39well cant we take the circles to be all lying inside the big circle along the circumference ?
If inside a big circle exactly 24 small circles each of radius 2 can be drawn in such a way that each small circle touches the big circle and also touch both its adjacent small circles. Then the radius of the big circle is.........
a) 2(1+ cosec Î /24) B) [(1+tan Î /24) / cos Î /24]
c) 2(1+cosec Î /12) d) 2(sin Î /48 + cos Î /48)2 / sin Î /24
no need to solve it completely, juzz give d figure n procedure... (i'm a bit confused on what is d quick approach..!!!?)
well, for ur cross-checking, d ans is A,D
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391yeah, dats what d question means...
give some initial steps plz or atleast a diagram...!!!
1aagaya :)
c d cenral angle for each equlatral tringle formed is 2pi/n for n sides polygon nw here u hav 24 sides
so itz 2pi/24
nw
(consider one of those trngles )
drwn a perpendicular frm d center to d side join d two centres
nw sin(pi/24) =2/hypo
d radius of bigger required circle is hypo+2
so itz
2cosec(pi/24)+2
1arey nahi mereko neend ayi/aya hai ......... [dunno hindi grammar :P]
1thanx for d solution Ms. X (whats in name :P)... :D
well, derive option D from A... for one more pink post... ;) :P ..... (kiddin, i'll do it myself)...
thanx... :)
