Triangle calculator

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

You have entered side a, b, c and angle β.

Acute scalene triangle.

Sides: a = 50   b = 111.8   c = 100

Area: T = 25009.99999278
Perimeter: p = 261.8
Semiperimeter: s = 130.9

Angle ∠ A = α = 26.56659220332° = 26°33'57″ = 0.46436628083 rad
Angle ∠ B = β = 89.99656455208° = 89°59'44″ = 1.57107203268 rad
Angle ∠ C = γ = 63.4388432446° = 63°26'18″ = 1.10772095185 rad

Height: ha = 100.9999997112
Height: hb = 44.72327190122
Height: hc = 509.9999998556

Median: ma = 103.0765797353
Median: mb = 55.90333988233
Median: mc = 70.70879910618

Inradius: r = 19.09985484552
Circumradius: R = 55.99000001614

Vertex coordinates: A[100; 0] B[0; 0] C[0.00438; 509.9999998556]
Centroid: CG[33.33546; 16.66766666185]
Coordinates of the circumscribed circle: U[50; 24.99662000722]
Coordinates of the inscribed circle: I[19.1; 19.09985484552]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 153.4344077967° = 153°26'3″ = 0.46436628083 rad
∠ B' = β' = 90.00443544792° = 90°16″ = 1.57107203268 rad
∠ C' = γ' = 116.5621567554° = 116°33'42″ = 1.10772095185 rad

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How did we calculate this triangle?

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 = 50 ; ; b = 111.8 ; ; c = 100 ; ;

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

p = a+b+c = 50+111.8+100 = 261.8 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 261.8 }{ 2 } = 130.9 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 130.9 * (130.9-50)(130.9-111.8)(130.9-100) } ; ; T = sqrt{ 6249999.96 } = 2500 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2500 }{ 50 } = 100 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2500 }{ 111.8 } = 44.72 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2500 }{ 100 } = 50 ; ;

5. 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{ 50**2-111.8**2-100**2 }{ 2 * 111.8 * 100 } ) = 26° 33'57" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 111.8**2-50**2-100**2 }{ 2 * 50 * 100 } ) = 89° 59'44" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 100**2-50**2-111.8**2 }{ 2 * 111.8 * 50 } ) = 63° 26'18" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 2500 }{ 130.9 } = 19.1 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 50 }{ 2 * sin 26° 33'57" } = 55.9 ; ;




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