Triangle calculator SAS

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


Acute isosceles triangle.

Sides: a = 100   b = 100   c = 51.76438090205

Area: T = 2500
Perimeter: p = 251.764380902
Semiperimeter: s = 125.882190451

Angle ∠ A = α = 75° = 1.3098996939 rad
Angle ∠ B = β = 75° = 1.3098996939 rad
Angle ∠ C = γ = 30° = 0.52435987756 rad

Height: ha = 50
Height: hb = 50
Height: hc = 96.59325826289

Median: ma = 61.96656837464
Median: mb = 61.96656837464
Median: mc = 96.59325826289

Inradius: r = 19.8659883831
Circumradius: R = 51.76438090205

Vertex coordinates: A[51.76438090205; 0] B[0; 0] C[25.88219045103; 96.59325826289]
Centroid: CG[25.88219045103; 32.1987527543]
Coordinates of the circumscribed circle: U[25.88219045103; 44.82987736084]
Coordinates of the inscribed circle: I[25.88219045103; 19.8659883831]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 105° = 1.3098996939 rad
∠ B' = β' = 105° = 1.3098996939 rad
∠ C' = γ' = 150° = 0.52435987756 rad

Calculate another triangle




How did we calculate this triangle?

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

a = 100 ; ; b = 100 ; ; gamma = 30° ; ; ; ; c**2 = a**2+b**2 - 2ab cos gamma ; ; c = sqrt{ a**2+b**2 - 2ab cos gamma } ; ; c = sqrt{ 100**2+100**2 - 2 * 100 * 100 * cos(30° ) } ; ; c = 51.76 ; ;


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 = 100 ; ; b = 100 ; ; c = 51.76 ; ;

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

p = a+b+c = 100+100+51.76 = 251.76 ; ;

3. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 251.76 }{ 2 } = 125.88 ; ;

4. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 125.88 * (125.88-100)(125.88-100)(125.88-51.76) } ; ; T = sqrt{ 6250000 } = 2500 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2500 }{ 100 } = 50 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2500 }{ 100 } = 50 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2500 }{ 51.76 } = 96.59 ; ;

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{ 100**2-100**2-51.76**2 }{ 2 * 100 * 51.76 } ) = 75° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 100**2-100**2-51.76**2 }{ 2 * 100 * 51.76 } ) = 75° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 51.76**2-100**2-100**2 }{ 2 * 100 * 100 } ) = 30° ; ;

7. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 2500 }{ 125.88 } = 19.86 ; ;

8. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 100 }{ 2 * sin 75° } = 51.76 ; ;




Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.