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Is it possible to draw $85 ^{\circ} $ and $110 ^{\circ}$ angles by compass and straightedge construction?If yes then how?

I thought about it and the only thing coming to mind other than a protractor is to use the concept of trigonometry functions but that still needs calculator,and the other ides coming to my mind involves trisecting an angle which is impossible,any other feasible ideas?

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We cannot construct these arcs by straightedge and compass. For if we can draw an $85^\circ$ arc, we can draw a $5^\circ$ arc by subtraction from an easily constructible $90^\circ$ arc, and then construct a $20^\circ$ angle by repeated duplication, again easily done by straightedge and compass. And if we can draw a $110^\circ$ arc, we can draw a $20^\circ$ angle by subtracting a $90^\circ$ arc.

However, it was proved by Wantzel, almost two centuries ago, that the $20^\circ$ angle cannot be drawn with straightedge and compass. Equivalently, we cannot trisect the $60^\circ$ angle with straightedge and compass. For details, please see the Wikipedia article on Angle Trisection.

Comment: If we allow other tools than the classical Euclidean straightedge and compass, the answer changes. There are $3$-legged compass analogues (linkages) that can do trisection. We can also trisect angles by Origami (the Japanese art of paper-folding). From antiquity on, various correct trisection methods have been described. As they must, they all go beyond straightedge and compass.

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    With respect to the comment, we can trisect angles is we use Meccano! In fact, using Meccano gives us a strictly stronger geometry than just straight-edge and compass. Gerard 't Hooft has written some papers on this. Although I do like the origami one!2011-08-04
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It is a theorem of Gauss (and... Wantzel? Gauss did sufficient, someone else did necessary) that we can construct a regular $n$-gon via a straight edge and compass if and only if $n$ is of the form $p_1p_2\ldots p_i2^j$ where the $p_k$ are distinct Fermat Primes.

Now, if you can construct an angle of $d$-degrees then you can use this to construct a regular $n$-gon, where $n$ is such that $d.n=0 \text{ mod }360$ (draw your angle on a circle, follow where your lines hit the edge and repeat - basically, work out the order of $d$ in the cyclic group of order $360$).

If $d=110$ then $n=36=2^{2}3^{2}$.

If $d=85$ then $n=72=2^{3}3^{2}$.

Your Fermat Primes are $3$, $5$, $17$, $257$ and $65537$ (although I believe it is an open question if there are any more). However, they each must occur no more than once in the prime factorisation of $n$. $3$ appears twice in both $36$ and $72$ (aka $9$ divides them both), so we cannot construct a $36$-gon nor a $72$-gon. Thus, we cannot construct angles of either 110 or 85 degrees.

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Depending on the accuracy you might need, it is fairly easy to work out (by hand) the length of he perimeter of a 5 degree arc of a given radius. You can then measure a chord on a circle using the compass to give you a 5 degree angle. Not necesserily accurate for 20/110 degrees but might be good enough. Should be surprisingly accurate for 5/85 degrees.

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    @Jonas Meyer:It's "he" and you are right,I have updated the question along your lines.2011-08-04
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Use the ruler and compass to draw two lines at 90 degrees. Then measure up the two lines the right lengths to form the two sides of a right triangle with an 85 degree angle on one corner and a 5 degree angle on the other. Joint the two measured lengths which gives you a line at 85 degrees and then draw the arc from the corner of the right triangle with the 85 degree angle.

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    Maybe this answer relies on the OP saying "ruler" and not "straightedge"?2011-08-04