Question
Construct with ruler and compass, angle of measure $30^\circ .$
✓
Answer
$i.\ $Draw a line $\overleftrightarrow {PQ}$ and mark a point $O$ on it.
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line $PQ$ at a point say $A.$
$iii.\ $With the pointer at $A ($as centre$)$, now draw an arc that passes through $O.$
$iv.\ $Let the two arcs intersect at $B.$ Join $OB.$ We get $\angle BOA$ whose measure is $60^\circ .$
$v.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle BOA.$ Label the points of intersection as $D$ and $C.$
$vi.\ $With $C$ as centre, draw $($in the interior of $\angle BOA)$ an arc whose radius is more than half the length $CD.$
$vii.\ $With the same radius and with $D$ as centre, draw another arc in the interior of $\angle BOA.$ Let the two intersect at $E.$ Then, $\overline {OE}$ is the bisector of $\angle BOA,$ i.e. $\angle BOE = \angle EOA = 30^\circ .$
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line $PQ$ at a point say $A.$
$iii.\ $With the pointer at $A ($as centre$)$, now draw an arc that passes through $O.$
$iv.\ $Let the two arcs intersect at $B.$ Join $OB.$ We get $\angle BOA$ whose measure is $60^\circ .$
$v.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle BOA.$ Label the points of intersection as $D$ and $C.$
$vi.\ $With $C$ as centre, draw $($in the interior of $\angle BOA)$ an arc whose radius is more than half the length $CD.$
$vii.\ $With the same radius and with $D$ as centre, draw another arc in the interior of $\angle BOA.$ Let the two intersect at $E.$ Then, $\overline {OE}$ is the bisector of $\angle BOA,$ i.e. $\angle BOE = \angle EOA = 30^\circ .$
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