16 18 30 triangle

Obtuse scalene triangle.

Sides: a = 16 b = 18 c = 30

Area: T = 119.7333036377
Perimeter: p = 64
Semiperimeter: s = 32

Angle ∠ A = α = 26.32545765375° = 26°19'28″ = 0.45994505348 rad
Angle ∠ B = β = 29.92664348666° = 29°55'35″ = 0.52223148218 rad
Angle ∠ C = γ = 123.7498988596° = 123°44'56″ = 2.1659827297 rad

Height: h_a = 14.96766295471
Height: h_b = 13.30436707085
Height: h_c = 7.98222024251

Median: m_a = 23.40993998214
Median: m_b = 22.29334968096
Median: m_c = 8.06222577483

Inradius: r = 3.74216573868
Circumradius: R = 18.04401338291

Vertex coordinates: A[30; 0] B[0; 0] C[13.86766666667; 7.98222024251]
Centroid: CG[14.62222222222; 2.66107341417]
Coordinates of the circumscribed circle: U[15; -10.02222965717]
Coordinates of the inscribed circle: I[14; 3.74216573868]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 153.6755423463° = 153°40'32″ = 0.45994505348 rad
∠ B' = β' = 150.0743565133° = 150°4'25″ = 0.52223148218 rad
∠ C' = γ' = 56.25110114041° = 56°15'4″ = 2.1659827297 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.

$a = 16 ; ; b = 18 ; ; c = 30 ; ;$

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

$p = a+b+c = 16+18+30 = 64 ; ;$

2. Semiperimeter of the triangle

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

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 32 * (32-16)(32-18)(32-30) } ; ; T = sqrt{ 14336 } = 119.73 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 119.73 }{ 16 } = 14.97 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 119.73 }{ 18 } = 13.3 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 119.73 }{ 30 } = 7.98 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 16**2-18**2-30**2 }{ 2 * 18 * 30 } ) = 26° 19'28" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 18**2-16**2-30**2 }{ 2 * 16 * 30 } ) = 29° 55'35" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-16**2-18**2 }{ 2 * 18 * 16 } ) = 123° 44'56" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 119.73 }{ 32 } = 3.74 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 26° 19'28" } = 18.04 ; ;$

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