|May6-11, 06:49 PM||#18|
"Simple" geometry problem
It doesn't equal 2r, my comment was if it does then that's the answer. What makes this problem "impossible" is that H or J is not fixed. phi becomes some angle but we don't know where the vertex is. AD can be thought of as a radius going through an angle of zero with AC through theta/2 to AI. Without H or J fixed at some distance this can't be solved. We were given a two stipulation fact: JD bisects delta AND meets AE at angle phi.
|May6-11, 06:57 PM||#19|
|May7-11, 04:29 PM||#20|
You have 10 (relevant) variables in a quadrilateral.
4 side lengths
4 vertex angles
2 diagonal lengths
You say you know four of these -one angle, two sides and one diagonal.
Attached are six equations (10 including the knowns) for the 6 unknowns.
These equations apply to a general quadrilateral. They cna be greatly simplified if the vertices can be shown to lie on a circle as I have already mentioned.
I have relabelled them to conform to conventional order. Theta is conventionally used for the intersection angle of the diagonals.
You can solve them at your leisure.
|May8-11, 06:34 PM||#21|
|Similar Threads for: "Simple" geometry problem|
|CM of a Irregular Object||Introductory Physics Homework||2|
|Irregular Cone Geometry||Precalculus Mathematics Homework||1|
|LQC into irregular cosmo||Beyond the Standard Model||2|
|Pretty basic chem/density problem: finding density of irregular shaped soluble solid||Chemistry||1|
|centroid of irregular polygons||General Math||1|