Triangle calculator SAS

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


Right scalene triangle.

Sides: a = 2.52   b = 3.6   c = 4.39443600217

Area: T = 4.536
Perimeter: p = 10.51443600217
Semiperimeter: s = 5.25771800108

Angle ∠ A = α = 34.99220201986° = 34°59'31″ = 0.61107259644 rad
Angle ∠ B = β = 55.00879798014° = 55°29″ = 0.96600703624 rad
Angle ∠ C = γ = 90° = 1.57107963268 rad

Height: ha = 3.6
Height: hb = 2.52
Height: hc = 2.06444644397

Median: ma = 3.81441316181
Median: mb = 3.09768370961
Median: mc = 2.19771800108

Inradius: r = 0.86328199892
Circumradius: R = 2.19771800108

Vertex coordinates: A[4.39443600217; 0] B[0; 0] C[1.44551251078; 2.06444644397]
Centroid: CG[1.94664950432; 0.68881548132]
Coordinates of the circumscribed circle: U[2.19771800108; -0]
Coordinates of the inscribed circle: I[1.65771800108; 0.86328199892]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 145.0087979801° = 145°29″ = 0.61107259644 rad
∠ B' = β' = 124.9922020199° = 124°59'31″ = 0.96600703624 rad
∠ C' = γ' = 90° = 1.57107963268 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 = 2.52 ; ; b = 3.6 ; ; gamma = 90° ; ; ; ; c**2 = a**2+b**2 - 2ab cos( gamma ) ; ; c = sqrt{ a**2+b**2 - 2ab cos( gamma ) } ; ; c = sqrt{ 2.52**2+3.6**2 - 2 * 2.52 * 3.6 * cos(90° ) } ; ; c = 4.39 ; ;


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 = 2.52 ; ; b = 3.6 ; ; c = 4.39 ; ;

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

p = a+b+c = 2.52+3.6+4.39 = 10.51 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 10.51 }{ 2 } = 5.26 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 5.26 * (5.26-2.52)(5.26-3.6)(5.26-4.39) } ; ; T = sqrt{ 20.58 } = 4.54 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 4.54 }{ 2.52 } = 3.6 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 4.54 }{ 3.6 } = 2.52 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 4.54 }{ 4.39 } = 2.06 ; ;

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{ 2.52**2-3.6**2-4.39**2 }{ 2 * 3.6 * 4.39 } ) = 34° 59'31" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 3.6**2-2.52**2-4.39**2 }{ 2 * 2.52 * 4.39 } ) = 55° 29" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 4.39**2-2.52**2-3.6**2 }{ 2 * 3.6 * 2.52 } ) = 90° ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 4.54 }{ 5.26 } = 0.86 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2.52 }{ 2 * sin 34° 59'31" } = 2.2 ; ;




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