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

Solution:  

To construct an angle 90°, the given steps are followed:

Step I: A ray OA is drawn.

Step II: O is taken as a centre with any radius, an arc DCB is that cuts OA at B is drawn.

Step III: With B as a centre with the same radius, a point C on the arc DCB is marked.

Step IV: With C as a centre and the same radius, a point D on the arc DCB is marked.

Step V: Taking C and D as centre, two arcs which intersect each other with the same radius at P are drawn.

Step VI: Finally, the ray OP is joined, which makes a 90° angle with OP.

To prove ∠POA = 90°

In order to prove this, draw a dotted line from the point O to C and O to D and the angles formed are:

From the construction, it is observed that

OB = BC = OC

Therefore, OBC is an equilateral triangle

So that, ∠BOC = 60°.

Similarly,

OD = DC = OC

Therefore, DOC is an equilateral triangle

So that, ∠DOC = 60°.

From SSS triangle congruence rule

△OBC \( \cong \) OCD

So, ∠BOC = ∠DOC [By C.P.C.T]

Therefore, ∠COP = ½ ∠DOC = ½ (60°).

∠COP = 30°

To find the ∠POA = 90°:

∠POA = ∠BOC+∠COP

∠POA = 60°+30°

∠POA = 90°

Hence, justified.

2. Construct an angle of 45° at the initial point of a given ray and justify the construction.

Solution: 

The steps of construction are as follows:

(i) Taking the given ray PQ, an arc of some radius with point P as its centre is drawn, which intersects PQ at R.

(ii) Taking R as centre and with the same radius as before, an arc intersecting the previously drawn arc at S is made.

(iii) Taking S as centre and with the same radius as before, an arc intersecting the arc at T is drawn. 

(iv) Taking S and T as centre, an arc of the same radius to intersect each other at U are drawn.

(v) Joining PU, let it intersect the arc at point V.

(vi) Now from R and V draw arcs with other at W with a radius more than 1/2RV to intersect each other. PW is the required ray making 45o with PQ.

To justify the construction, we have to prove <WPQ = 45o;  PS and PT are joint

We have SPQ = TPS = 60o. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.

UPS =   TPS= 

Now, UPQ = SPQ + UPS

                  = 60 + 30

                  = 90°

In step (vi) of this construction, we constructed PW as the bisector of UPQ

WPQ = UPQ  = 45°

3. Construct the angles of the following measurements:

(i) 30° (ii) 22 1/ 2° (iii) 15°

Solution: 

(i) \( 30^{\circ} \)

The steps of construction are as follows:

Step I: Draw the given ray PQ. Now taking P as centre and with some radius, draw an arc of a circle that intersects PQ at R.

Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.

Step III: Now taking R and S as centre and with radius more than ½ RS, draw arcs to intersect each other at T. Join PT, which is the required ray, making 30o with the given ray PQ.

(ii) 22 1/2°

Step I:  Draw a ray PQ. To construct an angle of 60°, with P as a centre and any radius, draw a wide arc to intersect.

Step II: PQ at R. With R as a centre and same radius draw an arc to intersect the initial arc at S. Then, ∠SPR = 60°

Step III: To construct an adjacent angle of 60°, with S as the centre and the same radius as before, draw an arc to intersect the initial arc at T. Then, ∠TPS = 60°

Step IV: To bisect ∠TPS, with T and S as centres and radius greater than half of TS , draw arcs to intersect each other at Z. Join P and Z. Then, ∠ZPQ = 90°e) To bisect ∠ZPQ, with R and U as centres and radius more than half of RU, draw arcs to intersect each other at V. Join P and V. Then, ∠VPQ = 45°

Step V: To bisect ∠VPQ = 45°, with W and R as centres and radius greater than half of WR, draw arcs to intersect each other at X. Join P and X. PX bisects ∠VPQ.

Hence, ∠XPQ = 1/2 ∠WPQ

= 1/2 × 45°

=22 1/2°

^{iii) 15°}

Construction procedure:

Step I: An angle ∠DOA = 60° is drawn.

Step II: Taking O as centre with any radius, an arc BC is drawn, which cuts OA at B and OD at C.

Step III: Now, drawing the bisector from the point B and C where it intersects at the point E such that it makes an angle ∠EOA = 30°.

Step IV: Again, ∠EOA is bisected such that ∠FOA is formed which makes an angle of 15° with OA.

Step V:Thus, ∠FOA is the required angle making 15° with OA.

4. Construct the following angles and verify by measuring them by a protractor:

(i) 75° (ii) 105° (iii) 135

Solution:

(i) 75°

Construction procedure:

Step I: A ray OA is drawn.

Step II: With O as centre draw an arc of any radius and intersect at the point B on the ray OA.

Step III: With B as centre draw an arc C and C as centre draw an arc D.

Step IV: With D and C as centre draw an arc, that intersect at the point P.

Step V: Join the points O and P

Step VI: The point that arc intersect the ray OP is taken as Q.

Step VII: With Q and C as centre draw an arc, that intersect at the point R.

Step VIII: Join the points O and R

Step IX: Thus, ∠AOE is the required angle making 75° with OA.

(ii) 105°

Construction procedure:

Step I: A ray OA is drawn.

Step II: With O as centre drawing an arc of any radius and intersect at the point B on the ray OA.

Step III: With B as centre an arc C and C as centre draw an arc D is drawn.

Step IV: With D and C as centre drawing an arc, that intersect at the point P.

Step V: Joining the points O and P

Step VI: The point that arc intersect the ray OP is taken as Q.

Step VII: With Q and D as centre draw an arc that intersects at the point R.

Step VIII: Join the points O and R

Step IX: Thus, ∠AOR is the required angle making 105° with OA.

(iii) 135°

Construction procedure:

Step I: Draw a line AOA‘

Step II: Drawing an arc of any radius that cuts the line AOA‘ at the point B and B‘

Step III:  With B as centre, an arc of same radius at the point C is made.

Step IV:  With C as centre, an arc of same radius at the point D is drawn. 

Step V:  With D and C as centre, draw an arc that intersects at the point P

Step VI: Join OP

Step VII: The point that the arc intersects the ray OP is taken as Q and it forms an angle 90°

Step VIII: With B‘ and Q as centre, draw an arc that intersects at the point R

Step IX: Thus, ∠AOR is the required angle making 135° with OA.

5. Construct an equilateral triangle, given its side, and justify the construction

Solution:

An equilateral triangle has three equal sides and three angles equal to 60º each.

Steps of construction:

Step I: Draw a ray AB.

Step II: With A as centre and radius equal to 3 cm, draw an arc to cut ray AB at C such that AC = 3 cm is made. 

Step III: With C as the centre and radius equal to AC (3cm), draw an arc intersecting the initial arc at D.

Step IV: Join AD and DC.

Triangle ADC is an equilateral triangle with sides of 3cm each.

Justification:

AC = AD (By construction since the radius of the arc is the same)

AC = CD (By construction since the same radius was used again)

Hence AC = AD = CD = 3cm

Thus, ADC is an equilateral triangle.