5 12 15 triangle

Obtuse scalene triangle.

Sides: a = 5   b = 12   c = 15

Area: T = 26.53329983228
Perimeter: p = 32
Semiperimeter: s = 16

Angle ∠ A = α = 17.14662099989° = 17°8'46″ = 0.29992578187 rad
Angle ∠ B = β = 45.03656507165° = 45°2'8″ = 0.78660203858 rad
Angle ∠ C = γ = 117.8188139285° = 117°49'5″ = 2.05663144491 rad

Height: ha = 10.61331993291
Height: hb = 4.42221663871
Height: hc = 3.53877331097

Median: ma = 13.35110299228
Median: mb = 9.43439811321
Median: mc = 5.31550729064

Inradius: r = 1.65883123952
Circumradius: R = 8.48800065662

Vertex coordinates: A[15; 0] B[0; 0] C[3.53333333333; 3.53877331097]
Centroid: CG[6.17877777778; 1.17992443699]
Coordinates of the circumscribed circle: U[7.5; -3.95773363976]
Coordinates of the inscribed circle: I[4; 1.65883123952]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 162.8543790001° = 162°51'14″ = 0.29992578187 rad
∠ B' = β' = 134.9644349284° = 134°57'52″ = 0.78660203858 rad
∠ C' = γ' = 62.18218607153° = 62°10'55″ = 2.05663144491 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 = 5 ; ; b = 12 ; ; c = 15 ; ;

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

p = a+b+c = 5+12+15 = 32 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 32 }{ 2 } = 16 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 16 * (16-5)(16-12)(16-15) } ; ; T = sqrt{ 704 } = 26.53 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 26.53 }{ 5 } = 10.61 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 26.53 }{ 12 } = 4.42 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 26.53 }{ 15 } = 3.54 ; ;

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{ 5**2-12**2-15**2 }{ 2 * 12 * 15 } ) = 17° 8'46" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 12**2-5**2-15**2 }{ 2 * 5 * 15 } ) = 45° 2'8" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 15**2-5**2-12**2 }{ 2 * 12 * 5 } ) = 117° 49'5" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 26.53 }{ 16 } = 1.66 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 5 }{ 2 * sin 17° 8'46" } = 8.48 ; ;




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

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