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Construct the angles of the following measurements:
(i) $ 30^{\circ} $
(ii) $ 22 \frac{1}{2} $
(iii) $ 15^{\circ} $.
To do:
We have to construct the given angles.
Solution:
(i)
Steps of construction :
(a) Draw a ray $AB$.
(b) With centre $A$ and a suitable radius draw an arc meeting $AB$ at $C$.
(c) With centre $C$ and the same radius as above draw another arc meeting the above arc at $D$.
(d) Extend $AD$ to form the ray $AX$
Therefore,
$\angle BAX= 60^o$.
(e) With $C$ and $D$ as centres and radius more than $\frac{1}{2}CD$ draw two arcs meeting each other at $E$.
(f) Join $A$ and $E$ and extend it to form ray $AE$.
Therefore, $BAE=30^o$
(ii)
Steps of construction:
(a) Draw a ray $BC$.
(b) With centre $B$ and a suitable radius, draw an arc meeting $BC$ at $E$.
(c) With centre $E$ and the same radius as above, draw an arc meeting the above arc at $F$.
(d) With centre $F$ and the same radius as above, draw an arc meeting the first arc at $G$.
(c) With $F$ and $G$ as centres and radius more than $\frac{1}{2}FG$, draw two arcs meeting each other at $H$.
(d) Join $BH$
$\angle HBC = 90^o$.
(e) Let $BH$ intersects the first arc at $M$.
(f) With $E$ and $M$ as centres and radius greater than $\frac{1}{2}EM$, draw two arcs meeting each other at $K$.
(g) Join $B$ and $K$ and extend it to form ray $BY$
$\angle CBK=45^o$
(h) Let $L$ be the point where ray $BY$ intersects the first arc.
(i) With $E$ and $L$ as centres and radius greater than $\frac{1}{2}EL$, draw two arcs meeting each other at $J$.
(j) Join $LJ$ and extend it to form ray $BA$
(k) Therefore, $\angle CBA=22\frac{1}{2}^o$
(iii)
Steps of construction :
(a) Draw a ray $AB$.
(b) With centre $A$ and a suitable radius draw an arc meeting $AB$ at $C$.
(c) With centre $C$ and the same radius as above draw another arc meeting the above arc at $D$.
(d) Extend $AD$ to form the ray $AX$
Therefore,
$\angle BAX= 60^o$.
(e) With $C$ and $D$ as centres and radius more than $\frac{1}{2}CD$ draw two arcs meeting each other at $E$.
(f) Join $A$ and $E$ and extend it to form ray $AE$.
Therefore, $BAE=30^o$
(g) Let $F$ be the point of intersection of ray $AE$ and the first arc.
(h) With $C$ and $F$ as centres and radius more than $\frac{1}{2}CF$ draw two arcs meeting each other at $Y$.
(i) Join $AY$
Therefore, $BAY=15^o$.
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