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March 19th, 2019, 04:50 PM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  How can I find the angle between two intersecting segments coming from a trapezoid?
I'm confused about this problem, since I don't know which Identity or theorem can be used to solve it. I tried to pull all the tricks under my sleeve but yet I can't find an answer, can somebody help me with this? The following figure represents a sugar whose atoms are labeled as $A$, $B$, $C$, $D$, $E$, $F$. $AE = 5 \sqrt{2} \, \overset{\circ}{A}$ , $EF = 5 \sqrt{3} \, FD = 5\, \overset{\circ}{A}$. If it is known that $BC=AB+CD$ and $\angle EBC = \angle FCB = 90^{\circ}$. Calculate the angle formed by the extension of the lines $AE$ and $FD$.See the figure from below which corresponds to the problem: The existing alternatives are: $\begin{array}{ll} 1.& 60^{\circ}\\ 2.& 53^{\circ}\\ 3.& 75^{\circ}\\ 4.& 90^{\circ}\\ 5.& 105^{\circ}\\ \end{array}$ So far what I could come up with was that what it was intended to be found is the angle which is colored by a blueberry shade. But I don't know how to relate it with the known information, hence what can be done?. If possible I would appreciate that the proposed answer could come up with some drawing or sketch as to better visualize what's trying to be meant. I have not posted it Last edited by Chemist116; March 19th, 2019 at 05:06 PM. Reason: added information missing from drawing 
March 19th, 2019, 09:17 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,942 Thanks: 1545 
Place point G on segment BC such that AB = BG. Since AB + CD = BC, then GC = CD. Connect both E and F to G forming two isosceles triangles, AEG and GFD. Note triangle EGF has $EG^2 + GF^2 = EF^2$ See what you can do from here ... 
March 20th, 2019, 05:29 AM  #3  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038  Quote:
Trying to make up a similar problem where triangle EFG is a 345 right triangle; FG=3, EG=4, EF=5. Any tips? Thanks. Last edited by Denis; March 20th, 2019 at 05:49 AM.  
March 20th, 2019, 02:27 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038 
Duhhhh...works for any right triangle with sides similarly given... With 345, fits inside isosceles right triangle with equal sides = 7. Thus solution always 90 degrees. 
March 20th, 2019, 03:22 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,194 Thanks: 897 Math Focus: Wibbly wobbly timeywimey stuff.  
March 20th, 2019, 04:57 PM  #6  
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  Just there but I still need a bit of help Quote:
See figure for more details: $\left(5 \sqrt{3}\right)^{2}= \left(5 \sqrt{2}\right)^{2} + \left(5 \right)^{2}  2 \left(5^{2} \sqrt{2}\right)\left(\sqrt{3}\right)\cos\angle EGF$ From this is reduced to: $ 2 \left(5^{2} \sqrt{2}\right)\left(\sqrt{3}\right) \cos\angle EGF = 0$ Therefore and assuming the angle $0 \leq \angle EGF \leq \frac{\pi}{2}$ then: $\angle EGF = 90^{\circ}$ Now the thing which I am still stuck at is how to prove or justify that the $\angle EHF = 90^{\circ}$ for now I am guessing it is going to be that value but I don't know how to establish that. I've tried to use all sorts of lines and parallel lines to do so but still I can't find a way. Can you further help me?. Last edited by Chemist116; March 20th, 2019 at 05:02 PM.  
March 20th, 2019, 05:43 PM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 2,942 Thanks: 1545 
Pythagoras? Since angle EGF = 90, then angle EGB and angle FGC are complementary $\implies$ angle A and angle D are complementary $\implies$ angle H = 90. Last edited by skeeter; March 20th, 2019 at 06:01 PM. 
March 20th, 2019, 05:46 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038  
March 21st, 2019, 04:15 PM  #9  
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  So at last So, next time I get a triangle whose I know its sides should I try to look if it follows Pythagoras theorem?. It looks like guessing, or just trial and error. I felt that using cosines law seemed a bit more rigorus proof. Quote:
It was after looking to the figure several times and prior reading at your answer that the only thing that I could come up with to justify the $90^{\circ}$ comes by drawing a parallel line to $AD$ so that interior angles summed up with the one which is being asked equate to a right angle. To reflect this I made a new drawing. Therefore from the figure: $\omega+\phi+\psi=180^{\circ}$ $\psi=180\left(\omega+\phi\right)=18090=90^{\circ}$ Again I'm not sure if this was what you meant but so far I cannot find other explanation. Did I do well?.  
March 21st, 2019, 05:03 PM  #10  
Math Team Joined: Jul 2011 From: Texas Posts: 2,942 Thanks: 1545  Quote:
Recommend you make yourself familiar with using the converse of the Pythagorean Theorem. With it one may determine if the longest side of a triangle is opposite a right angle, an obtuse angle, or an acute angle. If not a right angle, then the law of cosines is appropriate. I used the converse to determine $\angle{EGF} = 90$ ... not trial & error. Quote:
angle A and angle D are complementary $\implies \angle{A} + \angle{D}= 90$ $90 + \angle{H} = 180 \implies \angle{H} = 90$ You did ok, but it seems you make problems like this more difficult than necessary.  

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