# Construct the angles of the following measurements:(i) $30^{\circ}$(ii) $22 \frac{1}{2}$(iii) $15^{\circ}$.

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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, $BAY=15^o$.

Updated on 10-Oct-2022 13:41:42