Question

construct an angle of 90at the intinial point of giving ray and justify the construction​

Answer 1

Answer:

Step-by-step explanation:First draw a ray OA with initial point O. 2. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B. ... Taking C as a centre and with the same radius as before, draw an arc intersecting the arc drawn in step 2, at D.

First draw a ray OA with initial point O.

2.Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.

 

3.Taking B as centre and with the same radius as before, draw an arc intersecting the previous arc, say at a point C.

 

4. Taking C as a centre and with the same radius as before, draw an arc intersecting the arc drawn in step 2,  at D.

 

5.Draw the ray OE passing through C. Then ∠EOA = 60∘ & the ray OF passing through D. Then ∠FOE =60∘.

 

6.Next taking C and D as centres and with the radius more than 1/2 CD, draw arcs to intersect each other, at a point  G.

 

7.Draw the ray 0G, which is the angle bisector of the  ∠FOE,

i.e., ∠FOG = ∠EOG = 1/2 ∠FOE =1/2 (60∘) = 30∘.

Thus, ∠GOA = ∠GOE + ∠EOA = 30∘ +60∘= 90∘.

 

 

Justification:

 

(i) Join BC.

Then, OC = OB = BC (By construction)

∴   ΔCOB is an equilateral triangle.

∴   ∠COB =60∘.

∴   ∠ EOA = 60∘.

ii)   Join CD.

Then, OD = OC = CD (By construction)

So, ΔDOC is an equilateral triangle.

∴     ∠DOC = 60∘.

∴     ∠ FOE = 60∘.    

     

(iii)   Join CG and DG.

In ΔODG and ΔOCG,

OD = OC

[ Radii of the same arc]

DG = CG

[ Arcs of equal radii]

OG = OG   [Common]  

∴ ΔODG≅ Δ OCG   [SSS Rule]

∴ ∠DOG =∠COG [CPCT]

 

∴   ∠FOG = ∠EOG =1/2 ∠FOE = 1/2 (60∘) = 30∘

 

Thus, ∠GOA = ∠GOE + ∠EOA = 30∘ + 60∘= 90∘.

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