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it is not possible to construct angles which are not divisible by 7.5 degree using a compass. but you are free to use a protractor for that purpose:) Posted by Srusti Dash(BUXI JAGABANDHU ENGLISH MEDIUM SCH) on 28/10/12

it is not possible to construct angles which are not divisible by 7.5 degree using a compass. but you are free to use a protractor for that purpose:) Posted by Srusti Dash(BUXI JAGABANDHU ENGLISH MEDIUM SCH) on 28/10/12

it is not possible to construct angles which are not divisible by 7.5 degree using a compass. but you are free to use a protractor for that purpose:) Posted by Srusti Dash(BUXI JAGABANDHU ENGLISH MEDIUM SCH) on 28/10/12

but if i could make with a protactor i wouldnt hav asked this q u mean my book has wrong information..............yeah ....hummm. any more answers anyways 10x fr the ans..!!!!!!!!!!! Posted by Abhilasha Rathi(ST ANTHONY'S SR SEC SCHOOL) on 28/10/12

but if i could make with a protactor i wouldnt hav asked this q u mean my book has wrong information..............yeah ....hummm. any more answers anyways 10x fr the ans..!!!!!!!!!!! Posted by Abhilasha Rathi(ST ANTHONY'S SR SEC SCHOOL) on 28/10/12

but if i could make with a protactor i wouldnt hav asked this q u mean my book has wrong information..............yeah ....hummm. any more answers anyways 10x fr the ans..!!!!!!!!!!! Posted by Abhilasha Rathi(ST ANTHONY'S SR SEC SCHOOL) on 28/10/12

Draw a point, and a long straight line through it. Put the centre of the compass over the point, with north on the compass (or zero degrees) on the line. Then draw a second line out from the point at 80 degrees on the compass. You 'll then have a 80 degree angle between the lines. It 's usually a trick that teachers decide to do on their pupils. A compass is what you use to find north and a pair of compasses is used to draw circles. Posted by Shivani(dav international school kharghar) on 28/10/12

Draw a point, and a long straight line through it. Put the centre of the compass over the point, with north on the compass (or zero degrees) on the line. Then draw a second line out from the point at 80 degrees on the compass. You 'll then have a 80 degree angle between the lines. It 's usually a trick that teachers decide to do on their pupils. A compass is what you use to find north and a pair of compasses is used to draw circles. Posted by Shivani(dav international school kharghar) on 28/10/12

Draw a point, and a long straight line through it. Put the centre of the compass over the point, with north on the compass (or zero degrees) on the line. Then draw a second line out from the point at 80 degrees on the compass. You 'll then have a 80 degree angle between the lines. It 's usually a trick that teachers decide to do on their pupils. A compass is what you use to find north and a pair of compasses is used to draw circles. Posted by Shivani(dav international school kharghar) on 28/10/12

totally rong ans miss shivani ........ Posted by Abhilasha Rathi(ST ANTHONY'S SR SEC SCHOOL) on 2/11/12

totally rong ans miss shivani ........ Posted by Abhilasha Rathi(ST ANTHONY'S SR SEC SCHOOL) on 2/11/12

totally rong ans miss shivani ........ Posted by Abhilasha Rathi(ST ANTHONY'S SR SEC SCHOOL) on 2/11/12

you cant. omly the multiples of 15 can be constructed in compass Posted by Sanjana Godolkar(JANADEEPA SCHOOL) on 18/1/13

you cant. omly the multiples of 15 can be constructed in compass Posted by Sanjana Godolkar(JANADEEPA SCHOOL) on 18/1/13

you cant. omly the multiples of 15 can be constructed in compass Posted by Sanjana Godolkar(JANADEEPA SCHOOL) on 18/1/13

Any angle that is not divisible by 15,is to be drawn with a protractor. Posted by Chippu(R.N. PODAR SCHOOL) on 7/3/15

Any angle that is not divisible by 15,is to be drawn with a protractor. Posted by Chippu(R.N. PODAR SCHOOL) on 7/3/15

Any angle that is not divisible by 15,is to be drawn with a protractor. Posted by Chippu(R.N. PODAR SCHOOL) on 7/3/15

Hey guys, the method I will be telling isn 't something which can be justified, at least by me. However, it yields desired results, so you can go ahead with it you must be knowing how to construct a 60 ^{o} angle, so I won 't be writing down steps for that.problem arises when we have to construct 20 ^{o} angle, so here are the steps : draw any line PQ. Mark a point O anywhere between P and Q on the line. With O as centre, and any suitable radius, draw an arc intersecting PQ at A and B . With B as centre, and he same radius, cut an arc C on previously drawn arc AB. This is the normal 60 ^{o} angle. With B and C as centres, cut arcs intersecting at L. This is the bisected 60^{o} ie 30 ^{o} BUT DO NOT join L to O, instead join L to A. Angle LAQ is the required 20 ^{o} triangle. So now you can construct 60^{o} on it . HOPE IT HELPS :D Posted by Sana(DPS) on 9/3/15

Hey guys, the method I will be telling isn 't something which can be justified, at least by me. However, it yields desired results, so you can go ahead with it you must be knowing how to construct a 60 ^{o} angle, so I won 't be writing down steps for that.problem arises when we have to construct 20 ^{o} angle, so here are the steps : draw any line PQ. Mark a point O anywhere between P and Q on the line. With O as centre, and any suitable radius, draw an arc intersecting PQ at A and B . With B as centre, and he same radius, cut an arc C on previously drawn arc AB. This is the normal 60 ^{o} angle. With B and C as centres, cut arcs intersecting at L. This is the bisected 60^{o} ie 30 ^{o} BUT DO NOT join L to O, instead join L to A. Angle LAQ is the required 20 ^{o} triangle. So now you can construct 60^{o} on it . HOPE IT HELPS :D Posted by Sana(DPS) on 9/3/15

Hey guys, the method I will be telling isn 't something which can be justified, at least by me. However, it yields desired results, so you can go ahead with it you must be knowing how to construct a 60 ^{o} angle, so I won 't be writing down steps for that.problem arises when we have to construct 20 ^{o} angle, so here are the steps : draw any line PQ. Mark a point O anywhere between P and Q on the line. With O as centre, and any suitable radius, draw an arc intersecting PQ at A and B . With B as centre, and he same radius, cut an arc C on previously drawn arc AB. This is the normal 60 ^{o} angle. With B and C as centres, cut arcs intersecting at L. This is the bisected 60^{o} ie 30 ^{o} BUT DO NOT join L to O, instead join L to A. Angle LAQ is the required 20 ^{o} triangle. So now you can construct 60^{o} on it . HOPE IT HELPS :D Posted by Sana(DPS) on 9/3/15