Question

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

Answer 1

Step-by-step explanation:

1. First, draw a ray OA with intial point O.

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

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

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

5.Draw the ray OE passing through C. Then ∠EOA = 60∘ .

6.Draw the ray OF passing through D. Then ∠ FOE = 60∘ .

7. Next, taking C and D as centres and with radius more than ½ CD, draw arcs to intersect each other, at G.

8. Draw the ray OG, which is the bisector of the angle FOE, i.e., ∠FOG = ∠EOG = 1/2 ∠FOE = 1/2 (60∘ ) = 30∘ .

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

9.Now taking O as centre and any radius more than OB, draw an arc to intersect the rays OA and OG, at H and I.

10. Next, taking H and I as centres and with the radius more than 1/2 HI, draw arcs to ntersect each other, at J.

11. Draw the ray OJ. This ray OJ is the required bisector of the ∠ GOA. Thus, ∠GOJ = ∠AOJ = 1/2 ∠GOA = 1/2(90∘) = 45∘.

Justification:

(i)Join BC.

Then, OC = OB = BC triangle. (By construction)

∴ ∠COB is an equilateral triangle.

∴ ∠COB = 60∘.

∴ ∠EOA = 60∘.

(ii)Join CD.

Then, OD = OC = CD (By construction)

∆DOC 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]

∴ Δ ODG = ΔOCG [SSS Rule]

∴ ∠ DOG= ∠COG [CPCT]

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

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

∴ ∠AOJ= ∠GOJ= 1/2 ∠GOA = ½(90°)=45