15 16 29 triangle

Obtuse scalene triangle.

Sides: a = 15 b = 16 c = 29

Area: T = 79.37325393319
Perimeter: p = 60
Semiperimeter: s = 30

Angle ∠ A = α = 20.00662724699° = 20°23″ = 0.34991753257 rad
Angle ∠ B = β = 21.40333496394° = 21°24'12″ = 0.37435589222 rad
Angle ∠ C = γ = 138.5990377891° = 138°35'25″ = 2.41988584058 rad

Height: h_a = 10.58330052443
Height: h_b = 9.92215674165
Height: h_c = 5.47439682298

Median: m_a = 22.18767077323
Median: m_b = 21.65664078277
Median: m_c = 5.5

Inradius: r = 2.64657513111
Circumradius: R = 21.92219394345

Vertex coordinates: A[29; 0] B[0; 0] C[13.96655172414; 5.47439682298]
Centroid: CG[14.32218390805; 1.82546560766]
Coordinates of the circumscribed circle: U[14.5; -16.44114545759]
Coordinates of the inscribed circle: I[14; 2.64657513111]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 159.994372753° = 159°59'37″ = 0.34991753257 rad
∠ B' = β' = 158.5976650361° = 158°35'48″ = 0.37435589222 rad
∠ C' = γ' = 41.41096221093° = 41°24'35″ = 2.41988584058 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.

$a = 15 ; ; b = 16 ; ; c = 29 ; ;$

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

$p = a+b+c = 15+16+29 = 60 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 60 }{ 2 } = 30 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30 * (30-15)(30-16)(30-29) } ; ; T = sqrt{ 6300 } = 79.37 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 79.37 }{ 15 } = 10.58 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 79.37 }{ 16 } = 9.92 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 79.37 }{ 29 } = 5.47 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 15**2-16**2-29**2 }{ 2 * 16 * 29 } ) = 20° 23" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 16**2-15**2-29**2 }{ 2 * 15 * 29 } ) = 21° 24'12" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-15**2-16**2 }{ 2 * 16 * 15 } ) = 138° 35'25" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 79.37 }{ 30 } = 2.65 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 20° 23" } = 21.92 ; ;$

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

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