Satish followed the steps given below

Step $1$: Construct an angle of $90°$.

Step $2$: Bisect the $90°$ angle.

Step $3$: Bisect one of the angles obtained in step $2$.

Which step is not required to construct a $45°$ angle?

The angle which cannot be constructed using a ruler and compass only is:

In the first square paper, the sides $AB, AD$ are folded about the corner point $A$, in a way as shown in figure $2$, so that the two folded parts overlaps each other perfectly.

When it is unfolded, we shall find the mark as in figure $3$

We find $\angle DAE=\angle EAF=\angle FAB=$_____$°$

Based on the construction, which statement is true?

Steps of construction:

2. With point $P$ as centre and the same radius, mark an arc and name it as $F$. Again put your compass at $F$ and with the same radius, draw an arc which intersects the initial arc at $E$.

3. Fix your compass at $E$, draw an arc with a convenient radius. Repeat the process by putting the compasses at $F$ and let the two arcs intersect at $B$. Join $OB$. $\angle BOA=90°$.

4. Now, bisect the obtained angle between $BO$ and $OA$. Mark it as $OC$. $\angle COA=$ _____$°$

Step 1: Let us draw any line segment $AB$ with scale pencil.

Step 2: Now, with centre $A$ and with any radius, lets draw an arc with compass, so that it cuts the line segment $AB$ at $C$.

Step 3: Next with centre $C$ and with the same radius, lets draw an arc which cuts the previous arc at $D$.

Step 4: Now, the points $D$ and $A$ are joined and extended. The angle is formed $\angle DAB=$ _____$°$.

The angle which cannot be constructed using a ruler and compass only is

Steps of construction:

1. Take any suitable radius and draw an arc which cuts ray $OA$ at point $P$. 

2. With point $P$ as centre and the same radius, mark an arc and name it as $F$. Again put your compasses at $F$ and with the same radius, draw an arc which intersects the initial arc at $E$.

3. Fix your compasses at $E$, draw an arc with a convenient radius. Repeat the process by putting the compasses at $F$ and let the two arcs intersect at $B$. Join $OB$. $\angle BOA=90°$.

4. Now, Bisect the obtained angle $BO$ and $OA$. Mark it as $OC$. $\angle COA=45°$.

5. Bisect $45°$ to get the required angle. The angles $OC$ and $OA$ are bisected and mark it as $OD$.

6. Hence, the required angle is $\angle DOA=$_____$°$. (Express in decimal)

In the first square paper, the sides $AB, AD$ are folded about the corner point $A$, in a way as shown in figure $2$, so that the two folded parts overlaps each other perfectly.

When it is unfolded, the folds are marked as in figure $3$.

It is found that, $\angle DAE=\angle EAF=\angle FAB=30°$

$\angle BAE=\angle DAF=60°$

Again if the side $AD$ is folded about the point $A$ such that $AD$ coincides with $AE$, it is unfolded

Then, $\angle DAG=$_____$°$