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I)\(75^\circ\)

II) \(105^\circ\)

III)\(135^\circ\)

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. 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.

- 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 angles of the following measurements:I)\(30^\circ\)II) \(22\frac{1}{2} )^\circ\)III)\(15^\circ\)
- Construct an equilateral triangle, given its side and justify the construction.

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