Math Question

[bizmark]

Originally posted by: heliomphalodon

We're given a square ABCD together with a point E in the square such that angle(EAB) = angle(EBA) = 15.
To prove: triangle DEC is equilateral.

Consider the triangle AEB. Reflect it in the line AC so that A->A, B->D and E->E'.
Now consider triangle AEE'. It is equilateral.
Thus triangles AE'D and DE'E are congruent, so
triangle ADE is isosceles and thus
triangle DEC is equilateral. QED

Now that's the sort of proof I like!

pic of solution

Very nice. I've got to admit I was skeptical at the beginning. How do you *know* that AEE' is equilateral, though? It's still not really apparant to me Of course it's obvious if you already know that DEC is equilateral, but....

P.S. it's *very* obvious that AEE' is isosceles, but how do we know that EE' is the same length as the other two sides? Again, you could do it if you knew that the angle ADE was 30 degrees, but then the problem is already solved.

 

[heliomphalodon]

Originally posted by: bizmark

Originally posted by: heliomphalodon

We're given a square ABCD together with a point E in the square such that angle(EAB) = angle(EBA) = 15.
To prove: triangle DEC is equilateral.

Consider the triangle AEB. Reflect it in the line AC so that A->A, B->D and E->E'.
Now consider triangle AEE'. It is equilateral.
Thus triangles AE'D and DE'E are congruent, so
triangle ADE is isosceles and thus
triangle DEC is equilateral. QED

Now that's the sort of proof I like!

pic of solution

Very nice. I've got to admit I was skeptical at the beginning. How do you *know* that AEE' is equilateral, though? It's still not really apparant to me Of course it's obvious if you already know that DEC is equilateral, but....

P.S. it's *very* obvious that AEE' is isosceles, but how do we know that EE' is the same length as the other two sides? Again, you could do it if you knew that the angle ADE was 30 degrees, but then the problem is already solved.

As you say, it's very obvious that triangle EAE' is isosceles. To see that it is in fact equilateral, observe that angle(EAE') = 60.

Thanks for putting up a picture!

 

[bizmark]

Originally posted by: heliomphalodon

As you say, it's very obvious that triangle EAE' is isosceles. To see that it is in fact equilateral, observe that angle(EAE') = 60.

Thanks for putting up a picture!

D'oh! I thought of that, but I didn't take the next logical step. Now I see that any isosceles triangle with an included angle of 60 must be an equilateral triangle b/c the two other angles must be the same, and 120/2=60.

 

[BumJCRules]

heliomphalodon,

You said... "Close, and probably the right approach, but once you introduce an approximation you have lost rigor. A correct proof must be exact (or symbolic). I think your use of the Law of Sines is probably the key to a clean result, but we're not there yet..."

Please give me the exact decimal notation of Pi.

3.1415926535 8979323846 2643383279 5028841971 6939937510
5820974944 5923078164 0628620899 8628034825 3421170679
8214808651 3282306647 0938446095 5058223172 5359408128
4811174502 8410270193 8521105559 6446229489 5493038196
4428810975 6659334461 2847564823 3786783165 2712019091
4564856692 3460348610 4543266482 1339360726 0249141273
7245870066 0631558817 4881520920 9628292540 9171536436
7892590360 0113305305 4882046652 1384146951 9415116094
3305727036 5759591953 0921861173 8193261179 3105118548
0744623799 6274956735 1885752724 8912279381 8301194912

9833673362 4406566430 8602139494 6395224737 1907021798
6094370277 0539217176 2931767523 8467481846 7669405132
0005681271 4526356082 7785771342 7577896091 7363717872
1468440901 2249534301 4654958537 1050792279 6892589235
4201995611 2129021960 8640344181 5981362977 4771309960
5187072113 4999999837 2978049951 0597317328 1609631859
5024459455 3469083026 4252230825 3344685035 2619311881
7101000313 7838752886 5875332083 8142061717 7669147303
5982534904 2875546873 1159562863 8823537875 9375195778
1857780532 1712268066 1300192787 6611195909 2164201989

3809525720 1065485863 2788659361 5338182796 8230301952
0353018529 6899577362 2599413891 2497217752 8347913151
5574857242 4541506959 5082953311 6861727855 8890750983
8175463746 4939319255 0604009277 0167113900 9848824012
8583616035 6370766010 4710181942 9555961989 4676783744
9448255379 7747268471 0404753464 6208046684 2590694912
9331367702 8989152104 7521620569 6602405803 8150193511
2533824300 3558764024 7496473263 9141992726 0426992279
6782354781 6360093417 2164121992 4586315030 2861829745
5570674983 8505494588 5869269956 9092721079 7509302955

3211653449 8720275596 0236480665 4991198818 3479775356
6369807426 5425278625 5181841757 4672890977 7727938000
8164706001 6145249192 1732172147 7235014144 1973568548
1613611573 5255213347 5741849468 4385233239 0739414333
4547762416 8625189835 6948556209 9219222184 2725502542
5688767179 0494601653 4668049886 2723279178 6085784383
8279679766 8145410095 3883786360 9506800642 2512520511
7392984896 0841284886 2694560424 1965285022 2106611863
0674427862 2039194945 0471237137 8696095636 4371917287
4677646575 7396241389 0865832645 9958133904 7802759009

9465764078 9512694683 9835259570 9825822620 5224894077
2671947826 8482601476 9909026401 3639443745 5305068203
4962524517 4939965143 1429809190 6592509372 2169646151
5709858387 4105978859 5977297549 8930161753 9284681382
6868386894 2774155991 8559252459 5395943104 9972524680
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0168427394 5226746767 8895252138 5225499546 6672782398
6456596116 3548862305 7745649803 5593634568 1743241125

1507606947 9451096596 0940252288 7971089314 5669136867
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9009714909 6759852613 6554978189 3129784821 6829989487
2265880485 7564014270 4775551323 7964145152 3746234364
5428584447 9526586782 1051141354 7357395231 1342716610
2135969536 2314429524 8493718711 0145765403 5902799344
0374200731 0578539062 1983874478 0847848968 3321445713
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7229109816 9091528017 3506712748 5832228718 3520935396
5725121083 5791513698 8209144421 0067510334 6711031412

6711136990 8658516398 3150197016 5151168517 1437657618
3515565088 4909989859 9823873455 2833163550 7647918535
8932261854 8963213293 3089857064 2046752590 7091548141
6549859461 6371802709 8199430992 4488957571 2828905923
2332609729 9712084433 5732654893 8239119325 9746366730
5836041428 1388303203 8249037589 8524374417 0291327656
1809377344 4030707469 2112019130 2033038019 7621101100
4492932151 6084244485 9637669838 9522868478 3123552658
2131449576 8572624334 4189303968 6426243410 7732269780
2807318915 4411010446 8232527162 0105265227 2111660396

6655730925 4711055785 3763466820 6531098965 2691862056
4769312570 5863566201 8558100729 3606598764 8611791045
3348850346 1136576867 5324944166 8039626579 7877185560
8455296541 2665408530 6143444318 5867697514 5661406800
7002378776 5913440171 2749470420 5622305389 9456131407
1127000407 8547332699 3908145466 4645880797 2708266830
6343285878 5698305235 8089330657 5740679545 7163775254
2021149557 6158140025 0126228594 1302164715 5097925923
0990796547 3761255176 5675135751 7829666454 7791745011
2996148903 0463994713 2962107340 4375189573 5961458901

9389713111 7904297828 5647503203 1986915140 2870808599
0480109412 1472213179 4764777262 2414254854 5403321571
8530614228 8137585043 0633217518 2979866223 7172159160
7716692547 4873898665 4949450114 6540628433 6639379003
9769265672 1463853067 3609657120 9180763832 7166416274
8888007869 2560290228 4721040317 2118608204 1900042296
6171196377 9213375751 1495950156 6049631862 9472654736
4252308177 0367515906 7350235072 8354056704 0386743513
6222247715 8915049530 9844489333 0963408780 7693259939
7805419341 4473774418 4263129860 8099888687 4132604721

5695162396 5864573021 6315981931 9516735381 2974167729
4786724229 2465436680 0980676928 2382806899 6400482435
4037014163 1496589794 0924323789 6907069779 4223625082
2168895738 3798623001 5937764716 5122893578 6015881617
5578297352 3344604281 5126272037 3431465319 7777416031
9906655418 7639792933 4419521541 3418994854 4473456738
3162499341 9131814809 2777710386 3877343177 2075456545
3220777092 1201905166 0962804909 2636019759 8828161332
3166636528 6193266863 3606273567 6303544776 2803504507
7723554710 5859548702 7908143562 4014517180 6246436267

9456127531 8134078330 3362542327 8394497538 2437205835
3114771199 2606381334 6776879695 9703098339 1307710987
0408591337 4641442822 7726346594 7047458784 7787201927
7152807317 6790770715 7213444730 6057007334 9243693113
8350493163 1284042512 1925651798 0694113528 0131470130
4781643788 5185290928 5452011658 3934196562 1349143415
9562586586 5570552690 4965209858 0338507224 2648293972
8584783163 0577775606 8887644624 8246857926 0395352773
4803048029 0058760758 2510474709 1643961362 6760449256
2742042083 2085661190 6254543372 1315359584 5068772460

2901618766 7952406163 4252257719 5429162991 9306455377
9914037340 4328752628 8896399587 9475729174 6426357455
2540790914 5135711136 9410911939 3251910760 2082520261
8798531887 7058429725 9167781314 9699009019 2116971737
2784768472 6860849003 3770242429 1651300500 5168323364
3503895170 2989392233 4517220138 1280696501 1784408745
1960121228 5993716231 3017114448 4640903890 6449544400
6198690754 8516026327 5052983491 8740786680 8818338510
2283345085 0486082503 9302133219 7155184306 3545500766
8282949304 1377655279 3975175461 3953984683 3936383047

4611996653 8581538420 5685338621 8672523340 2830871123
2827892125 0771262946 3229563989 8989358211 6745627010
2183564622 0134967151 8819097303 8119800497 3407239610
3685406643 1939509790 1906996395 5245300545 0580685501
9567302292 1913933918 5680344903 9820595510 0226353536
1920419947 4553859381 0234395544 9597783779 0237421617
2711172364 3435439478 2218185286 2408514006 6604433258
8856986705 4315470696 5747458550 3323233421 0730154594
0516553790 6866273337 9958511562 5784322988 2737231989
8757141595 7811196358 3300594087 3068121602 8764962867

4460477464 9159950549 7374256269 0104903778 1986835938
1465741268 0492564879 8556145372 3478673303 9046883834
3634655379 4986419270 5638729317 4872332083 7601123029
9113679386 2708943879 9362016295 1541337142 4892830722
0126901475 4668476535 7616477379 4675200490 7571555278
1965362132 3926406160 1363581559 0742202020 3187277605
2772190055 6148425551 8792530343 5139844253 2234157623
3610642506 3904975008 6562710953 5919465897 5141310348
2276930624 7435363256 9160781547 8181152843 6679570611
0861533150 4452127473 9245449454 2368288606 1340841486

3776700961 2071512491 4043027253 8607648236 3414334623
5189757664 5216413767 9690314950 1910857598 4423919862
9164219399 4907236234 6468441173 9403265918 4044378051
3338945257 4239950829 6591228508 5558215725 0310712570
1266830240 2929525220 1187267675 6220415420 5161841634
8475651699 9811614101 0029960783 8690929160 3028840026
9104140792 8862150784 2451670908 7000699282 1206604183
7180653556 7252532567 5328612910 4248776182 5829765157
9598470356 2226293486 0034158722 9805349896 5022629174
8788202734 2092222453 3985626476 6914905562 8425039127

5771028402 7998066365 8254889264 8802545661 0172967026
6407655904 2909945681 5065265305 3718294127 0336931378
5178609040 7086671149 6558343434 7693385781 7113864558
7367812301 4587687126 6034891390 9562009939 3610310291
6161528813 8437909904 2317473363 9480457593 1493140529
7634757481 1935670911 0137751721 0080315590 2485309066
9203767192 2033229094 3346768514 2214477379 3937517034
4366199104 0337511173 5471918550 4644902636 5512816228
8244625759 1633303910 7225383742 1821408835 0865739177
1509682887 4782656995 9957449066 1758344137 5223970968

3408005355 9849175417 3818839994 4697486762 6551658276
5848358845 3142775687 9002909517 0283529716 3445621296
4043523117 6006651012 4120065975 5851276178 5838292041
9748442360 8007193045 7618932349 2292796501 9875187212
7267507981 2554709589 0455635792 1221033346 6974992356
3025494780 2490114195 2123828153 0911407907 3860251522
7429958180 7247162591 6685451333 1239480494 7079119153
2673430282 4418604142 6363954800 0448002670 4962482017
9289647669 7583183271 3142517029 6923488962 7668440323
2609275249 6035799646 9256504936 8183609003 2380929345

9588970695 3653494060 3402166544 3755890045 6328822505
4525564056 4482465151 8754711962 1844396582 5337543885
6909411303 1509526179 3780029741 2076651479 3942590298
9695946995 5657612186 5619673378 6236256125 2163208628
6922210327 4889218654 3648022967 8070576561 5144632046
9279068212 0738837781 4233562823 6089632080 6822246801
2248261177 1858963814 0918390367 3672220888 3215137556
0037279839 4004152970 0287830766 7094447456 0134556417
2543709069 7939612257 1429894671 5435784687 8861444581
2314593571 9849225284 7160504922 1242470141 2147805734

5510500801 9086996033 0276347870 8108175450 1193071412
2339086639 3833952942 5786905076 4310063835 1983438934
1596131854 3475464955 6978103829 3097164651 4384070070
7360411237 3599843452 2516105070 2705623526 6012764848
3084076118 3013052793 2054274628 6540360367 4532865105
7065874882 2569815793 6789766974 2205750596 8344086973
5020141020 6723585020 0724522563 2651341055 9240190274
2162484391 4035998953 5394590944 0704691209 1409387001
2645600162 3742880210 9276457931 0657922955 2498872758
4610126483 6999892256 9596881592 0560010165 5256375679

How accurate do you want a figure? Once you move out toward the millionths and beyond the number is only getting minutely closer to the exact but never quite close enough. When using trig, calculus, dify q, etc to do math you will round a lot. Spend some time in an engineering or science lab for any length of time and rounding is normal. What kind of error limits do you want?


BTW: What eductaion level are you at? Junior High, High School, University, Grad School, Doctorate, Post Doctorate?

If you want me to leave the answers in fractional notation, that would be 100% accurate. However human brains do not think that abstractly. Unless you can think of Pi, fractals,l and all other forms of repeating digit numerals to their fullest decimal extent.

 

[bizmark]

BumJCRules

It's traditional in abstract (i.e. non-engineering, non-scientific) mathematics to leave everything symbolic. For example you don't write sqrt(2) as 1.414.... etc.; rather you write it as sqrt(2). If you're going to end up with a whole number, all of the sqrt(2)'s are going to cancel out (or be squared) in the end somehow. There's no reason to approximate. The same with pi. You write pi as pi, and leave it that way, even in the answer. If your answer ends up being 3 + 7*sqrt(2) + 5*pi, you leave it as such instead of writing some decimal.

Please give me the exact decimal notation of Pi.

There is no exact decimal notation of pi. That's why we leave it as a symbol

 

[heliomphalodon]

Originally posted by: BumJCRules
heliomphalodon,

You said... "Close, and probably the right approach, but once you introduce an approximation you have lost rigor. A correct proof must be exact (or symbolic). I think your use of the Law of Sines is probably the key to a clean result, but we're not there yet..."

How accurate do you want a figure? Once you move out toward the millionths and beyond the number is only getting minutely closer to the exact but never quite close enough. When using trig, calculus, dify q, etc to do math you will round a lot. Spend some time in an engineering or science lab for any length of time and rounding is normal. What kind of error limits do you want?

BTW: What eductaion level are you at? Junior High, High School, University, Grad School, Doctorate, Post Doctorate?

If you want me to leave the answers in fractional notation, that would be 100% accurate. However human brains do not think that abstractly. Unless you can think of Pi, fractals,l and all other forms of repeating digit numerals to their fullest decimal extent.

Please. :disgust: We're talking about a proof here, not an engineering calculation. Pi is "the area of a circle of radius 1" and (in a proof like the one under discussion in this thread) must be represented symbolically -- not replaced with any kind of approximate expansion, in any base. Similarly sqrt(3) is represented symbolically, not replaced with some inexact approximation (however accurate).

When using trig, calculus, dify q, etc to do math you will round a lot.

On the contrary -- when using trig, calculus, differential equations, etc. to do math (not engineering) one rarely rounds at all, because the results are symbolic, not numeric.

What kind of error limits do you want?

No "error limit" is acceptable in a proof of the kind under consideration in this thread.

However human brains do not think that abstractly.

But of course they do. You merely muddy the waters with approximations and their decimal expansion representations. The point is that this is a proof, and as such is entirely abstract.

BTW: What eductaion level are you at?

I am a graduate student. My undergraduate degree is in mathematics. My concentrations of study are the foundations of mathematics, and the theory of computation. My undergraduate thesis was titled, "Latin Squares and Finite Projective Planes."

May I suggest that you examine the three proofs I have provided in this thread? The first is largely algebraic but still exact, the second is constructive and the third is purely geometric. All are correct. I would even claim that the third is elegant. The proof given by bizmark is nearly correct -- its only flaw is the arbitrary introduction of the number 10 for the length of a side of the square ABCD. Technically, that quantity must remain unspecified (abstract) and be represented symbolically.

I didn't mean to insult you, I'm sorry you seem to have taken offense. But the fact is that your calculation, while correct, lacks rigor and cannot be considered to be a proof.

 

[BumJCRules]

It is sarcasim... Sorry. I thought that I could use some after that comment about exact figures.

Like saying 400/3MHz. People don't understand it. So most people say 133MHz, even thought it is an aproximation. Most people need hard figures to make it compute in their minds. May most of you are the exception to the rule.

 

[heliomphalodon]

Originally posted by: BumJCRules
It is sarcasim... Sorry. I thought that I could use some after that comment about exact figures.

Like saying 400/3MHz. People don't understand it. So most people say 133MHz, even thought it is an aproximation. Most people need hard figures to make it compute in their minds. May most of you are the exception to the rule.

Point taken. I missed the sarcastic nuance

You're certainly right about the 400MHz/3 being commonly thought of as 133MHz.
I'm genuinely curious -- just how precise and consistent are the clocks in a modern PC? Since we're talking about a physical object there must be some point beyond which further precision is meaningless. Are the clocks precise enough to distinguish 133333333Hz from 133333332Hz ?

 

[bizmark]

Originally posted by: heliomphalodon

I'm genuinely curious -- just how precise and consistent are the clocks in a modern PC?

From PC to PC - not very. Within the same PC, pretty damn exact. Everything depends on crystal oscillators, which have a natural frequency that can't be changed. How 'exact' it is, is dependent only on how exactly we want (or are able) to measure it. I guess that PCs with the same oscillator (same crystal, factory, etc.) will be pretty close, but there's probably always some sort of minor variation/imperfection within the crystals that gives slightly different results.

In my machine at home, the processor is 850MHz -- 8.5 * the system bus speed. However, for whatever reason, my system bus isn't exactly 100MHz -- it's closer to 101MHz. So my processor speed is actually 856MHz.

Here at work, I've got a 1.9GHz PIV. WCPUID reports a system clock of 99.70MHz, leading to an actual processor speed of 1894.24MHz and a system bus speed of 398.79MHz.

BTW you don't have PM's enabled so I'll tell you here to look over this thread and tell me what you think Towards the end I really started to think that I shouldn't have said anything in the first place. BTW my username was wbwither back then.

 

[rbhawcroft]

Originally posted by: bizmark

Originally posted by: heliomphalodonWe're given a square ABCD together with a point E in the square such that angle(EAB) = angle(EBA) = 15.To prove: triangle DEC is equilateral.Consider the triangle AEB. Reflect it in the line AC so that A->A, B->D and E->E'.Now consider triangle AEE'. It is equilateral.Thus triangles AE'D and DE'E are congruent, sotriangle ADE is isosceles and thustriangle DEC is equilateral. QED Now that's the sort of proof I like!

pic of solutionVery nice. I've got to admit I was skeptical at the beginning. How do you *know* that AEE' is equilateral, though? It's still not really apparant to me Of course it's obvious if you already know that DEC is equilateral, but....P.S. it's *very* obvious that AEE' is isosceles, but how do we know that EE' is the same length as the other two sides? Again, you could do it if you knew that the angle ADE was 30 degrees, but then the problem is already solved.

just use SOH CAH TOA it only takes three calculations ffs. just divide the side triangle into two right angled triangle and the third calc is the length of the equilateral triangles middle spar, on the relevant side.

 

[bizmark]

Originally posted by: rbhawcroft
just use SOH CAH TOA it only takes three calculations ffs. just divide the side triangle into two right angled triangle and the third calc is the length of the equilateral triangles middle spar, on the relevant side.

No, that doesn't work (assuming I understand what you said). Dividing the triangle ADE into two right angled triangles (by a horizontal line through E) will get you nowhere. Call the point where the new horizontal line intersects AD F. Then you know that angle AEF is 15 degrees, but you have no idea what angle FED is. You also don't know what angle ADE is, so you're left with a right triangle for which you don't know the two non-right angles. You would if you *assumed* that DEC is a right triangle (i.e. you assumed that angle ADE is 30 degrees), but that's not a valid assumption (that's what we're trying to prove).

 

[Turkey]

How about this?

since tri(AEB) is isosceles & ABCD is a square:
ang(EAD) = ang(EBC) = 90 - 15 = 75
ang(EDC) = ang(ECD)
ang(DEA) = ang(BEC)


75 + ang(DEA) + ang(ADE) = 180

ang(DEC) + 2 * ang(EDC) = 180

150 + 2 * ang(DEA) + ang(DEC) = 360

ang(EDC) = 90 - ang(ADE)

4 equations, 4 unknowns... use your favorite reduction method to determine that ang(DEC) = ang(EDC) = ang(ECD) = 60.

 

[heliomphalodon]

Originally posted by: Turkey
How about this?

since tri(AEB) is isosceles & ABCD is a square:
ang(EAD) = ang(EBC) = 90 - 15 = 75
ang(EDC) = ang(ECD)
ang(DEA) = ang(BEC)


75 + ang(DEA) + ang(ADE) = 180

ang(DEC) + 2 * ang(EDC) = 180

150 + 2 * ang(DEA) + ang(DEC) = 360

ang(EDC) = 90 - ang(ADE)

4 equations, 4 unknowns... use your favorite reduction method to determine that ang(DEC) = ang(EDC) = ang(ECD) = 60.

Hmmm... I don't think so. Those four equations are not linearly independent. They generate a matrix M of coefficients:

1 1 0 0
0 0 1 2
2 0 1 0
0 1 0 1

But the determinant of M is zero, so the equations are not independent. This means that the system cannot be solved...

I haven't studied linear algebra in many years -- someone please correct me if I'm wrong!

 

[Turkey]

Oops... you are correct!

 

[Weyoun]

Here's a simple proof. I finished it up the other night, but since rbhawcroft's method has just been refuted, I'll prove it for him.

Let EF be a point parallel to the top side meeting BC at F
therefore, BEF and ECF are right angled triangles
^EBY = 75 (angle sum, 90)
^BEY = 15 (corresponding angles of parallel lines)

EY = L/2, where L is the length of one of the sides of the square
let BF = y
let CF = x
let THETA = ^ECY
Tan75 = L/2y --> y = L/(2Tan75)
Tan THETA = L/2x --> x = L/(2Tan THETA)
now, x + y = L
L/(2Tan75) + L/(2Tan THETA) = L
1/(2Tan75) + 1/(2Tan THETA) = 1
1/(2Tan THETA) = 1 - 1/(2Tan75)
= (2Tan75 - 1)/2Tan75
Tan THETA = Tan75 / (2Tan75 - 1)

Which you algebra buffs can simply or you calculator buffs can evaluate. Either way, THETA = 30 is the solution to this equation.

Hope this helps

 

[heliomphalodon]

Originally posted by: Weyoun
Here's a simple proof. I finished it up the other night, but since rbhawcroft's method has just been refuted, I'll prove it for him.

Let EF be a point parallel to the top side meeting BC at F
therefore, BEF and ECF are right angled triangles
^EBY = 75 (angle sum, 90)
^BEY = 15 (corresponding angles of parallel lines)

EY = L/2, where L is the length of one of the sides of the square
let BF = y
let CF = x
let THETA = ^ECY
Tan75 = L/2y --> y = L/(2Tan75)
Tan THETA = L/2x --> x = L/(2Tan THETA)
now, x + y = L
L/(2Tan75) + L/(2Tan THETA) = L
1/(2Tan75) + 1/(2Tan THETA) = 1
1/(2Tan THETA) = 1 - 1/(2Tan75)
= (2Tan75 - 1)/2Tan75
Tan THETA = Tan75 / (2Tan75 - 1)

Which you algebra buffs can simply or you calculator buffs can evaluate. Either way, THETA = 30 is the solution to this equation.

Hope this helps

OK, but EF has to be a line segment (not a point) and you must mean F where you've written Y (that's a typo, I guess). Finally, to finish things off we still need an exact value for Tan75 (= 2 + Sqrt3) so your last equation becomes
Tan THETA = (2 + Sqrt3)/(3 + 2Sqrt3)
= 1/Sqrt3
So THETA = Atan(1/Sqrt3) = 30 QED

This is essentially the same proof I gave initially...

 

[Weyoun]

In the proof I did just before that, I used Y as the intersection point, not F. Thanks for picking up on those few mistakes As I said, you can either evaluate or simplify, but either way the solution is THETA = 30. While I can see your point about evaluation and infinite symbol accuracy for proofs (and I 100% agree with you), once we have reached the final stage of the proof, simple evaluation will suffice.