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The next Triangle challenge extends the https://www.talkgraphics.com/showthr...Triangles-S90S one.
Take the same Triangle (B318px/H412px) [S90S].
Now instead of the two lengths being at an angle of 90°, I want the simplest means to change this interior angle to one of 142° [S142S]
If you are keen, produce a second triangle with an interior angle of 17° as well. [S17S]
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Acorn
This one is harder to describe but easy to do.
It does require visualisation and some simple arithmetic.
Here is an aid that I use:
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Acorn
Sorry Acorn I do not understand either your aid post #2 nor post #1 could you elaborate a little please.
Larry, here is another view of what I want the Challenge to solve:Quote:
Originally Posted by wizard509
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The Red triangle is S90S from before; two known side lengths (the base and the height) with 90° between them.
The green triangle has the same short sides, the base is obviously the same length and the edge with the blue check marks is asserted to be the same length. The Green triangle's hypotenuse, the longer third side, necessarily has to change when the 90° is increased to 142°.
The Challenge was simply how to do this accurately and simply using a Xara function.
The clue is how the Shape tool presents the angle of any node, where 0° always points horizontally left.
The direction of increase is clockwise and above 180°, the sign flips so the value decreases back to 0°.
The original Blue triangle is the additional Challenge where the angle subtended is decreased from 90° to 17°.
Acorn
This is my approach to the problem...
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I almost forgot to mention that the angle of the nodes is (-180 -the amount chosen). :)
Marco, a good answer.Quote:
Originally Posted by Marco D
How do you measure the angle? [Your Edit covers my question]
Have a think about not using the circle at all.
Acorn
I've edited my previous post with the answer... -180 minus the angle chosen.
As did I.Quote:
Originally Posted by Marco D
Nice cross-posting,
Acorn
Another solution could be this...
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this way you don't need to do the math. ;)
Marco, I agree this is a neat solution.Quote:
Originally Posted by Marco D
There are a couple of implicit steps you need to tease out.
I would co-locate the left nodes of each line with both kept horizontal.
Then set the rotation point to the left end of the line and finally rotate it to the required angle.
Joining lines can be tricky.
Select both and switch to the Shape tool.
Click and drag one of the left nodes away and back again and release.
Finally, press the Tab key to reach an end of either line and press the Enter key.
It might not be that easy to get the second triangle of this Challenge from this position.
Acorn
OK, my approach is to create a S90S triangle (https://www.talkgraphics.com/showthr...Triangles-S90S).
At this point I have the Shape tool open so I tab to the top node and change the -90 angle (A) to -38.
Done.
For S17S, change -90 to -163.
The trick is the entered value is -(180 - Angle), as identified by Marco.
With a changed Angle, a new triangle is available at any time, but necessarily requires a bit of arithmetic.
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Acorn