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=441.59; b=125.9155131498; c=378.38 and a=441.59; b=411.6298873222; c=378.38.

#1 Obtuse scalene triangle.

Sides: a = 441.59   b = 125.9155131498   c = 378.38

Area: T = 22058.82198297
Perimeter: p = 945.8855131498
Semiperimeter: s = 472.9432565749

Angle ∠ A = α = 112.1821924892° = 112°10'55″ = 1.95879439506 rad
Angle ∠ B = β = 15.31097418076° = 15°18'35″ = 0.26772054022 rad
Angle ∠ C = γ = 52.50883333° = 52°30'30″ = 0.91664433008 rad

Height: ha = 99.90663376873
Height: hb = 350.3765996392
Height: hc = 116.5966119402

Median: ma = 175.3932674719
Median: mb = 406.3533197557
Median: mc = 263.885504717

Inradius: r = 46.64216462108
Circumradius: R = 238.4422167729

Vertex coordinates: A[378.38; 0] B[0; 0] C[425.9199092129; 116.5966119402]
Centroid: CG[268.1099697376; 38.8655373134]
Coordinates of the circumscribed circle: U[189.19; 145.1276879838]
Coordinates of the inscribed circle: I[347.0277434251; 46.64216462108]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 67.81880751076° = 67°49'5″ = 1.95879439506 rad
∠ B' = β' = 164.6990258192° = 164°41'25″ = 0.26772054022 rad
∠ C' = γ' = 127.49216667° = 127°29'30″ = 0.91664433008 rad




How did we calculate this triangle?

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

a = 441.59 ; ; c = 378.38 ; ; gamma = 52.508° ; ;

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

c**2 = a**2 + b**2 - 2a b cos gamma ; ; ; ; 378.38**2 = 441.59**2 + b**2 - 2 * 441.59 * b * cos(52° 30'30") ; ; ; ; ; ; b**2 -537.544b +51830.304 =0 ; ; a=1; b=-537.544; c=51830.304 ; ; D = b**2 - 4ac = 537.544**2 - 4 * 1 * 51830.304 = 81632.3422099 ; ; D>0 ; ; ; ; b_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 537.54 ± sqrt{ 81632.34 } }{ 2 } ; ; b_{1,2} = 268.77200236 ± 142.856870862 ; ; b_{1} = 411.628873222 ; ;
b_{2} = 125.915131498 ; ; ; ; (b -411.628873222) (b -125.915131498) = 0 ; ; ; ; b > 0 ; ; ; ; b = 411.629 ; ;


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 = 441.59 ; ; b = 125.92 ; ; c = 378.38 ; ;

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

p = a+b+c = 441.59+125.92+378.38 = 945.89 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 945.89 }{ 2 } = 472.94 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 472.94 * (472.94-441.59)(472.94-125.92)(472.94-378.38) } ; ; T = sqrt{ 486591532.28 } = 22058.82 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 22058.82 }{ 441.59 } = 99.91 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 22058.82 }{ 125.92 } = 350.38 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 22058.82 }{ 378.38 } = 116.6 ; ;

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{ 441.59**2-125.92**2-378.38**2 }{ 2 * 125.92 * 378.38 } ) = 112° 10'55" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 125.92**2-441.59**2-378.38**2 }{ 2 * 441.59 * 378.38 } ) = 15° 18'35" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 378.38**2-441.59**2-125.92**2 }{ 2 * 125.92 * 441.59 } ) = 52° 30'30" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 22058.82 }{ 472.94 } = 46.64 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 441.59 }{ 2 * sin 112° 10'55" } = 238.44 ; ;





#2 Acute scalene triangle.

Sides: a = 441.59   b = 411.6298873222   c = 378.38

Area: T = 72112.43882994
Perimeter: p = 1231.599887322
Semiperimeter: s = 615.7999436611

Angle ∠ A = α = 67.81880751076° = 67°49'5″ = 1.1843648703 rad
Angle ∠ B = β = 59.67435915924° = 59°40'25″ = 1.04215006498 rad
Angle ∠ C = γ = 52.50883333° = 52°30'30″ = 0.91664433008 rad

Height: ha = 326.6043583865
Height: hb = 350.3765996392
Height: hc = 381.1644111736

Median: ma = 327.9554943262
Median: mb = 355.9877350804
Median: mc = 382.6588036091

Inradius: r = 117.1043774398
Circumradius: R = 238.4422167729

Vertex coordinates: A[378.38; 0] B[0; 0] C[222.9770060825; 381.1644111736]
Centroid: CG[200.4550020275; 127.0554703912]
Coordinates of the circumscribed circle: U[189.19; 145.1276879838]
Coordinates of the inscribed circle: I[204.1710563389; 117.1043774398]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 112.1821924892° = 112°10'55″ = 1.1843648703 rad
∠ B' = β' = 120.3266408408° = 120°19'35″ = 1.04215006498 rad
∠ C' = γ' = 127.49216667° = 127°29'30″ = 0.91664433008 rad

Calculate another triangle

How did we calculate this triangle?

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

a = 441.59 ; ; c = 378.38 ; ; gamma = 52.508° ; ; : Nr. 1

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

c**2 = a**2 + b**2 - 2a b cos gamma ; ; ; ; 378.38**2 = 441.59**2 + b**2 - 2 * 441.59 * b * cos(52° 30'30") ; ; ; ; ; ; b**2 -537.544b +51830.304 =0 ; ; a=1; b=-537.544; c=51830.304 ; ; D = b**2 - 4ac = 537.544**2 - 4 * 1 * 51830.304 = 81632.3422099 ; ; D>0 ; ; ; ; b_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 537.54 ± sqrt{ 81632.34 } }{ 2 } ; ; b_{1,2} = 268.77200236 ± 142.856870862 ; ; b_{1} = 411.628873222 ; ; : Nr. 1
b_{2} = 125.915131498 ; ; ; ; (b -411.628873222) (b -125.915131498) = 0 ; ; ; ; b > 0 ; ; ; ; b = 411.629 ; ; : 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 = 441.59 ; ; b = 411.63 ; ; c = 378.38 ; ;

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

p = a+b+c = 441.59+411.63+378.38 = 1231.6 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1231.6 }{ 2 } = 615.8 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 615.8 * (615.8-441.59)(615.8-411.63)(615.8-378.38) } ; ; T = sqrt{ 5200203757.48 } = 72112.44 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 72112.44 }{ 441.59 } = 326.6 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 72112.44 }{ 411.63 } = 350.38 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 72112.44 }{ 378.38 } = 381.16 ; ;

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{ 441.59**2-411.63**2-378.38**2 }{ 2 * 411.63 * 378.38 } ) = 67° 49'5" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 411.63**2-441.59**2-378.38**2 }{ 2 * 441.59 * 378.38 } ) = 59° 40'25" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 378.38**2-441.59**2-411.63**2 }{ 2 * 411.63 * 441.59 } ) = 52° 30'30" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 72112.44 }{ 615.8 } = 117.1 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 441.59 }{ 2 * sin 67° 49'5" } = 238.44 ; ;




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