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A235509 Decimal expansion of arccos(4/5).

%I A235509 #46 Nov 21 2024 07:55:44
%S A235509 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,
%T A235509 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,
%U A235509 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
%N A235509 Decimal expansion of arccos(4/5).
%C A235509 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.
%C A235509 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
%C A235509 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
%H A235509 Jean-François Alcover, <a href="/A235509/a235509.gif">Figure showing square ABCD and angle BAM</a>.
%H A235509 Maths Challenge, <a href="https://mathschallenge.net/view/perpendicular_medians">Perpendicular medians</a>, Problem ID: 296, 10 Dec 2006.
%H A235509 <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F A235509 Cos(A235509) + cos(A195771) = 1.
%F A235509 Equals arcsin(3/5). - _Michel Marcus_, Feb 07 2019
%F A235509 Equals arctan(3/4). - _Amiram Eldar_, Jul 04 2023
%e A235509 0.64350110879328438680280922871732263804151059111531238286560611871351...
%e A235509 In degrees: 36.869897645844...°
%t A235509 RealDigits[ArcCos[4/5], 10, 100] // First
%o A235509 (PARI) asin(3/5) \\ _Michel Marcus_, Feb 07 2019
%Y A235509 Cf. A195771, A335034.
%K A235509 nonn,cons
%O A235509 0,1
%A A235509 _Jean-François Alcover_, Jan 14 2014