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I)\(30^\circ\)

II) \(22\frac{1}{2} )^\circ\)

III)\(15^\circ\)

Answer :

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 \(\angle{AOD}\).

\(\therefore \) \(\angle{DOG}\) = \(\angle{AOG}\) = \(\frac{1}{2} \)\(\angle{AOD}\) = \(\frac{1}{2} \)\(30^\circ\) = \(15^\circ\)

- Construct an angle of \(90^\circ\) at the initial point of a given ray and justify the construction.
- Construct an angle of \(45^\circ\) at the initial point of a given ray and justify the construction.
- Construct the following angles and verify by measuring then by a protractor: I)\(75^\circ\)II) \(105^\circ\)III)\(135^\circ\)
- Construct an equilateral triangle, given its side and justify the construction.

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