10 15 15 triangle

Acute isosceles triangle.

Sides: a = 10 b = 15 c = 15

Area: T = 70.71106781187
Perimeter: p = 40
Semiperimeter: s = 20

Angle ∠ A = α = 38.9422441269° = 38°56'33″ = 0.68796738189 rad
Angle ∠ B = β = 70.52987793655° = 70°31'44″ = 1.23109594173 rad
Angle ∠ C = γ = 70.52987793655° = 70°31'44″ = 1.23109594173 rad

Height: h_a = 14.14221356237
Height: h_b = 9.42880904158
Height: h_c = 9.42880904158

Median: m_a = 14.14221356237
Median: m_b = 10.3087764064
Median: m_c = 10.3087764064

Inradius: r = 3.53655339059
Circumradius: R = 7.95549512883

Vertex coordinates: A[15; 0] B[0; 0] C[3.33333333333; 9.42880904158]
Centroid: CG[6.11111111111; 3.14326968053]
Coordinates of the circumscribed circle: U[7.5; 2.65216504294]
Coordinates of the inscribed circle: I[5; 3.53655339059]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 141.0587558731° = 141°3'27″ = 0.68796738189 rad
∠ B' = β' = 109.4711220634° = 109°28'16″ = 1.23109594173 rad
∠ C' = γ' = 109.4711220634° = 109°28'16″ = 1.23109594173 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.

$a = 10 ; ; b = 15 ; ; c = 15 ; ;$

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

$p = a+b+c = 10+15+15 = 40 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 40 }{ 2 } = 20 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 20 * (20-10)(20-15)(20-15) } ; ; T = sqrt{ 5000 } = 70.71 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 70.71 }{ 10 } = 14.14 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 70.71 }{ 15 } = 9.43 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 70.71 }{ 15 } = 9.43 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 10**2-15**2-15**2 }{ 2 * 15 * 15 } ) = 38° 56'33" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-10**2-15**2 }{ 2 * 10 * 15 } ) = 70° 31'44" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 15**2-10**2-15**2 }{ 2 * 15 * 10 } ) = 70° 31'44" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 70.71 }{ 20 } = 3.54 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 38° 56'33" } = 7.95 ; ;$

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