10 27 27 triangle

Acute isosceles triangle.

Sides: a = 10 b = 27 c = 27

Area: T = 132.6654991614
Perimeter: p = 64
Semiperimeter: s = 32

Angle ∠ A = α = 21.34438585715° = 21°20'38″ = 0.37325206072 rad
Angle ∠ B = β = 79.32880707142° = 79°19'41″ = 1.38545360232 rad
Angle ∠ C = γ = 79.32880707142° = 79°19'41″ = 1.38545360232 rad

Height: h_a = 26.53329983228
Height: h_b = 9.82770364159
Height: h_c = 9.82770364159

Median: m_a = 26.53329983228
Median: m_b = 15.24397506541
Median: m_c = 15.24397506541

Inradius: r = 4.14657809879
Circumradius: R = 13.73876106373

Vertex coordinates: A[27; 0] B[0; 0] C[1.85218518519; 9.82770364159]
Centroid: CG[9.61772839506; 3.27656788053]
Coordinates of the circumscribed circle: U[13.5; 2.54440019699]
Coordinates of the inscribed circle: I[5; 4.14657809879]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 158.6566141428° = 158°39'22″ = 0.37325206072 rad
∠ B' = β' = 100.6721929286° = 100°40'19″ = 1.38545360232 rad
∠ C' = γ' = 100.6721929286° = 100°40'19″ = 1.38545360232 rad

Calculate another triangle

How did we calculate this triangle?

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.

$a = 10 ; ; b = 27 ; ; c = 27 ; ;$

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

$p = a+b+c = 10+27+27 = 64 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 64 }{ 2 } = 32 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 32 * (32-10)(32-27)(32-27) } ; ; T = sqrt{ 17600 } = 132.66 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 132.66 }{ 10 } = 26.53 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 132.66 }{ 27 } = 9.83 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 132.66 }{ 27 } = 9.83 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 10**2-27**2-27**2 }{ 2 * 27 * 27 } ) = 21° 20'38" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-10**2-27**2 }{ 2 * 10 * 27 } ) = 79° 19'41" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-10**2-27**2 }{ 2 * 27 * 10 } ) = 79° 19'41" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 132.66 }{ 32 } = 4.15 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 21° 20'38" } = 13.74 ; ;$

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