15 20 22 triangle

Acute scalene triangle.

Sides: a = 15   b = 20   c = 22

Area: T = 145.7999305554
Perimeter: p = 57
Semiperimeter: s = 28.5

Angle ∠ A = α = 41.50879617794° = 41°30'29″ = 0.72444505988 rad
Angle ∠ B = β = 62.08436609946° = 62°5'1″ = 1.0843564296 rad
Angle ∠ C = γ = 76.40883772261° = 76°24'30″ = 1.33435777587 rad

Height: ha = 19.44399074072
Height: hb = 14.58799305554
Height: hc = 13.25444823231

Median: ma = 19.64105193414
Median: mb = 15.95330561335
Median: mc = 13.83883525031

Inradius: r = 5.11657651072
Circumradius: R = 11.31769263306

Vertex coordinates: A[22; 0] B[0; 0] C[7.02327272727; 13.25444823231]
Centroid: CG[9.67442424242; 4.41881607744]
Coordinates of the circumscribed circle: U[11; 2.65994776877]
Coordinates of the inscribed circle: I[8.5; 5.11657651072]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 138.4922038221° = 138°29'31″ = 0.72444505988 rad
∠ B' = β' = 117.9166339005° = 117°54'59″ = 1.0843564296 rad
∠ C' = γ' = 103.5921622774° = 103°35'30″ = 1.33435777587 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 = 15 ; ; b = 20 ; ; c = 22 ; ;

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

p = a+b+c = 15+20+22 = 57 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 57 }{ 2 } = 28.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 28.5 * (28.5-15)(28.5-20)(28.5-22) } ; ; T = sqrt{ 21257.44 } = 145.8 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 145.8 }{ 15 } = 19.44 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 145.8 }{ 20 } = 14.58 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 145.8 }{ 22 } = 13.25 ; ;

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{ 15**2-20**2-22**2 }{ 2 * 20 * 22 } ) = 41° 30'29" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-15**2-22**2 }{ 2 * 15 * 22 } ) = 62° 5'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-15**2-20**2 }{ 2 * 20 * 15 } ) = 76° 24'30" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 145.8 }{ 28.5 } = 5.12 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 41° 30'29" } = 11.32 ; ;




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

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