 rad
Angle ∠ C = γ = 60.66002863319° = 60°36'1″ = 1.05876745241 rad

Height: ha = 105.4177167989
Height: hb = 161.2011246496
Height: hc = 118.9355066012

Median: ma = 110.4966443559
Median: mb = 164.032994215
Median: mc = 133.0966100325

Inradius: r = 41.49880873782
Circumradius: R = 94.12112915251

Vertex coordinates: A[164; 0] B[0; 0] C[141.7411292205; 118.9355066012]
Centroid: CG[101.9143764069; 39.64550220041]
Coordinates of the circumscribed circle: U[82; 46.20440855157]
Coordinates of the inscribed circle: I[114.0155058022; 41.49880873782]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 100.6600286332° = 100°36'1″ = 1.38657864286 rad
∠ B' = β' = 140° = 0.69881317008 rad
∠ C' = γ' = 119.4399713668° = 119°23'59″ = 1.05876745241 rad

Calculate another triangle

How did we calculate this triangle?

1. Input data entered: side b, c and angle β.

b = 121 ; ; c = 164 ; ; beta = 40° ; ; : Nr. 1

2. From angle β, c and side b we calculate a - by using the law of cosines and quadratic equation:

b**2 = c**2 + a**2 - 2c a cos beta ; ; ; ; 121**2 = 164**2 + a**2 - 2 * 164 * a * cos(40° ) ; ; ; ; ; ; a**2 -251.263a +12255 =0 ; ; a=1; b=-251.263; c=12255 ; ; D = b**2 - 4ac = 251.263**2 - 4 * 1 * 12255 = 14112.8827731 ; ; D>0 ; ; ; ; a_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 251.26 ± sqrt{ 14112.88 } }{ 2 } ; ; a_{1,2} = 125.63128867 ± 59.3988273728 ; ; a_{1} = 185.030116043 ; ; a_{2} = 66.2324612972 ; ; : Nr. 1
 ; ; (a -185.030116043) (a -66.2324612972) = 0 ; ; ; ; a > 0 ; ; ; ; a = 185.03 ; ; : 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 = 185.03 ; ; b = 121 ; ; c = 164 ; ;

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

p = a+b+c = 185.03+121+164 = 470.03 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 470.03 }{ 2 } = 235.02 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 235.02 * (235.02-185.03)(235.02-121)(235.02-164) } ; ; T = sqrt{ 95114677.71 } = 9752.68 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 9752.68 }{ 185.03 } = 105.42 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 9752.68 }{ 121 } = 161.2 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 9752.68 }{ 164 } = 118.94 ; ;

7. 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{ 185.03**2-121**2-164**2 }{ 2 * 121 * 164 } ) = 79° 23'59" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 121**2-185.03**2-164**2 }{ 2 * 185.03 * 164 } ) = 40° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 164**2-185.03**2-121**2 }{ 2 * 121 * 185.03 } ) = 60° 36'1" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 9752.68 }{ 235.02 } = 41.5 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 185.03 }{ 2 * sin 79° 23'59" } = 94.12 ; ;

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