Triangle calculator SAS

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


Acute isosceles triangle.

Sides: a = 2.5   b = 2.5   c = 1.08221980697

Area: T = 1.32106820679
Perimeter: p = 6.08221980697
Semiperimeter: s = 3.04110990348

Angle ∠ A = α = 77.5° = 77°30' = 1.35326301703 rad
Angle ∠ B = β = 77.5° = 77°30' = 1.35326301703 rad
Angle ∠ C = γ = 25° = 0.4366332313 rad

Height: ha = 1.05765456544
Height: hb = 1.05765456544
Height: hc = 2.44107400178

Median: ma = 1.46656317174
Median: mb = 1.46656317174
Median: mc = 2.44107400178

Inradius: r = 0.43442778886
Circumradius: R = 1.28803493929

Vertex coordinates: A[1.08221980697; 0] B[0; 0] C[0.54110990348; 2.44107400178]
Centroid: CG[0.54110990348; 0.81435800059]
Coordinates of the circumscribed circle: U[0.54110990348; 1.16603906249]
Coordinates of the inscribed circle: I[0.54110990348; 0.43442778886]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 102.5° = 102°30' = 1.35326301703 rad
∠ B' = β' = 102.5° = 102°30' = 1.35326301703 rad
∠ C' = γ' = 155° = 0.4366332313 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 = 2.5 ; ; b = 2.5 ; ; gamma = 25° ; ; ; ; c**2 = a**2+b**2 - 2ab cos( gamma ) ; ; c = sqrt{ a**2+b**2 - 2ab cos( gamma ) } ; ; c = sqrt{ 2.5**2+2.5**2 - 2 * 2.5 * 2.5 * cos(25° ) } ; ; c = 1.08 ; ;


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 = 2.5 ; ; b = 2.5 ; ; c = 1.08 ; ;

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

p = a+b+c = 2.5+2.5+1.08 = 6.08 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 6.08 }{ 2 } = 3.04 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 3.04 * (3.04-2.5)(3.04-2.5)(3.04-1.08) } ; ; T = sqrt{ 1.74 } = 1.32 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 1.32 }{ 2.5 } = 1.06 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 1.32 }{ 2.5 } = 1.06 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 1.32 }{ 1.08 } = 2.44 ; ;

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{ 2.5**2-2.5**2-1.08**2 }{ 2 * 2.5 * 1.08 } ) = 77° 30' ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 2.5**2-2.5**2-1.08**2 }{ 2 * 2.5 * 1.08 } ) = 77° 30' ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 1.08**2-2.5**2-2.5**2 }{ 2 * 2.5 * 2.5 } ) = 25° ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 1.32 }{ 3.04 } = 0.43 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2.5 }{ 2 * sin 77° 30' } = 1.28 ; ;




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