15 16 17 triangle

Acute scalene triangle.

Sides: a = 15   b = 16   c = 17

Area: T = 109.9821816679
Perimeter: p = 48
Semiperimeter: s = 24

Angle ∠ A = α = 53.96881209275° = 53°58'5″ = 0.94219214013 rad
Angle ∠ B = β = 59.61100575507° = 59°36'36″ = 1.04403917716 rad
Angle ∠ C = γ = 66.42218215218° = 66°25'19″ = 1.15992794807 rad

Height: ha = 14.66442422239
Height: hb = 13.74877270849
Height: hc = 12.93990372563

Median: ma = 14.70554411699
Median: mb = 13.89224439894
Median: mc = 12.97111217711

Inradius: r = 4.5832575695
Circumradius: R = 9.2744260335

Vertex coordinates: A[17; 0] B[0; 0] C[7.58882352941; 12.93990372563]
Centroid: CG[8.19660784314; 4.31330124188]
Coordinates of the circumscribed circle: U[8.5; 3.7109704134]
Coordinates of the inscribed circle: I[8; 4.5832575695]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 126.0321879072° = 126°1'55″ = 0.94219214013 rad
∠ B' = β' = 120.3989942449° = 120°23'24″ = 1.04403917716 rad
∠ C' = γ' = 113.5788178478° = 113°34'41″ = 1.15992794807 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 = 16 ; ; c = 17 ; ;

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

p = a+b+c = 15+16+17 = 48 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 48 }{ 2 } = 24 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 24 * (24-15)(24-16)(24-17) } ; ; T = sqrt{ 12096 } = 109.98 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 109.98 }{ 15 } = 14.66 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 109.98 }{ 16 } = 13.75 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 109.98 }{ 17 } = 12.94 ; ;

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**2-16**2-17**2 }{ 2 * 16 * 17 } ) = 53° 58'5" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 16**2-15**2-17**2 }{ 2 * 15 * 17 } ) = 59° 36'36" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 17**2-15**2-16**2 }{ 2 * 16 * 15 } ) = 66° 25'19" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 109.98 }{ 24 } = 4.58 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 53° 58'5" } = 9.27 ; ;




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

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