16 16 29 triangle

Obtuse isosceles triangle.

Sides: a = 16   b = 16   c = 29

Area: T = 98.07661821239
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 25.00878332347° = 25°28″ = 0.43664690287 rad
Angle ∠ B = β = 25.00878332347° = 25°28″ = 0.43664690287 rad
Angle ∠ C = γ = 129.9844333531° = 129°59'4″ = 2.26986545961 rad

Height: ha = 12.26595227655
Height: hb = 12.26595227655
Height: hc = 6.76438746292

Median: ma = 22.01113607031
Median: mb = 22.01113607031
Median: mc = 6.76438746292

Inradius: r = 3.21656125287
Circumradius: R = 18.92440645364

Vertex coordinates: A[29; 0] B[0; 0] C[14.5; 6.76438746292]
Centroid: CG[14.5; 2.25546248764]
Coordinates of the circumscribed circle: U[14.5; -12.16601899072]
Coordinates of the inscribed circle: I[14.5; 3.21656125287]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 154.9922166765° = 154°59'32″ = 0.43664690287 rad
∠ B' = β' = 154.9922166765° = 154°59'32″ = 0.43664690287 rad
∠ C' = γ' = 50.01656664694° = 50°56″ = 2.26986545961 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 = 16 ; ; c = 29 ; ;

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

p = a+b+c = 16+16+29 = 61 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 61 }{ 2 } = 30.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30.5 * (30.5-16)(30.5-16)(30.5-29) } ; ; T = sqrt{ 9618.94 } = 98.08 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 98.08 }{ 16 } = 12.26 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 98.08 }{ 16 } = 12.26 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 98.08 }{ 29 } = 6.76 ; ;

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**2-16**2-29**2 }{ 2 * 16 * 29 } ) = 25° 28" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 16**2-16**2-29**2 }{ 2 * 16 * 29 } ) = 25° 28" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-16**2-16**2 }{ 2 * 16 * 16 } ) = 129° 59'4" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 98.08 }{ 30.5 } = 3.22 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 25° 28" } = 18.92 ; ;




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

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