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Construct the angles of the following measurements:I)$$30^\circ$$II) $$22\frac{1}{2} )^\circ$$III)$$15^\circ$$

I) Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Draw the ray OE passing through C. Then, $$\angle{EOA}$$ = $$60^\circ$$.

4. Taking B and C as centres and with the radius more than ($$\frac{1}{2}$$ )BC, draw arcs to intersect each other, say at D.

5. Draw the ray OD, this ray OD is the bisector of the $$\angle{EOA}$$ = $$60^\circ$$.
$$\Rightarrow$$ $$\angle{EOD}$$ = $$\angle{AOD}$$ = ($$\frac{1}{2}$$ )$$\angle{EOA}$$ = ($$\frac{1}{2}$$ )$$60^\circ$$ = $$30^\circ$$

II) Steps of construction:

1.Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, drawn an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then, $$\angle{EOA}$$ = $$60^\circ$$

5. Draw the ray OF passing through D. Then, $$\angle{FOE}$$ = $$60^\circ$$.

6. Next, taking C and D as centres and with the radius more then ($$\frac{1}{2}$$ )CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the $$\angle{FOE}$$
$$\Rightarrow$$ $$\angle{FOG}$$ = $$\angle{EOG}$$ = ($$\frac{1}{2}$$ )$$\angle{FOE}$$ = $$\frac{1}{2}$$$$60^\circ$$ = $$30^\circ$$
Thus, $$\angle{GOA}$$ = $$\angle{GOE}$$ + $$\angle{EOA}$$
$$\Rightarrow$$ $$\angle{GOA}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$.

8. Now taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I, respectively.

9. Next, taking H and I as centres and with the radius more than $$\frac{1}{2}$$HI, draw arcs to intersect each other, say at J.

10. Draw the raw OJ. This ray OJ is the required bisector of the $$\angle{GOA}$$.
Thus, $$\angle{GOJ}$$ = $$\angle{AOJ}$$ = $$\frac{1}{2}$$$$\angle{GOA}$$ = $$\frac{1}{2}$$$$90^\circ$$ = $$45^\circ$$.

11. Now, taking O as centre and any radius, drawn an arc to intersect the rays OA and OJ, say at K and L, respectively.

12. Next, taking K and L as centres and with the radius more than $$\frac{1}{2}$$KL, draw arcs to intersect each other, say at H.

13. Draw the ray OM. This ray OM is the bisector of the $$\angle{AOJ}$$
$$\Rightarrow$$ $$\angle{JOM}$$ = $$\angle{AOM}$$ = (1/2) $$\angle{AOJ}$$ = $$\frac{1}{2}$$× $$45^\circ$$ = $$22\frac{1}{2} ^\circ$$

III) Steps of construction:

1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an intersecting the previously drawn arc, say at a point c.

3. Draw the ray OE passing through C. Then, $$\angle{EOA}$$ = $$60^\circ$$

4. Next, taking B and C as centres and with the radius more then $$\frac{1}{2}$$BC, draw arcs to intersect each other, say at D.

5. Draw the ray OD intersecting the arc drawn in step 1 at F. This ray OD is the bisector of the the $$\angle{EOA}$$.
$$\Rightarrow$$ $$\angle{EOD}$$ = $$\angle{AOD}$$ = $$\frac{1}{2}$$$$\angle{EOA}$$ = $$\frac{1}{2}$$$$60^\circ$$ = $$30^\circ$$

6. Now, taking B and F as centers and with the radius more then $$\frac{1}{2}$$ BF, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the $$\angle{AOD}$$.
$$\therefore$$ $$\angle{DOG}$$ = $$\angle{AOG}$$ = $$\frac{1}{2}$$$$\angle{AOD}$$ = $$\frac{1}{2}$$$$30^\circ$$ = $$15^\circ$$