15 15 20 triangle

Acute isosceles triangle.

Sides: a = 15   b = 15   c = 20

Area: T = 111.8033398875
Perimeter: p = 50
Semiperimeter: s = 25

Angle ∠ A = α = 48.19896851042° = 48°11'23″ = 0.84110686706 rad
Angle ∠ B = β = 48.19896851042° = 48°11'23″ = 0.84110686706 rad
Angle ∠ C = γ = 83.62106297916° = 83°37'14″ = 1.45994553125 rad

Height: ha = 14.907711985
Height: hb = 14.907711985
Height: hc = 11.18803398875

Median: ma = 16.00878105936
Median: mb = 16.00878105936
Median: mc = 11.18803398875

Inradius: r = 4.4722135955
Circumradius: R = 10.06223058987

Vertex coordinates: A[20; 0] B[0; 0] C[10; 11.18803398875]
Centroid: CG[10; 3.72767799625]
Coordinates of the circumscribed circle: U[10; 1.11880339887]
Coordinates of the inscribed circle: I[10; 4.4722135955]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 131.8110314896° = 131°48'37″ = 0.84110686706 rad
∠ B' = β' = 131.8110314896° = 131°48'37″ = 0.84110686706 rad
∠ C' = γ' = 96.37993702084° = 96°22'46″ = 1.45994553125 rad

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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 = 15 ; ; b = 15 ; ; c = 20 ; ;

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

p = a+b+c = 15+15+20 = 50 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 50 }{ 2 } = 25 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 25 * (25-15)(25-15)(25-20) } ; ; T = sqrt{ 12500 } = 111.8 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 111.8 }{ 15 } = 14.91 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 111.8 }{ 15 } = 14.91 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 111.8 }{ 20 } = 11.18 ; ;

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{ 15**2-15**2-20**2 }{ 2 * 15 * 20 } ) = 48° 11'23" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-15**2-20**2 }{ 2 * 15 * 20 } ) = 48° 11'23" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20**2-15**2-15**2 }{ 2 * 15 * 15 } ) = 83° 37'14" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 111.8 }{ 25 } = 4.47 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 48° 11'23" } = 10.06 ; ;




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