Need hlep with a triangle question (Trigonometry)

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The discussion revolves around solving a trigonometry problem involving an equilateral triangle ABC that has been folded, with vertex A resting on line BC at point D. The lengths BD and DC are given as 1 and 2 units, respectively. Participants analyze the angles formed by the folding, confirming that angle PDQ is 60 degrees due to the properties of the equilateral triangle. They deduce that lengths DQ and AQ are both 2 units, while lengths QC and BP are 1 unit each. The conversation highlights the need for clarity in the problem's details and the possibility of visualizing the scenario on a coordinate plane for better understanding.
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I'm doing some trigonometry, and right now I'm given this equilateral triangle,

http://i.imgur.com/WOPCk.png
So how would I go about to find side AP, AQ, and PQ? I think I have a grasp of what it is, but not sure how to execute. I know you have 60 degree angle for ABC and all sides are 3, but not sure how to go about finding the rest of the stuff. Would I maybe make side AP 3 - x and BP x and then AQ as 3-y and QC as y?
 
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Lengths AP, AQ, PQ could be pretty much anything, for the information you have given. Have you missed some of the details of the question out?
 
So the whole question is...

An equilateral triangle ABC has been creased and folded so that its vertex A now rests on BC at D, such that BD = 1 and DC = 2. Find the length of

A) AP B) AQ C) PQ

That's all it says, so I drew a diagram of it, since u know the angles of ABC and the outer line.
 
Isn't angle PDQ 60 degrees because it was formed by folding the original triangle?

Line PA and line PD are also the same, as should AQ and DQ for the same reason above.
 
djpailo said:
Isn't angle PDQ 60 degrees because it was formed by folding the original triangle?

Line PA and line PD are also the same, as should AQ and DQ for the same reason above.

Yeah, should be since it's an equilateral, so now you know the inner angles and the outer angles, plus the bottom side.
 
I visualize the triangle PAQ spinning on an axis that intersects point Q and that is parallel to line BC. So PAQ is the vertical flip of QDP. I think, both angles are equal in every way. I picture a quadrilateral, where all the angles that make up the two unknown, bigger, angles in the quadrilateral are equal to 60 deg. I see that all the angles in the triangles in ABC are 60 degrees now.

DQ = 2 units

DQ = AQ = 2 units

As I continue it seems that all the unknown lines' lengths except QC and BP, which are 1 unit each, are equal to 2 units.

In my attempt to answer the posed question, I suspected that this thread might be a slightly elaborate prank. However, I continued to work through the problem in case it was not. I think it's possible that the validity of the question and given info by the original poster can be verified by someone who can translate the given info into a representation on the xy plane. If my work here is erroneous, feel free to inform me.
 
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