8 15 16 triangle

Acute scalene triangle.

Sides: a = 8 b = 15 c = 16

Area: T = 59.4330106007
Perimeter: p = 39
Semiperimeter: s = 19.5

Angle ∠ A = α = 29.68662952314° = 29°41'11″ = 0.51881235945 rad
Angle ∠ B = β = 68.21769125021° = 68°13'1″ = 1.19106097287 rad
Angle ∠ C = γ = 82.09767922665° = 82°5'48″ = 1.43328593304 rad

Height: h_a = 14.85875265017
Height: h_b = 7.92440141343
Height: h_c = 7.42987632509

Median: m_a = 14.98333240638
Median: m_b = 10.18657743937
Median: m_c = 8.97221792225

Inradius: r = 3.04876977439
Circumradius: R = 8.07767145181

Vertex coordinates: A[16; 0] B[0; 0] C[2.969875; 7.42987632509]
Centroid: CG[6.32329166667; 2.4766254417]
Coordinates of the circumscribed circle: U[8; 1.11105482462]
Coordinates of the inscribed circle: I[4.5; 3.04876977439]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 150.3143704769° = 150°18'49″ = 0.51881235945 rad
∠ B' = β' = 111.7833087498° = 111°46'59″ = 1.19106097287 rad
∠ C' = γ' = 97.90332077335° = 97°54'12″ = 1.43328593304 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.

$a = 8 ; ; b = 15 ; ; c = 16 ; ;$

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

$p = a+b+c = 8+15+16 = 39 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 39 }{ 2 } = 19.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 19.5 * (19.5-8)(19.5-15)(19.5-16) } ; ; T = sqrt{ 3531.94 } = 59.43 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 59.43 }{ 8 } = 14.86 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 59.43 }{ 15 } = 7.92 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 59.43 }{ 16 } = 7.43 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 8**2-15**2-16**2 }{ 2 * 15 * 16 } ) = 29° 41'11" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-8**2-16**2 }{ 2 * 8 * 16 } ) = 68° 13'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 16**2-8**2-15**2 }{ 2 * 15 * 8 } ) = 82° 5'48" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 59.43 }{ 19.5 } = 3.05 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 8 }{ 2 * sin 29° 41'11" } = 8.08 ; ;$

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