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Help I dont know how to do this.. :S


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Posted
My triangle is very close to the one shown....you have overlooked the instructions on changing the view to plan.

 

I haven't overlooked that and you assume too much my friend, as I said without the drawing you have no idea which leg is longest.

 

Well this is enough of "when geeks go wild" for me.

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Posted (edited)
I haven't overlooked that and you assume too much my friend, as I said without the drawing you have no idea which leg is longest.

 

 

Indeed, there is no way to predict the real dimensions of the triangle in the screencap if the view is not already perpendicular to the plane the triangle is drawn on.

 

Furthermore, the hint says:

 

"OBTAIN TRUE VIEW BY UCS AND NEW COMMAND. SELECT OB OPTION AND PICK THE TRIANGLE. TYPE PLAN ENTER. USING XLINE AND THE BISECT OPTION DRAW A LINE FROM THE TOP... "

 

The instructions do not specify which side of the triangle to "pick" to align the UCS with, so unless it's and equilateral triangle, "top" is relative, and the bisecting angle depends on which where the user "picks" the triangle.

Edited by nestly
Posted

The answer is 21½ degrees approximately.

Posted

I did as instructed and only one angle.

I think the problem has been described in full.

 

(There is a real trick in the description - follow it exaclty, otherwise you will be mislead by what you "think" it says.)

 

trick = distractor

Posted
My triangle is very close to the one shown....you have overlooked the instructions on changing the view to plan.

 

You cannot possible draw your triangle very close to the one shown - you have no way of knowing the view angle of rotation.

Doesn't really matter to the solution - it is a general solution, you just can't give an exact answer without the geometry.

Posted

How do you obtain "true view by UCS" ?

Posted

The triangle is a close polyline - follow the instructions.

It appears (OK I'm making a couple of assumptions here based on my experience with teachers) that the view was rotated and a polyline triangle drawn. Then the view was set back to UCS world to confuse the unaware (first of two distractors).

Posted

How are the people without experience with teachers and not assuming anything supposed to solve this?

Posted
How are the people without experience with teachers and not assuming anything supposed to solve this?

 

The instructions in the screencap are not well written, however it's unclear whether those instructions are exactly the same as those provided by the teacher.

Posted
You cannot possible draw your triangle very close to the one shown - you have no way of knowing the view angle of rotation.

Doesn't really matter to the solution - it is a general solution, you just can't give an exact answer without the geometry.

 

My triangle IS drawn close to the one shown, as in it has three sides. That is all that is needed to answer the question asked by the OP

How do i do this??
.

 

HE NEVER ASKED FOR THE ANSWER TO THE PROBLEM NOR CAN ONE BE GIVEN.

Posted
I wonder if this thread will develop like the Perimeter one :?

 

:lol:Have your answer yet?:lol:

Posted
My triangle IS drawn close to the one shown, as in it has three sides. That is all that is needed to answer the question asked by the OP .

 

HE NEVER ASKED FOR THE ANSWER TO THE PROBLEM NOR CAN ONE BE GIVEN.

 

Yep, I agree, well the exact answer could be given if the dwg file was provided.

Posted
the exact answer could be given if the dwg file was provided.

 

Which side of the triangle would you choose to align the UCS with? :wink:

Posted
Which side of the triangle would you choose to align the UCS with? :wink:

 

Doesn't matter, and in any case the instructions say to use the Oject option which doesn't give you the choice - it does it for you based on the original object.

Posted
Yeah, I'm not smart enough to understand the instructions. What's the point of drawing a xline if you're using the option.

 

Ambiguous, IMO

 

You are bisecting a particular angle and distance (hint - not the angle you probably think and probably not the distance you think) - try it.

Posted
another says to draw to the mid-point of the shortest leg.

 

The second distractor. Don't just read it - do it and observe the results (even though you select the midpoint as instructed - that is not where the line is drawn).

Posted

I am afraid that this thread has descended into the realms of fantasy.

 

This is the OP's message

 

How do i do this??

 

In the image attached, the first three lines seem to me to be asking to quantify an angle formed in a particular way. In the Hint section, it also says to use a dimension with the angle option.

 

Now, to me, if you use an angular dimension, you are going to see an angle.

 

And I posted my answer in post #23. o:)

Posted
The second distractor. Don't just read it - do it and observe the results (even though you select the midpoint as instructed - that is not where the line is drawn).

 

OK, I "did"it, and I continue to believe the question, and the solution are ambiguous and/or subjective. What part of the instructions am I not following correctly, or not understanding? I could have picked the "mid-point" and the endpoint of the longest line to define the xline, but that is not what the initial question suggests.

 

 

http://www.screencast.com/t/yXunjRzGerse

Posted
.... but that is not what the initial question suggests.

 

 

Like dimensions on a drawing - I would interpret the step-by-step instructions having precedence.

Posted

Posts by OP: 1.

 

Posts by all others: 38.

 

Either the OP gave up or he dropped the class. Meanwhile, back on the merry-go-round.............................:lol:

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