Question

Construct an angle of $90°$ at the initial point of a given ray and justify the construction.

Answer 1

Constructing an angle of $90°$

Let us draw a ray $\mathrm{OA}$ and considering $\mathrm{O}$ as a centre with any radius, draw an arc that cuts $\mathrm{OA}$ at $\mathrm{B}$.

Considering $\mathrm{B}$ as a centre with the same radius as before, mark a point $\mathrm{C}$ and then considering $\mathrm{C}$ as a centre and the same radius, mark a point $\mathrm{D}$ on the previously drawn arc.

Now let us take $\mathrm{C}$ and $\mathrm{D}$ as the centre and radius more than $\frac{1}{2}\mathrm{CD}$, draw two arcs that intersect each other at $\mathrm{P}$.

Now the ray $\mathrm{OP}$ is joined which makes an angle of $90°$.

Thus, $\angle \mathrm{AOP}=90°$ .

Justification

To prove $\angle \mathrm{AOP}=90°$

To prove this, draw a dotted line from the point $O$ to $\mathrm{C}$ and $O$ to $\mathrm{D}$:

From the construction, we note that

$O\mathrm{B}=\mathrm{BC}=\mathrm{OC}$

Hence, $O\mathrm{BC}$ is an equilateral triangle

So that, $\angle \mathrm{BOC}=60°$.

Likewise

$O\mathrm{D}=\mathrm{D}C=\mathrm{OC}$

Hence, $\angle \mathrm{DOC}$ is an equilateral triangle

So that, $\angle \mathrm{DOC}=60°$.

From $\mathrm{SSS}$ triangle congruence rule

$∆\mathrm{OBC}\cong \mathrm{OCD}$

So, $\angle \mathrm{BOC}=\angle DOC$ [By C.P.C.T]

Hence, $\angle \mathrm{COP}=\frac{1}{2}\angle \mathrm{DOC}=\frac{1}{2}\left(60°\right)$.

$\angle \mathrm{COP}=30°$

To determine $\angle \mathrm{AOP}=90°$:

$\angle \mathrm{AOP}=$$\angle \mathrm{BOC}+\angle \mathrm{COP}$

$\angle \mathrm{AOP}=$ $60°+30°$

$\angle \mathrm{AOP}=90°$

Hence, Proved