Triangle calculator SAS

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


Acute isosceles triangle.

Sides: a = 40   b = 40   c = 30.61546745892

Area: T = 565.6855424949
Perimeter: p = 110.6154674589
Semiperimeter: s = 55.30773372946

Angle ∠ A = α = 67.5° = 67°30' = 1.17880972451 rad
Angle ∠ B = β = 67.5° = 67°30' = 1.17880972451 rad
Angle ∠ C = γ = 45° = 0.78553981634 rad

Height: ha = 28.28442712475
Height: hb = 28.28442712475
Height: hc = 36.95551813005

Median: ma = 29.47325151642
Median: mb = 29.47325151642
Median: mc = 36.95551813005

Inradius: r = 10.22880357837
Circumradius: R = 21.64878440058

Vertex coordinates: A[30.61546745892; 0] B[0; 0] C[15.30773372946; 36.95551813005]
Centroid: CG[15.30773372946; 12.31883937668]
Coordinates of the circumscribed circle: U[15.30773372946; 15.30773372946]
Coordinates of the inscribed circle: I[15.30773372946; 10.22880357837]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 112.5° = 112°30' = 1.17880972451 rad
∠ B' = β' = 112.5° = 112°30' = 1.17880972451 rad
∠ C' = γ' = 135° = 0.78553981634 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 = 40 ; ; b = 40 ; ; gamma = 45° ; ; ; ; c**2 = a**2+b**2 - 2ab cos gamma ; ; c = sqrt{ a**2+b**2 - 2ab cos gamma } ; ; c = sqrt{ 40**2+40**2 - 2 * 40 * 40 * cos(45° ) } ; ; c = 30.61 ; ;


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 = 40 ; ; b = 40 ; ; c = 30.61 ; ;

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

p = a+b+c = 40+40+30.61 = 110.61 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 110.61 }{ 2 } = 55.31 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 55.31 * (55.31-40)(55.31-40)(55.31-30.61) } ; ; T = sqrt{ 320000 } = 565.69 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 565.69 }{ 40 } = 28.28 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 565.69 }{ 40 } = 28.28 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 565.69 }{ 30.61 } = 36.96 ; ;

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{ 40**2-40**2-30.61**2 }{ 2 * 40 * 30.61 } ) = 67° 30' ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 40**2-40**2-30.61**2 }{ 2 * 40 * 30.61 } ) = 67° 30' ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30.61**2-40**2-40**2 }{ 2 * 40 * 40 } ) = 45° ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 565.69 }{ 55.31 } = 10.23 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 40 }{ 2 * sin 67° 30' } = 21.65 ; ;




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