
Geometry Geometry Math Forum 
 LinkBack  Thread Tools  Display Modes 
March 19th, 2019, 05:50 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 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 06:06 PM. Reason: added information missing from drawing 
March 19th, 2019, 10:17 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 
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, 06:29 AM  #3  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  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 06:49 AM.  
March 20th, 2019, 03:27 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
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, 04:22 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,340 Thanks: 983 Math Focus: Wibbly wobbly timeywimey stuff.  
March 20th, 2019, 05:57 PM  #6  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 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 06:02 PM.  
March 20th, 2019, 06:43 PM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 
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 07:01 PM. 
March 20th, 2019, 06:46 PM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  
March 21st, 2019, 05:15 PM  #9  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 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, 06:03 PM  #10  
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677  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.  

Tags 
angle, coming, find, intersecting, segments, trapezoid 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Find SV in the trapezoid  Math3ka  Geometry  4  November 10th, 2015 03:49 AM 
Find angle for max area of Trapezoid  coder  Calculus  3  July 23rd, 2015 12:09 PM 
Find the center of overlap between two line segments  1101  Algebra  4  February 22nd, 2011 12:43 PM 
Find the height of a trapezoid with basis 6 ft and 10 ft and  tobymac  Algebra  1  July 21st, 2009 11:45 PM 
find the dimension for isosceles trapezoid  dai_lo  Calculus  1  February 27th, 2008 12:45 PM 