5 15 15 triangle

Acute isosceles triangle.

Sides: a = 5 b = 15 c = 15

Area: T = 36.97554986444
Perimeter: p = 35
Semiperimeter: s = 17.5

Angle ∠ A = α = 19.18881364537° = 19°11'17″ = 0.33548961584 rad
Angle ∠ B = β = 80.40659317731° = 80°24'21″ = 1.40333482476 rad
Angle ∠ C = γ = 80.40659317731° = 80°24'21″ = 1.40333482476 rad

Height: h_a = 14.79901994577
Height: h_b = 4.93300664859
Height: h_c = 4.93300664859

Median: m_a = 14.79901994577
Median: m_b = 8.29215619759
Median: m_c = 8.29215619759

Inradius: r = 2.11328856368
Circumradius: R = 7.60663882926

Vertex coordinates: A[15; 0] B[0; 0] C[0.83333333333; 4.93300664859]
Centroid: CG[5.27877777778; 1.64333554953]
Coordinates of the circumscribed circle: U[7.5; 1.26877313821]
Coordinates of the inscribed circle: I[2.5; 2.11328856368]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 160.8121863546° = 160°48'43″ = 0.33548961584 rad
∠ B' = β' = 99.59440682269° = 99°35'39″ = 1.40333482476 rad
∠ C' = γ' = 99.59440682269° = 99°35'39″ = 1.40333482476 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.

$a = 5 ; ; b = 15 ; ; c = 15 ; ;$

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

$p = a+b+c = 5+15+15 = 35 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 35 }{ 2 } = 17.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 17.5 * (17.5-5)(17.5-15)(17.5-15) } ; ; T = sqrt{ 1367.19 } = 36.98 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 36.98 }{ 5 } = 14.79 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 36.98 }{ 15 } = 4.93 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 36.98 }{ 15 } = 4.93 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 5**2-15**2-15**2 }{ 2 * 15 * 15 } ) = 19° 11'17" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-5**2-15**2 }{ 2 * 5 * 15 } ) = 80° 24'21" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 15**2-5**2-15**2 }{ 2 * 15 * 5 } ) = 80° 24'21" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 36.98 }{ 17.5 } = 2.11 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 5 }{ 2 * sin 19° 11'17" } = 7.61 ; ;$

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