Triangle calculator

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

You have entered side a, c and angle γ.

Triangle has two solutions: a=380; b=252.6633368508; c=150 and a=380; b=482.4660123602; c=150.

#1 Obtuse scalene triangle.

Sides: a = 380   b = 252.6633368508   c = 150

Area: T = 12181.91440423
Perimeter: p = 782.6633368508
Semiperimeter: s = 391.3321684254

Angle ∠ A = α = 139.9955074621° = 139°59'42″ = 2.44333749887 rad
Angle ∠ B = β = 25.30549253788° = 25°18'18″ = 0.44216542648 rad
Angle ∠ C = γ = 14.7° = 14°42' = 0.25765634 rad

Height: ha = 64.11553370645
Height: hb = 96.42880189422
Height: hc = 162.4265520563

Median: ma = 84.0879658021
Median: mb = 259.7898963495
Median: mc = 313.8388157165

Inradius: r = 31.12993834167
Circumradius: R = 295.5577248947

Vertex coordinates: A[150; 0] B[0; 0] C[343.537740738; 162.4265520563]
Centroid: CG[164.5122469127; 54.14218401878]
Coordinates of the circumscribed circle: U[75; 285.8832996006]
Coordinates of the inscribed circle: I[138.6688315746; 31.12993834167]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 40.00549253788° = 40°18″ = 2.44333749887 rad
∠ B' = β' = 154.6955074621° = 154°41'42″ = 0.44216542648 rad
∠ C' = γ' = 165.3° = 165°18' = 0.25765634 rad




How did we calculate this triangle?

1. Input data entered: side a, c and angle γ.

a = 380 ; ; c = 150 ; ; gamma = 14.7° ; ;

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

c**2 = a**2 + b**2 - 2a b cos gamma ; ; ; ; 150**2 = 380**2 + b**2 - 2 * 380 * b * cos 14° 42' ; ; ; ; ; ; b**2 -735.123b +121900 =0 ; ; p=1; q=-735.123; r=121900 ; ; D = q**2 - 4pr = 735.123**2 - 4 * 1 * 121900 = 52806.5486515 ; ; D>0 ; ; ; ; b_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 735.12 ± sqrt{ 52806.55 } }{ 2 } ; ; b_{1,2} = 367.56174605 ± 114.898377547 ; ; b_{1} = 482.460123597 ; ; b_{2} = 252.663368503 ; ;
 ; ; text{ Factored form: } ; ; (b -482.460123597) (b -252.663368503) = 0 ; ; ; ; b > 0 ; ; ; ; b = 482.46 ; ;
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 = 380 ; ; b = 252.66 ; ; c = 150 ; ;

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

p = a+b+c = 380+252.66+150 = 782.66 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 782.66 }{ 2 } = 391.33 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 391.33 * (391.33-380)(391.33-252.66)(391.33-150) } ; ; T = sqrt{ 148399029.73 } = 12181.91 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 12181.91 }{ 380 } = 64.12 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 12181.91 }{ 252.66 } = 96.43 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 12181.91 }{ 150 } = 162.43 ; ;

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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 252.66**2+150**2-380**2 }{ 2 * 252.66 * 150 } ) = 139° 59'42" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 380**2+150**2-252.66**2 }{ 2 * 380 * 150 } ) = 25° 18'18" ; ; gamma = 180° - alpha - beta = 180° - 139° 59'42" - 25° 18'18" = 14° 42' ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 12181.91 }{ 391.33 } = 31.13 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 380 }{ 2 * sin 139° 59'42" } = 295.56 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 252.66**2+2 * 150**2 - 380**2 } }{ 2 } = 84.08 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 150**2+2 * 380**2 - 252.66**2 } }{ 2 } = 259.789 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 252.66**2+2 * 380**2 - 150**2 } }{ 2 } = 313.838 ; ;







#2 Obtuse scalene triangle.

Sides: a = 380   b = 482.4660123602   c = 150

Area: T = 23261.33769688
Perimeter: p = 1012.46601236
Semiperimeter: s = 506.2330061801

Angle ∠ A = α = 40.00549253788° = 40°18″ = 0.69882176649 rad
Angle ∠ B = β = 125.2955074621° = 125°17'42″ = 2.18768115887 rad
Angle ∠ C = γ = 14.7° = 14°42' = 0.25765634 rad

Height: ha = 122.4288089309
Height: hb = 96.42880189422
Height: hc = 310.1511159583

Median: ma = 302.5465674953
Median: mb = 158.9287836717
Median: mc = 427.7376934848

Inradius: r = 45.95501296427
Circumradius: R = 295.5577248947

Vertex coordinates: A[150; 0] B[0; 0] C[-219.5599236219; 310.1511159583]
Centroid: CG[-23.1866412073; 103.3843719861]
Coordinates of the circumscribed circle: U[75; 285.8832996006]
Coordinates of the inscribed circle: I[23.77699381992; 45.95501296427]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 139.9955074621° = 139°59'42″ = 0.69882176649 rad
∠ B' = β' = 54.70549253788° = 54°42'18″ = 2.18768115887 rad
∠ C' = γ' = 165.3° = 165°18' = 0.25765634 rad

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How did we calculate this triangle?

1. Input data entered: side a, c and angle γ.

a = 380 ; ; c = 150 ; ; gamma = 14.7° ; ; : Nr. 1

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

c**2 = a**2 + b**2 - 2a b cos gamma ; ; ; ; 150**2 = 380**2 + b**2 - 2 * 380 * b * cos 14° 42' ; ; ; ; ; ; b**2 -735.123b +121900 =0 ; ; p=1; q=-735.123; r=121900 ; ; D = q**2 - 4pr = 735.123**2 - 4 * 1 * 121900 = 52806.5486515 ; ; D>0 ; ; ; ; b_{1,2} = fraction{ -q ± sqrt{ D } }{ 2p } = fraction{ 735.12 ± sqrt{ 52806.55 } }{ 2 } ; ; b_{1,2} = 367.56174605 ± 114.898377547 ; ; b_{1} = 482.460123597 ; ; b_{2} = 252.663368503 ; ; : Nr. 1
 ; ; text{ Factored form: } ; ; (b -482.460123597) (b -252.663368503) = 0 ; ; ; ; b > 0 ; ; ; ; b = 482.46 ; ; : 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 = 380 ; ; b = 482.46 ; ; c = 150 ; ;

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

p = a+b+c = 380+482.46+150 = 1012.46 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1012.46 }{ 2 } = 506.23 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 506.23 * (506.23-380)(506.23-482.46)(506.23-150) } ; ; T = sqrt{ 541089797.57 } = 23261.34 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 23261.34 }{ 380 } = 122.43 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 23261.34 }{ 482.46 } = 96.43 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 23261.34 }{ 150 } = 310.15 ; ;

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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 482.46**2+150**2-380**2 }{ 2 * 482.46 * 150 } ) = 40° 18" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 380**2+150**2-482.46**2 }{ 2 * 380 * 150 } ) = 125° 17'42" ; ; gamma = 180° - alpha - beta = 180° - 40° 18" - 125° 17'42" = 14° 42' ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 23261.34 }{ 506.23 } = 45.95 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 380 }{ 2 * sin 40° 18" } = 295.56 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 482.46**2+2 * 150**2 - 380**2 } }{ 2 } = 302.546 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 150**2+2 * 380**2 - 482.46**2 } }{ 2 } = 158.928 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 482.46**2+2 * 380**2 - 150**2 } }{ 2 } = 427.737 ; ;
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