Question

Construct an angle of ${45}^{\circ }$ at the initial point of a given ray and justify the construction.

Answer 1

Steps to construct an angle of ${45}^{\circ }$
(i) Let us take a ray PQ with initial point P and draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as the centre and with the same radius as before, draw an arc intersecting the arc at T(see figure).
(iv) Taking S and T as the centre, draw an arc of the same radius to intersect each other at U.
(v) Join PU, let it intersects the arc at point V.
(vi) From R and V, draw arcs with radius more than $\frac{1}{2}$ RV to intersect each other at W.
(vii) Join PW, which is the required ray that makes ${45}^{\circ }$ with PQ.


Justification
We can justify the construction, if we can prove $\angle WPQ={45}^{\circ }$.
For this, join PS and PT.

We have, $\angle SPQ=\angle TPS={60}^{\circ }$.
In (iii) and (iv) steps of this construction, PU was drawn as the bisector of $\angle TPS$.
$\therefore \angle UPS=\frac{1}{2}\angle TPS=\frac{{60}^{\circ }}{2}={30}^{\circ }$
Also, $\angle UPQ=\angle SPQ+\angle UPS$
$={60}^{\circ }+{30}^{\circ }$
$={90}^{\circ }$
In step (vi) of this construction, PW was constructed as the bisector of $\angle UPQ$
$\therefore \angle WPQ=\frac{1}{2}\angle UPQ=\frac{{90}^{\circ }}{2}={45}^{\circ }$