Construct an angle of 45° at the initial point of a given ray and justify the construction.
Answers
Steps of Construction :-
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 :-
Now, Join BC.
Then,
- OC = OB = BC triangle. (By construction)
- ∴ ∠COB is an equilateral triangle.
- ∴ ∠COB = 60°.
- ∴ ∠EOA = 60°.
Now, Join CD.
Then,
- OD = OC = CD (By construction)
So,
- ∆DOC is an equilateral triangle.
- ∴ ∠DOC = 60°.
- ∴ ∠ FOE = 60°.
Now, 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°
Hence, justified.
