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

Solution:

Construction:

- Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
- Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚

Justification:

By construction, we have AP = AQ = PQ, therefore, triangle APQ is an equilateral triangle. Thus, ∠APQ = ∠AQP = ∠QAP = 60˚

Also, we have, AQ = AR =RQ , therefore, triangle ARQ is also an equilateral triangle. Thus, ∠ARQ = ∠AQR = ∠QAR = 60˚

Since, AS bisects ∠QAR then we have ∠QAS = ∠SAR = ^{1}/_{2}∠QAR = ^{1}/_{2} x 60˚ = 30˚

Thus, ∠PAS = ∠PAQ + ∠QAS = 60˚ + 30˚ = 90˚

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

Solution:

Construction:

- Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
- Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
- Again taking S and P as centres with any radius draw two arcs intersecting each other at L.
- Join AL . Therefore, ∠PAL = 45 ˚

**Justification:**

Join GH and CH

In ∆AHG and ∆AHC, we have [arcs of equal radii]

HG = HC [radii of the same arc]

AG = AC [common]

⸫∆AHG ≅ ∆AHC [SSS congruence]

⇒∠HAG = ∠HAC [CPCT] ……………(i)

But ∠HAG + ∠HAC = 90˚ [By construction] ……………….(ii)

⇒∠HAG = ∠HAC = 45˚ [from (i) and (ii)]

**Construct the angles of the following measurements:**

**(i) 30° (ii) 22 ^{1}/_{2}° (iii) 15°**

Solution:

(i) 30° = ^{60}^{ ˚}/_{2}

Construction:

- Draw a ray AB with initial point A
- Draw an arc XY with any radius by taking A as centre which cuts the ray AB at C.
- Draw an arc by taking C as centre with same radius which intersects arc XY at D. Since AC = AD = DC then, ∆ADC is an equilateral triangle.
- ∠ADC = ∠ACD = ∠DAC = 60°
- Taking C and D as centres with any radius draw two arcs intersecting one another at E. Join AE.
- Thus,
^{1}/_{2}∠EAC = ∠DAC ;^{1}/_{2}∠EAC = 60˚ ⇒∠EAC = 30˚

(ii)22^{1}/_{2}˚ = ^{90˚}/_{2} = 45˚

Construction:

- Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
- Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
- Taking S and P as centres draw two arcs intersecting each other at T. Join AT. i.e., ∠TAP =
^{1}/_{2 }∠SAP

∠TAP = ^{1}/_{2} x 90 ˚ = 45 ˚ = 22^{1}/_{2}˚

(iii) 15°

Construction:

- Draw a ray AB with initial point A
- Draw an arc XY with any radius by taking A as centre which cuts the ray AB at C.
- Draw an arc by taking C as centre with same radius which intersects arc XY at D. Since AC = AD = DC then, ∆ADC is an equilateral triangle.
- ∠ADC = ∠ACD = ∠DAC = 60°
- Taking C and D as centres with any radius draw two arcs intersecting one another at E. Join AE.
- Thus,
^{1}/_{2}∠EAC = ∠DAC ;^{1}/_{2}∠EAC = 60˚ ⇒∠EAC = 30˚ - Taking E and C as centres draw two arcs with any radius intersecting each other at F. Join AF
- Thus,
^{1}/_{2}∠FAC = ∠EAC ;^{1}/_{2}∠FAC = 30˚ ⇒∠FAC = 15˚

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

**(i) 75° (ii) 105° (iii) 135°**

Solution:

(i) 75°

Construction:

- Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Since AQ = AP = PQ, ∆APQ is an equilateral triangle. Therefore, ∠PAQ = 60 ˚
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
- ∠SAB = ∠PAQ + ∠QAS ⇒90 ˚ = 60 ˚+ ∠QAS ⇒∠QAS = 30˚
- Taking S and Q as centres draw two arcs intersecting each other at M.
- Join Then ∠MAQ =
^{1}/_{2}∠QAS =^{1}/_{2}x 30˚ = 15˚ - Therefore, ∠PAM = ∠MAQ + ∠PAQ = 60˚ + 15˚ = 75˚

(ii) 105°

Construction:

- Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Since AQ = AP = PQ, ∆APQ is an equilateral triangle. Therefore, ∠PAQ = 60 ˚
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
- Taking S and R as centre with any radius draw two arcs intersecting each other at T.
- Join AT. Then, ∠SAT =
^{1}/_{2}∠SAR =^{1}/_{2}x 30 ˚ = 15 ˚ - Thus, ∠PAT = ∠PAS + ∠SAT = 90 ˚ + 15 ˚ = 105 ˚

(iii) 135°

Construction:

- Draw a line ray AB with a point O.
- Taking O as a centre draw an arc XY with any radius intersecting the ray AB at C.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at D.
- Taking D as centre with same radius draw another arc which intersects the arc XY at E. Draw a ray joining OF.
- Now, taking D and E as centres with any radius draw two arc intersecting one another at F.
- Join OF. Then angle ∟FOC = 90˚
- Taking O as radius draw an arc interesting the ray AB at G. ∟FOG = 90˚
- Now, taking G and F as centres with any radius draw two arc intersecting one another at H.
- Join OH. ∟HOF =
^{1}/_{2}∟FOG =^{1}/_{2 }x 90˚ = 45˚ - ∠HOC = ∠FOC + ∠HOF = 90˚ + 45˚ = 135˚

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

Solution:

Steps of Construction:

- Draw a line segment AB of the given length
- Taking A and B as centres with radius same as that of the length of AB draw two arcs intersecting each other at C
- Join AC and BC.
- Then triangle ABC is an equilateral triangle.

Justification:

- By construction AB= BC = AC
- Then triangle ABC is an equilateral triangle.

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