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How to make angle of 45° with compass ?
Given :
The given angle is 45°
To do :
We have to explain how to make 45°
Solution :
Steps of Construction:
1. First, draw a ray OA with initial point O.
2. Taking O as a centre and same 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 the 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 intersect 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∘.
Iv) Join HJ and IJ.
In ΔOIJ and ΔOHJ,
OI = OH
[Radii of the same arc]
IJ = HJ
[Arcs of equal radii]
OJ = OJ
[Common]
∴ ΔOIJ ≅ ΔOHJ
[SSS Rule]
∴ ∠IOJ = ∠ HOJ
[CPCT]
∴ ∠AOJ= ∠GOJ= 1/2 ∠GOA = ½(90°)=45
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