What is the proof of constructing $30^{\circ}, 60^{\circ}, 120^{\circ}$ and $135^{\circ}$ angles with ruler and compass? I can prove $90^{\circ}$ by proving that the line joining point of intersection of two circle is perpendicular to the line joining their radius. But I can't prove these angles. Thank You.
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The proof may depend on the actual construction you use. – Hagen von Eitzen Nov 01 '17 at 10:34
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The more interesting construction is that of an angle of 36 degrees (or for that matter any multiple of 18 degrees). – Paramanand Singh Nov 01 '17 at 10:40
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+1 for showing your interest in proofs. Most students learning geometric constructions just try to remember it as a routine procedure. – Paramanand Singh Nov 02 '17 at 08:38
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You recieved 2 answers to your question. If any of them is what you need, you should accept it by clicking the grey tick mark below the question score. Otherwise, explain what is still missing in the answers! – 5xum Nov 02 '17 at 10:18
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@5xum sorry sir I did not knew that before, i will do that now. – Ram Keswani Nov 02 '17 at 13:32
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@RamKeswanim You should do that on all the questions you posted... – 5xum Nov 03 '17 at 07:36
2 Answers
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For constructing $60^{\circ}$ what we do is, simply construct a equilateral triangle. Since each interior angle of an equilateral triangle is $60^{\circ}$, we're done.
Here $PA=AB=PB$
Rest is just bisecting this angle to get $30^{\circ}$, and reconstruction of this angle on the side $PA$, to double the angle, thus resulting into $120^{\circ}$
What remains is $135^{\circ}$, for this we construct $90^{\circ}$, again $90^{\circ}$ on previous one, and then bisect later one, getting an angle $90^{\circ}+\frac{90^{\circ}}{2}=135^{\circ}$
Jaideep Khare
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@Jaideej Khare, This consrtuction engages the compass. If I understand OP's question, only the ruler is allowed. – szw1710 Nov 01 '17 at 10:39
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1@RamKeswani Remember that while constructing all the arcs, the length of compass remains constant. Thus, when $P$ is the base, $A$ and $B$ are equidistant from$P$ and thus $PA=PB$. In Step 2, we choose $B$ as the base, and keep the length constant, and hence $AB=PA=PB$. – Jaideep Khare Nov 01 '17 at 10:43
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@JaideepKhare right sir, I got the proof of it through circles. now how do we prove bisected angles? – Ram Keswani Nov 01 '17 at 10:48
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1@RamKeswani Bisecting angles is just finding the midpoint of angles, which will lie on the line joining midpoint of chord, and center of circle. Hence, Bisecting angles is basically, Bisecting chords. – Jaideep Khare Nov 01 '17 at 10:52
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@JaideepKhare Sir why while constructing a perpendicular bisector on a line, the radius should be smaller than the line? – Ram Keswani Nov 01 '17 at 16:12
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2@RamKeswani: radius need not be smaller than the line segment. In fact it should be greater than half the length of the line segment. But note that choosing a large radius will make your image larger and perhaps some part of it might go outside the page. – Paramanand Singh Nov 01 '17 at 19:34
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@ParamanandSingh yes sir I saw that. and also saw that it cannot be less than half the length because then the circles(arc) wint intersect. But what about when it is exactly half the length sir? – Ram Keswani Nov 02 '17 at 08:17
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@RamKeswani: at exactly half the length the circles will meet at exactly one point (namely the mid point of the segment). The problem however is that you can't find half the length just by thinking. Drawing a perpendicular bisector is one of the easiest ways to find the mid point and thus half the length of the segment. – Paramanand Singh Nov 02 '17 at 08:34
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@RamKeswani If the radius is exactly equal to half the length of line segment, then the circles will intersect tangentially at the mid point. – Jaideep Khare Nov 02 '17 at 08:34
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@JaideepKhare then sir this seems a much easier and faster way. why do schools do not teach us this? – Ram Keswani Nov 02 '17 at 09:07
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@ParamanandSingh sir we can measure and then divide, or when we are just asked to make any line we can make it a multiple of 2. – Ram Keswani Nov 02 '17 at 09:09
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@RamKeswani : I learnt this in school. And measuring involves a marked ruler and this is not allowed in geometric constructions. – Paramanand Singh Nov 02 '17 at 10:33
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See the picture above: if we have $60^{\circ}$, it is easy to draw a right angle. When we have this, $135^{\circ}$ is easy because you can draw a square and its diagonal.
szw1710
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