construct the following angles at the initial point of a given Ray and justify the construction 90° in 45°
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The below given steps will be followed to construct an angle of 45°.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PU. Let it intersect the arc at point V.
(vi) From R and V, draw arcs with radius more than 1/2 RV to intersect each other at W. Join PW.
PW is the required ray making 45° with PQ.
Justification of Construction:
We can justify the construction, if we can prove ∠WPQ = 45°.
For this, join PS and PT.
We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.
∴ ∠UPS = 1/2 ∠TPS = 60°/2 = 30°
Also, ∠UPQ = ∠SPQ + ∠UPS
= 60° + 30° = 90°
In step (vi) of this construction, PW was constructed as the bisector of ∠UPQ.
∴ ∠WPQ = 1/2 ∠UPQ = 90°/2 = 45°
The below given steps will be followed to construct an angle of 90°.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PU, which is the required ray making 90° with the given ray PQ.

Justification of Construction:
We can justify the construction, if we can prove ∠UPQ = 90°.
For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.
∴ ∠UPS = 1/2 ∠TPS = 1/2 x 60° = 30°
Also, ∠UPQ = ∠SPQ + ∠UPS
= 60° + 30°
= 90°
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PU. Let it intersect the arc at point V.
(vi) From R and V, draw arcs with radius more than 1/2 RV to intersect each other at W. Join PW.
PW is the required ray making 45° with PQ.
Justification of Construction:
We can justify the construction, if we can prove ∠WPQ = 45°.
For this, join PS and PT.
We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.
∴ ∠UPS = 1/2 ∠TPS = 60°/2 = 30°
Also, ∠UPQ = ∠SPQ + ∠UPS
= 60° + 30° = 90°
In step (vi) of this construction, PW was constructed as the bisector of ∠UPQ.
∴ ∠WPQ = 1/2 ∠UPQ = 90°/2 = 45°
The below given steps will be followed to construct an angle of 90°.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PU, which is the required ray making 90° with the given ray PQ.

Justification of Construction:
We can justify the construction, if we can prove ∠UPQ = 90°.
For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.
∴ ∠UPS = 1/2 ∠TPS = 1/2 x 60° = 30°
Also, ∠UPQ = ∠SPQ + ∠UPS
= 60° + 30°
= 90°
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