Triangle calculator

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

You have entered height hc, angle α and angle β.

Acute isosceles triangle.

Sides: a = 17.32105080757   b = 17.32105080757   c = 17.32105080757

Area: T = 129.9043810568
Perimeter: p = 51.96215242271
Semiperimeter: s = 25.98107621135

Angle ∠ A = α = 60° = 1.04771975512 rad
Angle ∠ B = β = 60° = 1.04771975512 rad
Angle ∠ C = γ = 60° = 1.04771975512 rad

Height: ha = 15
Height: hb = 15
Height: hc = 15

Median: ma = 15
Median: mb = 15
Median: mc = 15

Inradius: r = 5
Circumradius: R = 10

Vertex coordinates: A[17.32105080757; 0] B[0; 0] C[8.66602540378; 15]
Centroid: CG[8.66602540378; 5]
Coordinates of the circumscribed circle: U[8.66602540378; 5]
Coordinates of the inscribed circle: I[8.66602540378; 5]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 120° = 1.04771975512 rad
∠ B' = β' = 120° = 1.04771975512 rad
∠ C' = γ' = 120° = 1.04771975512 rad

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How did we calculate this triangle?

1. Input data entered: angle α, angle β and height hc.

 alpha = 60° ; ; beta = 60° ; ; hc = 15 ; ;

2. From angle α and angle β we calculate γ:

 alpha + beta + gamma = 180° ; ; gamma = 180° - alpha - beta = 180° - 60 ° - 60 ° = 60 ° ; ;

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

b**2 = a**2 + c**2 - 2a c cos beta ; ; ; ; 17.321**2 = 17.321**2 + c**2 - 2 * 17.321 * c * cos(60° ) ; ; ; ; ; ; c**2 -17.321c =0 ; ; a=1; b=-17.321; c=0 ; ; D = b**2 - 4ac = 17.321**2 - 4 * 1 * 0 = 300 ; ; D>0 ; ; ; ; c_{1,2} = fraction{ -b ± sqrt{ D } }{ 2a } = fraction{ 17.32 ± sqrt{ 300 } }{ 2 } ; ; c_{1,2} = 8.66025404 ± 8.66025403784 ; ; c_{1} = 10 sqrt{ 3} = 17.3205080778 ; ; c_{2} = 2.15561080097E-9 ; ;
 ; ; (c -17.3205080778) (c -2.15561080097E-9) = 0 ; ; ; ; c > 0 ; ; ; ; c = 17.321 ; ;


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 = 17.32 ; ; b = 17.32 ; ; c = 17.32 ; ;

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

p = a+b+c = 17.32+17.32+17.32 = 51.96 ; ;

5. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 51.96 }{ 2 } = 25.98 ; ;

6. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 25.98 * (25.98-17.32)(25.98-17.32)(25.98-17.32) } ; ; T = sqrt{ 16875 } = 129.9 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 129.9 }{ 17.32 } = 15 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 129.9 }{ 17.32 } = 15 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 129.9 }{ 17.32 } = 15 ; ;

8. 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{ 17.32**2-17.32**2-17.32**2 }{ 2 * 17.32 * 17.32 } ) = 60° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 17.32**2-17.32**2-17.32**2 }{ 2 * 17.32 * 17.32 } ) = 60° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 17.32**2-17.32**2-17.32**2 }{ 2 * 17.32 * 17.32 } ) = 60° ; ;

9. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 129.9 }{ 25.98 } = 5 ; ;

10. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 17.32 }{ 2 * sin 60° } = 10 ; ;




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