Triangle calculator SAS

Please enter two sides of the triangle and the included angle
°


Acute isosceles triangle.

Sides: a = 31.94   b = 31.94   c = 26.997685456

Area: T = 390.7455328426
Perimeter: p = 90.877685456
Semiperimeter: s = 45.438842728

Angle ∠ A = α = 65° = 1.13444640138 rad
Angle ∠ B = β = 65° = 1.13444640138 rad
Angle ∠ C = γ = 50° = 0.8732664626 rad

Height: ha = 24.46774595132
Height: hb = 24.46774595132
Height: hc = 28.9477470718

Median: ma = 24.88988725752
Median: mb = 24.88988725752
Median: mc = 28.9477470718

Inradius: r = 8.59994465878
Circumradius: R = 17.62109453658

Vertex coordinates: A[26.997685456; 0] B[0; 0] C[13.498842728; 28.9477470718]
Centroid: CG[13.498842728; 9.6499156906]
Coordinates of the circumscribed circle: U[13.498842728; 11.32765253521]
Coordinates of the inscribed circle: I[13.498842728; 8.59994465878]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 115° = 1.13444640138 rad
∠ B' = β' = 115° = 1.13444640138 rad
∠ C' = γ' = 130° = 0.8732664626 rad

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How did we calculate this triangle?

1. Calculation of the third side c of the triangle using a Law of Cosines

a = 31.94 ; ; b = 31.94 ; ; gamma = 50° ; ; ; ; c**2 = a**2+b**2 - 2ab cos( gamma ) ; ; c = sqrt{ a**2+b**2 - 2ab cos( gamma ) } ; ; c = sqrt{ 31.94**2+31.94**2 - 2 * 31.94 * 31.94 * cos(50° ) } ; ; c = 27 ; ;


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 = 31.94 ; ; b = 31.94 ; ; c = 27 ; ;

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

p = a+b+c = 31.94+31.94+27 = 90.88 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 90.88 }{ 2 } = 45.44 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 45.44 * (45.44-31.94)(45.44-31.94)(45.44-27) } ; ; T = sqrt{ 152681.91 } = 390.75 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 390.75 }{ 31.94 } = 24.47 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 390.75 }{ 31.94 } = 24.47 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 390.75 }{ 27 } = 28.95 ; ;

6. 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{ 31.94**2-31.94**2-27**2 }{ 2 * 31.94 * 27 } ) = 65° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 31.94**2-31.94**2-27**2 }{ 2 * 31.94 * 27 } ) = 65° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-31.94**2-31.94**2 }{ 2 * 31.94 * 31.94 } ) = 50° ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 390.75 }{ 45.44 } = 8.6 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 31.94 }{ 2 * sin 65° } = 17.62 ; ;




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