Little explorations with HP calculators (no Prime)

03302017, 06:36 PM
(This post was last modified: 03302017 07:04 PM by Han.)
Post: #80




RE: Little explorations with the HP calculators
(03302017 05:47 PM)Gerson W. Barbosa Wrote: That's what I'd done too, but the solution is very simple, I realize now. No need to solve any linear system. I suspected there had to be a simple approach. I'll try your hints and also rework my solution to see where I may have made mistakes. EDIT (after some initial thoughts): Quote:PS: From E draw a perpendicular line to line AB, which intercepts it at F. Now we have two similar righttriangles: BEF and EFD. Angle DEF = angle EBF = 18 degrees, as we already know. Angle BDE, which you have named 'u' is its complement, 72 degrees. Finally, angle EDC, or your 'x', can be easily determined as 180  72  48 = 60 degrees. Are you saying draw EF so that \( \angle BFE \) is 90 degrees? If so, how does it follow that triangle BEF is similar to triangle EFD without also knowing that \( \angle BED \) is 90 degrees (which does not seem obvious to me). Or are you suggesting we make \( \angle BEF \) equal to 90 degrees? This would make it even less obvious how triangle EFD is even a right triangle. EDIT 2: Your solution seems to suggest that we only need angles A and C. Am I missing another simple observation ? Graph 3D  QPI  SolveSys 

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