Exercise - 13.1
construct the following angles of the initial point of a given
ray and Justify the construction.
(@) 90° (b) 45°
Answers
a)
Join OC and BC
Thus,
OB=BC=OC [Radius of equal arcs]
∴△OCB is an equilateral triangle
∴∠BOC=60°
Join OD,OC and CD
Thus,
OD=OC=DC [Radius of equal arcs]
∴△DOC is an equilateral triangle
∴∠DOC=60°
Join PD and PC
Now,
In △ODP and △OCP
OD=OC [Radius of same arcs]
DP=CP [Arc of same radii]
OP=OP [Common]
∴△ODP≅△OCP [SSS congruency]
∴∠DOP=∠COP [CPCT]
So, we can say that
∠DOP=∠COP= ½∠DOC
∠DOP=∠COP= ½×60=30°
Now,
∠AOP=∠BOC+∠COP
∠AOP=60+30
∠AOP=90°
Hence justified✅
b)
(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
Hence justified✅
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