Stuck on trig

  • Thread starter Natasha1
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Could someone help me find the length in cm of the lines between the points in the picture attached.

The 3 angles around point 6 are 120 degrees each. This is also the case for point 8 which is linked to point 2, 3 and 7.

Now the 3 angles formed by the lines leaving point 7 are also 120 degrees.

If the distance between point 1 and 2, 2 and 3, 3 and 4, 4 and 5 is 1 cm.

What is the total distance of the lines?
 

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  • #2
Tom Mattson
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Natasha1 said:
The 3 angles around point 6 at 120 degrees each. This is also the case for point 8 which is linked to point 2, 3 and 7.

Now the 3 angles formed by the lines leaving point 7 are also 120 degrees.

Well they can't all be at 120 degrees, or else they would be on top of each other.

Can you post the exact problem statement, as well as your attempted solution? Thanks.
 
  • #3
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Tom Mattson said:
Well they can't all be at 120 degrees, or else they would be on top of each other.

Can you post the exact problem statement, as well as your attempted solution? Thanks.

I can't get an answer that's the reason of my post. I made the question up. Are you sure they would be on one and each other :-(((.

I'm pretty sure you are not correct, I could have 3 points with 3 angles formed around each of them with angles of 120 degrees.
 
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  • #4
Tom Mattson
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Natasha1 said:
I can't get an answer that's the reason of my post.

I'm not asking for the answer, I'm asking for your attempted solution. Our site policy (which you agreed to before posting) requires that you show some work before receiving help.

I'm pretty sure you are not correct, I could have 3 points there of degree 3 with angles of 120 degrees.

OK, I see what you're saying. When I see "120 degree angle" I think "120 degree angle measured counterclockwise from the positive x axis". That would put your rays at 120, 240, and 360 degrees. But even so, your angles don't look like they are in those positions.
 
  • #5
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The distances of the lines between points 5,6,4 and 2,8,3 (which are 2 isoceles triangles) is 2 cm.

This is how much I get as I have 4 sin 30 = 2

I am missing the distances from points 1,7 and 6,7 and also 7,8 to get my total distance can someone help :-) please
 
  • #6
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Natasha? you made this question up?...an attempt at drawing the problem yields (at least for me) a contradiction.

All five perimeter points are 1cm away from each other and therefore they must surely form a regular pentagon. By drawing rays with the specified 120 deg angles you can see where it becomes impossible to continue.

http://img109.imageshack.us/img109/1306/pentagon0bu.gif [Broken]
 
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