Damn, I've been out of high school level math for so long, I must have forgotten some basic theorem. I must have done hundreds of these questions by the time I was 15. It would be awkward if I asked my professor this question so I need your help.
>>16539975If I recall correctly, this is a genuinely difficult problem that takes some time to "find the trick". I believe one way to do it is to extend some lines out, and wtv. Just wanted to comment before I go to sleep
>>16540038this is true. it can be solved by what is expected a primary school geometry class covers, but is tricky.
>>16539975Well, how I did it, I extended ED and BC until the meet on the left side at point F, then I used the law of sines three times (along with a simple a + b = c equation - so a system of four equations overall) to get angle EFB, which gave me angle BDE, which gives angle CDE. I was actually solving for BDE instead of CDE by accident, so there's prob a faster way. But the equations were very easy to solve (prob only for this problem - in general it'll def be harder to do a law of sines). I recommend looking for a better way.
acturally >>16540066, the law of sines generalizes very well, it looks like. If you want to know the sides I used for the law of sines..........>The three law of sines I used involved lines GB, EB, and DB, using the angles I already mentioned.
>>16540066Haha this is also a method but it sounds like using a nuclear bomb. I am going to ask my professor to solve it. I will upload the solution (no reason he doesn’t know it, right?) in case anyone wants to know too.
>>16540074Shit sorry, it doesn't work, I made a mistakeIgnore >>16540066 >>16540073Ill try again later
>>16540079Successfully solved. The professor was drunk but still solved it immediately. He said that this is indeed a medium-to-low-level problem that will appear in youth mathematics competitions, and he also showed that this kind of problem, even the most difficult one, can be solved using middle school mathematics.I guess I'm a midwit.
>>16540092When he is sober asking him why is DEF inline when F is the intersect of circle radius BD rotate round center B and ray CB
>>16540092Ooh, neat. I figure out how to use the law of sines that gives a unique solution, but it isn't pretty. Using it for a triangle will always lead to a loop on itself or recursion because constructing a relevant triangle isn't unique (im annoyed i didnt notice this before). Constructing a relevant quadrilateral definitely will be unique though, so you'd need to use four equations for the four sides. It seems for any EDCB, >sum of all angles equals 360>sin(EDB)*sin(DCE)*sin(CBD)*sin(BEC) = sin(DEC)*sin(EBD)*sin(BCE)*sin(CDB)In your case, >sin(x)*sin(30)*sin(60)*sin(50) = sin(110-x)*sin(20)*sin(50)*sin(40)So this problem is asking for x+40. This one is NOT trivial to solve though, so your way is better.>>16540134I feel this is just unique to the problem, no?, cause DBC = 60 and BDF = 30. Like, there are many angles that will do this too, but not all of them will.
>>16539975You can spend 15 min doing the math or you can take the superior engineer point of view and just draw the triangle and measure the angle.I got CDE = 70 degrees
>>16540134alpha is measured counterclockwise from axis AF to line CF and beta is measured counterclockwise from axis AF to line DFgamma is measured counterclockwise from axis AF to line AB and delta is measured counterclockwise from axis AF to line ACangle ABF = alpha - gammaangle ACF = alpha - deltaangle ADF = beta - deltaangle AEF = beta - gamma
>>16542603furthermore:angle ABF = alpha - gamma = 180 - (180 - 20)/2 = 100angle ACF = alpha - delta = (180 - 20)/2 = 80angle ADF = beta - delta = 180 - angle CDEangle AEF = beta - gamma = angle BEDandangle BAC = delta - gamma = 20
angle CFD = 30angle ABF = 100angle ACF = 80angle DBF = 120angle ECF = 50
