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Little explorations with HP calculators (no Prime)
03-30-2017, 05:47 PM (This post was last modified: 03-30-2017 07:55 PM by Gerson W. Barbosa.)
Post: #78
RE: Little explorations with the HP calculators
(03-30-2017 12:26 PM)Dieter Wrote:  
(03-29-2017 06:15 PM)Han Wrote:  So \(\angle BED\) , \(\angle BDE \), \( \angle CDE \) and \( \angle AED \) should be the only angles whose measure you cannot determine directly from the properties mentioned. What if you set up an appropriate system of equations involving these angles?

I have tried this as well. Let <BDE = u, <BED = v, <CDE = x and <AED = y and it is possible to set up several linear equations, for instance...

u + v = 162
x + y = 111
u + x = 132
v + y = 141

...but I cannot get four independent ones, so there is no solution for a linear equation system. What do you get?


That's what I'd done too, but the solution is very simple, I realize now. No need to solve any linear system.

u = 72
v = 90
x = 60
y = 51

Explanation later.


PS: From E draw a perpendicular line to line AB, which intercepts it at F. Now we have two similar right-triangles: 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.

PPS: Please do not consider this. Without any grounds I had assumed angle BED was a right angle. Fooled by my own drawing...
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RE: Little explorations with the HP calculators - Gerson W. Barbosa - 03-30-2017 05:47 PM

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