9 15 19 triangle

Obtuse scalene triangle.

Sides: a = 9 b = 15 c = 19

Area: T = 66.08546994394
Perimeter: p = 43
Semiperimeter: s = 21.5

Angle ∠ A = α = 27.63295013938° = 27°37'46″ = 0.482222577 rad
Angle ∠ B = β = 50.61768729893° = 50°37'1″ = 0.88334310907 rad
Angle ∠ C = γ = 101.7543625617° = 101°45'13″ = 1.77659357929 rad

Height: h_a = 14.68554887643
Height: h_b = 8.81112932586
Height: h_c = 6.95662841515

Median: m_a = 16.51551445649
Median: m_b = 12.8355497653
Median: m_c = 7.92114897589

Inradius: r = 3.07437069507
Circumradius: R = 9.70334564043

Vertex coordinates: A[19; 0] B[0; 0] C[5.71105263158; 6.95662841515]
Centroid: CG[8.23768421053; 2.31987613838]
Coordinates of the circumscribed circle: U[9.5; -1.97766300083]
Coordinates of the inscribed circle: I[6.5; 3.07437069507]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 152.3770498606° = 152°22'14″ = 0.482222577 rad
∠ B' = β' = 129.3833127011° = 129°22'59″ = 0.88334310907 rad
∠ C' = γ' = 78.24663743831° = 78°14'47″ = 1.77659357929 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.

$a = 9 ; ; b = 15 ; ; c = 19 ; ;$

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

$p = a+b+c = 9+15+19 = 43 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 43 }{ 2 } = 21.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 21.5 * (21.5-9)(21.5-15)(21.5-19) } ; ; T = sqrt{ 4367.19 } = 66.08 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 66.08 }{ 9 } = 14.69 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 66.08 }{ 15 } = 8.81 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 66.08 }{ 19 } = 6.96 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 9**2-15**2-19**2 }{ 2 * 15 * 19 } ) = 27° 37'46" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-9**2-19**2 }{ 2 * 9 * 19 } ) = 50° 37'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 19**2-9**2-15**2 }{ 2 * 15 * 9 } ) = 101° 45'13" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 66.08 }{ 21.5 } = 3.07 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 27° 37'46" } = 9.7 ; ;$

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