Create an arc where quadrant and tangent in same location tangent to angled line

[afisher]

See attachment

[Anonymous]

If I'm understanding what you are looking for correctly, then there are 2 solutions. I've re-attached your drawing with both solutions.

[nestly2 · Accepted solution]

I feel like I should know how to do this without any construction geometry, but at this moment, this is all I'm coming up with.

[afisher]

I was looking for the larger arc below the horizontal line.  How did you accomplish that circle?

[Kent1Cooper]

One way to go about it, not requiring any Circle(s) as middle-men:

Fillet the horizontal and angles Lines at any non-zero radius.

Scale the resulting Arc, using the now-Apparent Intersection of those two Lines as the base point, with the Reference option: use the distance from that Apparent Intersection to the left end of the Arc as the Reference dimension, and take that to the Tangent/Quadrant point.

Trim or Extend [depending on how the Fillet radius you used compares to the "answer" radius] the two Lines to the scaled Arc.

....

OR [this does involve an intermediate Circle] if you don't want a Fillet-like result, but want the Lines left as they are, draw a Circle using the TTR option, tangent to the horizontal and angled Lines at any non-zero radius, and Scale it similarly [except that the Intersection won't be Apparent], substituting "top quadrant point of the Circle" for "left end of the Arc" in the description above, and trim the Circle to the two Lines.

[Anonymous]

hi afisher,

I actually did it the same way nestly2 has posted - just did up a drawing to show you what I did and then saw that they had already posted a nice video of it

[GrantsPirate]

bisect the large angle, extend to the vertical, draw the arc from that intersection.