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