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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Whew, that's a beauty!

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm... it really is. :P I guess just to toss some ideas out there, but for the sphere to be placed inside that space, it would be tangent to: A) all three cones, and B) the plane containing the vertices of the three cones. Perhaps that could be useful somehow...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think that (B) ^ you mention helps a lot.

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1* And the distance to all four of those locations from the center are the same

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have a feeling similar triangle proportions could also be useful.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how about passing a plane containing the height of one cone and the center of the cylinder. on that plane is an isosceles triangle and a circle.

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm.. something like this? Although it is a sphere. I can see some angles we can find with trig here pretty easily also dw:1351910642408:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the projection of the center of the circle tn the plane containing the three centers of the bases is the centroid of the equilateral trianlge formed by the centers of the bases.

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm.. actually, I believe we could also determine the length of the segment from the vertex of this triangle to the sphere using that fact ^, and then use trigonometry with the angle to find the radius... dw:1351911155230:dw

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1351911813104:dw So if my Geometry skills aren't too rusty, 306090 says that length from the center to the vertex is 100/sqrt(3) or 100sqrt(3)/3. The angle adjacent to the kite figure within the tangents is determined by rightangle trig: tan t = opp/adj: opp = 50, adj = 120 tan t = 50/120 = 5/12 t = arctan(5/12) So the complementary angle is 90  arctan(5/12), or about 67.38. We break the angle in half for our right triangle because the diagonal bisects both angles. (90  arctan(5/12))/2, or about 33.69. So we have a triangle with one angle known and one side known, so trigonometry will help us here. :) dw:1351911929972:dw

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1So if we just plug that into a calculator (I'm going to use wolfram): http://www.wolframalpha.com/input/?i=100sqrt%283%29%2F3+*+tan%2833.69%29 It looks like we're at about 38.4899, which is pretty close to 38.5. :)

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1In summary: 1) Identify what tangents there are. (To the cones, to the plane containing the top of the cones) 2) Observe a planar section of the situation, namely the plane through the cone's vertical height and the circle's center. Problems are easier in 2D than 3D! 3) Use right angle trigonometry to find all of the information.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351914302686:dw i got 38.49002

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Yep, looks correct! I wouldn't have thought about that method! :) If I hadn't rounded the arctan(5/12) and done the calculations at the end with those rounded numbers, I would have obtained 38.49002 as well (doing the calculations with my computer's calculator is being more cooperative than wolfram). :P
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