16 24 28 triangle

Acute scalene triangle.

Sides: a = 16   b = 24   c = 28

Area: T = 191.6254633072
Perimeter: p = 68
Semiperimeter: s = 34

Angle ∠ A = α = 34.77219440319° = 34°46'19″ = 0.60768849107 rad
Angle ∠ B = β = 58.81113776665° = 58°48'41″ = 1.02664521779 rad
Angle ∠ C = γ = 86.41766783015° = 86°25' = 1.5088255565 rad

Height: ha = 23.9533079134
Height: hb = 15.96987194227
Height: hc = 13.68774737909

Median: ma = 24.8199347292
Median: mb = 19.39107194297
Median: mc = 14.83223969742

Inradius: r = 5.63660186198
Circumradius: R = 14.02774241203

Vertex coordinates: A[28; 0] B[0; 0] C[8.28657142857; 13.68774737909]
Centroid: CG[12.09552380952; 4.56224912636]
Coordinates of the circumscribed circle: U[14; 0.87767140075]
Coordinates of the inscribed circle: I[10; 5.63660186198]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 145.2288055968° = 145°13'41″ = 0.60768849107 rad
∠ B' = β' = 121.1898622333° = 121°11'19″ = 1.02664521779 rad
∠ C' = γ' = 93.58333216985° = 93°35' = 1.5088255565 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 = 24 ; ; c = 28 ; ;

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

p = a+b+c = 16+24+28 = 68 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 68 }{ 2 } = 34 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 34 * (34-16)(34-24)(34-28) } ; ; T = sqrt{ 36720 } = 191.62 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 191.62 }{ 16 } = 23.95 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 191.62 }{ 24 } = 15.97 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 191.62 }{ 28 } = 13.69 ; ;

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**2-24**2-28**2 }{ 2 * 24 * 28 } ) = 34° 46'19" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-16**2-28**2 }{ 2 * 16 * 28 } ) = 58° 48'41" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-16**2-24**2 }{ 2 * 24 * 16 } ) = 86° 25' ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 191.62 }{ 34 } = 5.64 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 34° 46'19" } = 14.03 ; ;




Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.