Goutham wrote: An angle is given in a plane. Using only a compass, one must find out $(a)$ if this angle is acute. Find the minimal number of circles one must draw to be sure. $(b)$ if this angle equals $31^{\circ}$.(One may draw as many circles as one needs). $(a)$ seems to be easy... Let the angle be $\angle ABC.$ Then you draw a circle with radius $CB$ and centre at $C$. If $BA$ is a secant to the circle in the original way $\overrightarrow{BA}$ is drawn, it will be acute; if it is tangent then it will be $90^{\circ}$ and if it is obtuse then it will be a secant when you produce $\overrightarrow{AB}$ to meet it. As for $(b)$ I have not figured it out yet.

listen,instead of drawing circles can we do it this way: at first,we find out $15$degree angle by at first finding out $60$degree then $30$ degree and then $15$ degree.Then,we find out $16$ degree in the following way:we know,$s=(T*2*pie*r/360)$ where,$T$=value of angle. $s$=length of arc $r$=radius Now,if we consider $T=90$,and $r=1 unit$,we get,$s=pie/4$. now,if we consider this equation,we let,$T=16$ degree,for $s=pie/4$ we will get a value of $r$. We take a straight line $AC$,with centre $A$ and radius $r$,we cut off an arc at $B$.from $B$,with radius $s$,we cut off an arc E on the previous arc.We join $A,E$.Then,$angle BAE=16$ degree.Then by drawing a triangle with $15$ degree and $16$ degree as interior angles,We get exterior angle $31$degree. If our angle is equal to this angle,then it is equal to $31$ degree,or,not.

For b, I wonder if we can do it this way, but do not try this at home. Draw a circle C centered at the angle, which intersects the 2 rays at 2 points. Measure their distance d and draw another circle centered at one of the points which intersect C at another point. Draw another circle centered at that point with radius d and intersects C at another point. Do this 360 times, and see if it revolves around C 31 times.

For b): Draw a circle centered at the vertex of the angle (we will name it $\alpha$). We can easily find out if $2\alpha$ is smaller, bigger or equal to $60^{\circ}$. If it is smaller or equal then we are done. Let $2\alpha>60^{\circ}$. We can find out if $2\alpha-60^{\circ}$=$2^{\circ}$ (note that $180\cdot2^{\circ}=360^{\circ}$).

Any angle of integer number of degrees divisible by 3 is constructible. Multiply the unknown angle by 3 and compare it to $93^\circ.$ One way to construct $93^\circ$ angle is $108^\circ - 15^\circ$ or regular pentagon interior angle take away $60^\circ / 4.$ Let $\angle BOA$ be the given angle, $OB = OA,$ and let $\angle COA = 3 \angle BOA.$ Perpendicular to $OA$ at $O$ cuts circle $(O, OA)$ at $D.$ $M$ is midpoint of $OD$ and circle $(M, MA)$ cuts the ray $(OD$ at $E$ $\Longrightarrow$ $\frac{OE}{OA}= g$ (golden ratio). $OAFE$ is a rectangle and circle $(A, AF)$ cuts $(O, OA)$ at $G$ $\Longrightarrow$ $\frac{AG}{OA}= \frac{AF}{OA}= \frac{OE}{OA}= g$ $\Longrightarrow$ $\angle GOA = 108^\circ.$ Circle $(G, GO)$ cuts $(O, OA)$ at $H$ $\Longrightarrow$ $\triangle GOH$ is equilateral. Bisector of $\angle GOH$ cuts $(O, OA)$ at $I$ and bisector of $\angle GOI$ cuts $(O, OA)$ at $J.$ Compare $\angle COA = 3 \angle BOA, \angle JOA = 93^\circ.$

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Can anyone tell me if there is anything wrong with this solution? Multilply the given angle 360 times and see if it makes exactly 31 complete rounds?

I don't know if I am authorized to tell that , but i think that your solution is correct. @yetti We are not allowed to use ruler.