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# Construct the following angles and verify by measuring then by a protractor: I)$$75^\circ$$II) $$105^\circ$$III)$$135^\circ$$

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. Taking C as centre and with the same radius as before, draw 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 intersecting the arc of step 1 at H.. This ray OG is the bisector of the $$\angle{FOE}$$
$$\Rightarrow$$ $$\angle{FOG}$$ = $$\angle{EOG}$$ = $$\frac{1}{2}$$$$\angle{FOE}$$ = $$\frac{1}{2}$$×$$60^\circ$$ = $$30^\circ$$

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

9. Draw the ray OI. This ray OI is the bisector of the $$\angle{GOE}$$
$$\Rightarrow$$ $$\angle{GOI}$$ = $$\angle{EOI}$$ = $$\frac{1}{2}$$$$\angle{GOE}$$ = $$\frac{1}{2}$$$$30^\circ$$ = $$15^\circ$$
Thus, $$\angle{IOA}$$ = $$\angle{IOE}$$ + $$\angle{EOA}$$
$$\Rightarrow$$ $$\angle{IOA}$$ = $$15^\circ$$ + $$60^\circ$$ = $$75^\circ$$

On measuring the IOA by protractor, we find that $$\angle{IOA}$$ = $$75^\circ$$.
Thus, the construction is verified.

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, draw 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 intersecting the arc of step 1 at H.. This ray OG is the bisector of the $$\angle{FOE}$$
i.e., $$\angle{FOG}$$ = $$\angle{EOG}$$ = $$\frac{1}{2}$$$$\angle{FOE}$$ = $$\frac{1}{2}$$×$$60^\circ$$ = $$30^\circ$$
Thus, $$\angle{GOA}$$ = $$\angle{GOE}$$ + $$\angle{EOA}$$
$$\Rightarrow$$ $$\angle{GOA}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$.

8. Next, taking H and D as centres and with the radius more then ($$\frac{1}{2}$$HD, draw arcs to intersect each other, say at I.

9. Draw the ray OI. This ray OI is the bisector of the $$\angle{FOG}$$
$$\Rightarrow$$ $$\angle{FOI}$$ = $$\angle{GOI}$$ = $$\frac{1}{2}$$$$\angle{FOG}$$ = $$\frac{1}{2}$$×$$30^\circ$$ = $$15^\circ$$
Thus, $$\angle{IOA}$$ = $$\angle{IOG}$$ + $$\angle{GOA}$$
$$\Rightarrow$$ $$\angle{IOA}$$ = $$15^\circ$$ + $$90^\circ$$ = $$105^\circ$$
On measuring the IOA by protractor, we find that $$\angle{FOA}$$ = $$105^\circ$$.
Thus, the construction is verified.

III) Steps of construction:

1. Produce AO to A' to form ray OA'.

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

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

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

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

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

7. 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.

8. Draw the ray OG intersecting the arc of step 1 at H.. This ray OG is the bisector of the $$\angle{FOE}$$
i.e., $$\angle{FOG}$$ = $$\angle{EOG}$$ = $$\frac{1}{2}$$$$\angle{FOE}$$ = $$\frac{1}{2}$$×$$60^\circ$$ = $$30^\circ$$
Thus, $$\angle{GOA}$$ = $$\angle{GOE}$$ + $$\angle{EOA}$$
$$\Rightarrow$$ $$\angle{GOA}$$ = $$60^\circ$$ + $$30^\circ$$ = $$90^\circ$$.
$$\Rightarrow$$ $$\angle{B'OH}$$ = $$90^\circ$$

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

10. Draw the ray OI. This ray OI is the bisector of the $$\angle{B'OG}$$
$$\Rightarrow$$ $$\angle{B'OI}$$ = $$\angle{GOI}$$ = $$\frac{1}{2}$$$$\angle{B'OG}$$ = $$\frac{1}{2}$$×$$90^\circ$$ = $$45^\circ$$
Thus, $$\angle{IOA}$$ = $$\angle{IOG}$$ + $$\angle{GOA}$$
$$\Rightarrow$$ $$\angle{IOA}$$ = $$45^\circ$$ + $$90^\circ$$ = $$135^\circ$$
On measuring the IOA by protractor, we find that $$\angle{IOA}$$ = $$135^\circ$$.
Thus, the construction is verified.