My Math Forum X,Y angles.

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December 6th, 2010, 08:45 AM   #1
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X,Y angles.

[color=#000000]Find the measure of the angles X and Y.
[/color]

[attachment=0:1m0lk1hg]angles.png[/attachment:1m0lk1hg]

[color=#000000]Where ? put Y .[/color]
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 angles.png (6.2 KB, 198 views)

 December 7th, 2010, 02:43 PM #2 Math Team   Joined: Apr 2010 Posts: 2,770 Thanks: 356 Re: X,Y angles. Hello ZardoZ, I'm not sure on how to phrase in English, but I'll try: First we'll solve for X: The angles of the Pentagon are $108^{\circ}$. constructing an isosceles triangle with top angle of $108^{\circ}$ gives a base angle of $(180^{\circ}-108^{\circ})/2=36^{\circ}$ (The "baseline" is the lengthened line from M to $\prod$. Now by using complementary angles, we have: $(180^{\circ}-36^{\circ})/2=144^{\circ}$ Now for Y. Lengthen the line to the other pentagon intersecting the triangle in . Now, we have a triangle, B. So we have $\triangle BO\prod$ The angles of a hexagon are $120^{\circ}$ $\angle \prod= (180^{\circ}-120^{\circ})/2=60^{\circ}$ Angle M = Angle B (do you see that in the picture?) So Angle B = $36^{\circ}$ Now, angle Y $= 180^{\circ}-36^{\circ}-60^{\circ}=84^{\circ}$ Hoempa
 December 7th, 2010, 03:17 PM #3 Global Moderator   Joined: Dec 2006 Posts: 16,591 Thanks: 1199 With "in ." changed to "in B." (which you probably intended) both calculations make sense and get the correct answers. The minimal annotation of the diagram makes it a bit awkward to explain things clearly. In particular, one needs to assume that the pentagons and hexagon are regular, so that their angles are known.
 December 7th, 2010, 06:10 PM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 131 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: X,Y angles. [color=#000000]The answers are correct, I must have mentioned though that the pentagons and hexagon are regular.[/color]
December 7th, 2010, 07:58 PM   #5
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Re: X,Y angles.

Quote:
 Originally Posted by Hoempa Hello ZardoZ, I'm not sure on how to phrase in English, but I'll try: First we'll solve for X: The angles of the Pentagon are $108^{\circ}$. constructing an isosceles triangle with top angle of $108^{\circ}$ gives a base angle of $(180^{\circ}-108^{\circ})/2=36^{\circ}$ (The "baseline" is the lengthened line from M to $\prod$. Now by using complementary angles, we have: $(180^{\circ}-36^{\circ})/2=144^{\circ}$ Now for Y. Lengthen the line to the other pentagon intersecting the triangle in . Now, we have a triangle, B. So we have $\triangle BO\prod$ The angles of a hexagon are $120^{\circ}$ $\angle \prod= (180^{\circ}-120^{\circ})/2=60^{\circ}$ Angle M = Angle B (do you see that in the picture?) So Angle B = $36^{\circ}$ Now, angle Y $= 180^{\circ}-36^{\circ}-60^{\circ}=84^{\circ}$ Hoempa
how come (180-36)/2=144 and (180-120)/2=60 ( doesnt "/2" mean divide by two or am i missing something?)

 December 7th, 2010, 08:59 PM #6 Global Moderator   Joined: Dec 2006 Posts: 16,591 Thanks: 1199 I think Hoempa simply forgot to omit "/2", which doesn't belong in that line, but was needed in the earlier calculation.
 December 8th, 2010, 12:46 PM #7 Math Team   Joined: Apr 2010 Posts: 2,770 Thanks: 356 Re: X,Y angles. What you say is correct, skipjack, I copy-pasted these calculations and forgot to eliminate "/2", and I see that I was to say "in B." I can't edit though. Glad, it made sense, afterwards. Hoempa

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