Triangle calculator

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

You have entered side a, angle β and angle γ.

Right scalene triangle.

Sides: a = 33   b = 16.5   c = 28.57988383249

Area: T = 235.775541618
Perimeter: p = 78.07988383249
Semiperimeter: s = 39.03994191624

Angle ∠ A = α = 90° = 1.57107963268 rad
Angle ∠ B = β = 30° = 0.52435987756 rad
Angle ∠ C = γ = 60° = 1.04771975512 rad

Height: ha = 14.28994191624
Height: hb = 28.57988383249
Height: hc = 16.5

Median: ma = 16.5
Median: mb = 29.74657980226
Median: mc = 21.82774483163

Inradius: r = 6.03994191624
Circumradius: R = 16.5

Vertex coordinates: A[28.57988383249; 0] B[0; 0] C[28.57988383249; 16.5]
Centroid: CG[19.05325588833; 5.5]
Coordinates of the circumscribed circle: U[14.28994191624; 8.25]
Coordinates of the inscribed circle: I[22.53994191624; 6.03994191624]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 90° = 1.57107963268 rad
∠ B' = β' = 150° = 0.52435987756 rad
∠ C' = γ' = 120° = 1.04771975512 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 = 33 ; ; b = 16.5 ; ; c = 28.58 ; ;

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

p = a+b+c = 33+16.5+28.58 = 78.08 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 78.08 }{ 2 } = 39.04 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 39.04 * (39.04-33)(39.04-16.5)(39.04-28.58) } ; ; T = sqrt{ 55590.05 } = 235.78 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 235.78 }{ 33 } = 14.29 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 235.78 }{ 16.5 } = 28.58 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 235.78 }{ 28.58 } = 16.5 ; ;

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{ 33**2-16.5**2-28.58**2 }{ 2 * 16.5 * 28.58 } ) = 90° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 16.5**2-33**2-28.58**2 }{ 2 * 33 * 28.58 } ) = 30° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28.58**2-33**2-16.5**2 }{ 2 * 16.5 * 33 } ) = 60° ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 235.78 }{ 39.04 } = 6.04 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 33 }{ 2 * sin 90° } = 16.5 ; ;




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