A235509 Decimal expansion of arccos(4/5).

6, 4, 3, 5, 0, 1, 1, 0, 8, 7, 9, 3, 2, 8, 4, 3, 8, 6, 8, 0, 2, 8, 0, 9, 2, 2, 8, 7, 1, 7, 3, 2, 2, 6, 3, 8, 0, 4, 1, 5, 1, 0, 5, 9, 1, 1, 1, 5, 3, 1, 2, 3, 8, 2, 8, 6, 5, 6, 0, 6, 1, 1, 8, 7, 1, 3, 5, 1, 2, 4, 7, 4, 8, 1, 1, 6, 2, 1, 0, 8, 8, 7, 1, 2, 8, 1, 6, 8, 4, 4, 7, 0, 1, 2, 8, 2, 7, 4, 8, 8

Offset

0,1

Comments

Given a square ABCD, there is one point M equidistant from A, B and the middle of CD. The measure of the angle BAM is arccos(4/5) (or arcsec(5/4)). This angle is the smallest angle of the well-known (3, 4, 5) Pythagorean triangle.

Also the polar angle phi of the viewing cone that cuts out exactly 10% of the celestial sphere; phi = arccos(1-2f), where f is the cut-out fraction of the full solid angle. - Stanislav Sykora, Feb 14 2016

Given a triangle ABC whose medians drawn from A and B are perpendicular in centroid G, then angle C <= arccos(4/5) (see Maths Challenge link with figure and proof). - Bernard Schott, Mar 29 2023

Links

Table of n, a(n) for n=0..99.

Jean-François Alcover, Figure showing square ABCD and angle BAM.

Maths Challenge, Perpendicular medians, Problem ID: 296, 10 Dec 2006.

Formula

Cos(A235509) + cos(A195771) = 1.

Equals arcsin(3/5). - Michel Marcus, Feb 07 2019

Equals arctan(3/4). - Amiram Eldar, Jul 04 2023

Example

0.64350110879328438680280922871732263804151059111531238286560611871351...
In degrees: 36.869897645844...°

Mathematica

RealDigits[ArcCos[4/5], 10, 100] // First

Prog

(PARI) asin(3/5) \\ Michel Marcus, Feb 07 2019

Crossrefs

Cf. A195771, A335034.

Sequence in context: A182618 A118227 A199429 * A346696 A224927 A200104

Adjacent sequences: A235506 A235507 A235508 * A235510 A235511 A235512

Keyword

nonn,cons

Author

Jean-François Alcover, Jan 14 2014

Status

approved