Triangle calculator - result

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

You have entered side b, c and area T.

Triangle has two solutions: a=5.76999996488; b=4; c=2 and a=2.74404386516; b=4; c=2.

#1 Obtuse scalene triangle.

Sides: a = 5.76999996488   b = 4   c = 2

Area: T = 2.5
Perimeter: p = 11.76999996488
Semiperimeter: s = 5.85499998244

Angle ∠ A = α = 141.3187812547° = 141°19'4″ = 2.46664611207 rad
Angle ∠ B = β = 26.01443677223° = 26°52″ = 0.45440363696 rad
Angle ∠ C = γ = 12.66878197312° = 12°40'4″ = 0.22110951634 rad

Height: ha = 0.87771930365
Height: hb = 1.25
Height: hc = 2.5

Median: ma = 1.37702193258
Median: mb = 3.77442546282
Median: mc = 4.82113066692

Inradius: r = 0.42773504402
Circumradius: R = 4.56599997191

Vertex coordinates: A[2; 0] B[0; 0] C[5.12224989992; 2.5]
Centroid: CG[2.37441663331; 0.83333333333]
Coordinates of the circumscribed circle: U[1; 4.44989995997]
Coordinates of the inscribed circle: I[1.85499998244; 0.42773504402]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 38.68221874535° = 38°40'56″ = 2.46664611207 rad
∠ B' = β' = 153.9865632278° = 153°59'8″ = 0.45440363696 rad
∠ C' = γ' = 167.3322180269° = 167°19'56″ = 0.22110951634 rad


How did we calculate this triangle?

1. Input data entered: side b, c and area T.

b = 4 ; ; c = 2 ; ; T = 2.5 ; ;

2. From area T, side b and side c we calculate side a - using Heron's formula for the area and solve of the bikvadratic equation:

s = fraction{ a+b+c }{ 2 } ; ; T**2 = s(s-a)(s-b)(s-c) ; ; ; ; s = fraction{ a+4+2 }{ 2 } = fraction{ a+6 }{ 2 } = a/2 + 3 ; ; ; ; T**2 = s(s-a)(s-b)(s-c) ; ; T**2 = ( a/2 + 3) ( a/2 + 3-a) ( a/2 + 3-4) ( a/2 + 3 - 2) ; ; ; ; 2.5**2 = ( a/2 + 3) ( 3-a/2) ( a/2 + (-1)) ( a/2 + 1) ; ; 100 = ( a + 6) ( 6-a) ( a + (-2)) ( a + 2) ; ;
 ; ; D = b**2 * c**2 - 4 * T**2 = 4**2 * 2**2 - 4 * 2.5**2 = 39 ; ; ; ; D_1 = -2 * sqrt{ D } + b**2 + c**2 = -2 * sqrt{ 39 } + 4**2 + 2**2 = 7.51 ; ; D_2 = 2 * sqrt{ D } + b**2 + c**2 = 2 * sqrt{ 39 } + 4**2 + 2**2 = 32.49 ; ; ; ; a_1 = sqrt{ D_1 } = sqrt{ 7.51 } = 2.74 ; ; a_2 = sqrt{ D_2 } = sqrt{ 32.49 } = 5.7 ; ;
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 = 5.7 ; ; b = 4 ; ; c = 2 ; ;

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

p = a+b+c = 5.7+4+2 = 11.7 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 11.7 }{ 2 } = 5.85 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 5.85 * (5.85-5.7)(5.85-4)(5.85-2) } ; ; T = sqrt{ 6.25 } = 2.5 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2.5 }{ 5.7 } = 0.88 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2.5 }{ 4 } = 1.25 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2.5 }{ 2 } = 2.5 ; ;

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{ 4**2+2**2-5.7**2 }{ 2 * 4 * 2 } ) = 141° 19'4" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 5.7**2+2**2-4**2 }{ 2 * 5.7 * 2 } ) = 26° 52" ; ; gamma = 180° - alpha - beta = 180° - 141° 19'4" - 26° 52" = 12° 40'4" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 2.5 }{ 5.85 } = 0.43 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 5.7 }{ 2 * sin 141° 19'4" } = 4.56 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 4**2+2 * 2**2 - 5.7**2 } }{ 2 } = 1.37 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 2**2+2 * 5.7**2 - 4**2 } }{ 2 } = 3.774 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 4**2+2 * 5.7**2 - 2**2 } }{ 2 } = 4.821 ; ;



#2 Obtuse scalene triangle.

Sides: a = 2.74404386516   b = 4   c = 2

Area: T = 2.5
Perimeter: p = 8.74404386516
Semiperimeter: s = 4.37702193258

Angle ∠ A = α = 38.68221874535° = 38°40'56″ = 0.67551315329 rad
Angle ∠ B = β = 114.1880061185° = 114°10'48″ = 1.99328180078 rad
Angle ∠ C = γ = 27.13877513611° = 27°8'16″ = 0.47436431128 rad

Height: ha = 1.82545254266
Height: hb = 1.25
Height: hc = 2.5

Median: ma = 2.85499998244
Median: mb = 1.32547648854
Median: mc = 3.27994819715

Inradius: r = 0.57220536691
Circumradius: R = 2.19223509213

Vertex coordinates: A[2; 0] B[0; 0] C[-1.12224989992; 2.5]
Centroid: CG[0.29325003336; 0.83333333333]
Coordinates of the circumscribed circle: U[1; 1.95110004003]
Coordinates of the inscribed circle: I[0.37702193258; 0.57220536691]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 141.3187812547° = 141°19'4″ = 0.67551315329 rad
∠ B' = β' = 65.82199388146° = 65°49'12″ = 1.99328180078 rad
∠ C' = γ' = 152.8622248639° = 152°51'44″ = 0.47436431128 rad

Calculate another triangle

How did we calculate this triangle?

1. Input data entered: side b, c and area T.

b = 4 ; ; c = 2 ; ; T = 2.5 ; ; : Nr. 1

2. From area T, side b and side c we calculate side a - using Heron's formula for the area and solve of the bikvadratic equation:

s = fraction{ a+b+c }{ 2 } ; ; T**2 = s(s-a)(s-b)(s-c) ; ; ; ; s = fraction{ a+4+2 }{ 2 } = fraction{ a+6 }{ 2 } = a/2 + 3 ; ; ; ; T**2 = s(s-a)(s-b)(s-c) ; ; T**2 = ( a/2 + 3) ( a/2 + 3-a) ( a/2 + 3-4) ( a/2 + 3 - 2) ; ; ; ; 2.5**2 = ( a/2 + 3) ( 3-a/2) ( a/2 + (-1)) ( a/2 + 1) ; ; 100 = ( a + 6) ( 6-a) ( a + (-2)) ( a + 2) ; ; : Nr. 1
 ; ; D = b**2 * c**2 - 4 * T**2 = 4**2 * 2**2 - 4 * 2.5**2 = 39 ; ; ; ; D_1 = -2 * sqrt{ D } + b**2 + c**2 = -2 * sqrt{ 39 } + 4**2 + 2**2 = 7.51 ; ; D_2 = 2 * sqrt{ D } + b**2 + c**2 = 2 * sqrt{ 39 } + 4**2 + 2**2 = 32.49 ; ; ; ; a_1 = sqrt{ D_1 } = sqrt{ 7.51 } = 2.74 ; ; a_2 = sqrt{ D_2 } = sqrt{ 32.49 } = 5.7 ; ; : 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 = 2.74 ; ; b = 4 ; ; c = 2 ; ;

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

p = a+b+c = 2.74+4+2 = 8.74 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 8.74 }{ 2 } = 4.37 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 4.37 * (4.37-2.74)(4.37-4)(4.37-2) } ; ; T = sqrt{ 6.25 } = 2.5 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2.5 }{ 2.74 } = 1.82 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2.5 }{ 4 } = 1.25 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2.5 }{ 2 } = 2.5 ; ;

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{ 4**2+2**2-2.74**2 }{ 2 * 4 * 2 } ) = 38° 40'56" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 2.74**2+2**2-4**2 }{ 2 * 2.74 * 2 } ) = 114° 10'48" ; ;
 gamma = 180° - alpha - beta = 180° - 38° 40'56" - 114° 10'48" = 27° 8'16" ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 2.5 }{ 4.37 } = 0.57 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 2.74 }{ 2 * sin 38° 40'56" } = 2.19 ; ;

10. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 4**2+2 * 2**2 - 2.74**2 } }{ 2 } = 2.85 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 2**2+2 * 2.74**2 - 4**2 } }{ 2 } = 1.325 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 4**2+2 * 2.74**2 - 2**2 } }{ 2 } = 3.279 ; ;
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