Trigonometry problem
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From: Francois Grieu on 9 Jul 2010 00:52 On 08/07/2010 23:52, Rob Johnson wrote: > In article<4c3638ac$0$10461$426a74cc(a)news.free.fr>, > Francois Grieu<fgrieu(a)gmail.com> wrote: >> Le 08/07/2010 21:19, Ludovicus a �crit : >>> On 8 jul, 13:10, Francois Grieu<fgr...(a)gmail.com> wrote: >>>> Le 08/07/2010 17:03, Ludovicus a �crit : >>>> >>>>> On 8 jul, 01:59, Fred Nurk<albert.xtheunkno...(a)gmail.com> wrote: >>>>>> A sphere of radius length 8 cm rests on the top of a hollow inverted cone >>>>>> of height 15 cm whose vertical angle is 60 degrees. Find the height of >>>>>> the centre of the sphere above the vertex of the cone. >>>> >>>>>> Where is the vertical angle located on a cone? >>>> >>>>> It is impossile that a sphere of 16 cms of diameter >>>>> can rest on the top of that cone. >>>> >>>> Not fully on top, but still that sounds on top to me. >>>> >>>>> Because the diameter on the top of a cone of 60 degrees >>>>> at 15 cms is 17.32 cms. >>>> >>>> Yes. >>>> >>>>> The height must be an unknown quantity, not 15. >>>> >>>> "height 15 cm" allow to compute an answer; it would be >>>> the same for "height 16 cm", but not for "height 5 cm". >>>> >>>> Francois Grieu >>> >>> It's not necessary the height nor the diameter to compute the answer. >>> The answer is: >>> In a cone of 6o degrees the center of a sphere in contact with the >>> inner surface, is at one diameter from the vertex of cone. >> >> That's true if the height of the cone is at least 3^(1/2) times >> the radius of the sphere; >> Below that limit, the contact circle between the sphere and the >> cone changes, and the sphere lowers. > > If, by "height of the cone", you mean the distance from the base of > the cone to its vertex (this is the usual meaning, and is the "h" > used in the formula V = 1/3 B h), then the height of the cone must > be 3/2 the radius of the sphere. Yes, I goofed. Was late. > In the extreme case, the radius of the base of the cone is sqrt(3)/2 > times the radius of the sphere, and the height of a 60 degree cone > is sqrt(3) times the radius of the base of the cone. Thus, the > height of the cone must be at least 3/2 the radius of the sphere. Thanks. Francois Grieu
From: Gerry Myerson on 9 Jul 2010 01:56 In article <rJeZn.385$Yv.32(a)viwinnwfe01.internal.bigpond.com>, Fred Nurk <albert.xtheunknown0(a)gmail.com> wrote: > Alois Steindl wrote: > > <snip> > > Hello, > > you should be able to find that out either from your textbook or by a > > google search with proper keywords. > > It's not in my textbook. I can't Google *that* well... Have you tried? Did you try typing vertical angle cone into Google to see what comes out? It might not work, but it's worth a try. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Fred Nurk on 9 Jul 2010 04:08 Ludovicus wrote: > <snip> > It's not necessary the height nor the diameter to compute the answer. > The answer is: > In a cone of 6o degrees the center of a sphere in contact with the inner > surface, is at one diameter from the vertex of cone. Agreed. I get it now. I did the problem by finding the length from the vertex of the cone to the chord that meets with the cone plus the length of from the chord to the centre of the circle. Thank you, Fred
From: cwldoc on 9 Jul 2010 07:26 > A sphere of radius length 8 cm rests on the top of a > hollow inverted cone > of height 15 cm whose vertical angle is 60 degrees. > Find the height of > the centre of the sphere above the vertex of the > cone. > > Where is the vertical angle located on a cone? > > TIA, > Fred Let h = height of cone and r = radius of sphere and H = height of center of sphere above bertex of cone A simple 2-D sketch (exploiting axial symmetry) shows that for h >= r/(2sqrt(3)) the answer is simple : H = (2/sqrt(3))r If h < r/(2sqrt(3)), the problem is slightly more complicated. Fortunately, for r = 8 and h = 15 we have the simple solution.
From: cwldoc on 9 Jul 2010 13:42
> > A sphere of radius length 8 cm rests on the top of > a > > hollow inverted cone > > of height 15 cm whose vertical angle is 60 > degrees. > > Find the height of > > the centre of the sphere above the vertex of the > > cone. > > > > Where is the vertical angle located on a cone? > > > > TIA, > > Fred > > Let h = height of cone and r = radius of sphere and H > = height of center of sphere above bertex of cone > > A simple 2-D sketch (exploiting axial symmetry) shows > that for > h >= r/(2sqrt(3)) > the answer is simple : > H = (2/sqrt(3))r > > If h < r/(2sqrt(3)), the problem is slightly more > complicated. Fortunately, for r = 8 and h = 15 we > have the simple solution. CORRECTION and clarification: Let h = height of cone r = radius of sphere H = height of center of sphere above bertex of cone Case I h >= [r/(sin(60/2 deg)] - r sin(60/2 deg) = (3/2)r H = r/(sin(60/2 deg)] = 2r (Just imagine the above sphere resting on another sphere of radius r centered at the vertex of the cone.) Case II h < (3/2)r ("ice cream cone") The equation for the sphere (in cylindrical coordinates) is p^2 + (z-H)^2 = r^2. At any point where the sphere is touching the cone, z = h and p = h tan(60/2 deg) = h/(sqrt(3)). Substitution gives (h^2)/3 + (h - H)^2 = r^2. Simplifying gives H - h = |h - H| = sqrt[r^2 - (1/3)h^2] H = sqrt[r^2 - (1/3)h^2] + h since H < h For h = 15 cm and r = 8 cm, h = 15 cm > 12 cm = (3/2) 8 cm = (3/2)r, so Case I applies and we have H = 2r = 16 cm |