# Q1.    Construct an angle of 90o at the initial point of a given ray and justify the construction.

The steps of construction to follow:

Step 1: Draw a ray OP.

Then, take O as the centre and any radius draw an arc cutting OP at Q.

Step 2: Now, taking Q as the centre and with the same radius as before draw an arc cutting the previous arc at R. Repeat the process with R to cut the previous arc at S.

Step 3: Take R and S as centre draw the arc of radius more than the half of RS and draw two arcs intersecting at A. Then, join OA.

Hence, $\angle POA = 90^{\circ }$.

Justification:

We need to justify, $\angle POA = 90^{\circ }$

So, join OR and OS and RQ. we obtain

By construction OQ = OS = QR.

So, $\triangle ROQ$ is an equilateral triangle. Similarly $\triangle SOR$ is an equilateral triangle.

So, $\angle SOR = 60^{\circ}$

Now, $\angle ROQ = 60^{\circ}$ that means $\angle ROP = 60^{\circ}$.

Then, join AS and AR:

Now, in triangles OSA and ORA:

$SR = SR$  (common)

$AS = AR$  (Radii of same arcs)

$OS = OR$  (radii of the same arcs)

So, $\angle SOA = \angle ROA = \frac{1}{2}\angle SOR$

Therefore, $\angle ROA = 30^{\circ}$

and $\angle POA = \angle ROA+\angle POR = 30^{\circ} +60^{\circ} =90^{\circ}$

Hence, justified.

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