Find the value of x and y from the circle with centre ‘O’ where
∠ACB = 65°, ∠OAB = x and ∠AOB = y.
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Answer:
Given,
(i)∠AOC=100
o
Now,
∠AOC+∠BOC=180
o
=>∠BOC=180
o
−100
o
=80
o
Now,
we know that the angle at the centre is twice the angle at the
circumference subtended by the same arc.
Therefore,
∠BDC=
2
1
∠BOC
=(
2
1
×80
o
)
=40
o
(ii)Given,
O is the centre of the circle.
∠AOD=40
o
and ∠BDC=100
o
Now,
we know that the angle at the centre is twice the angle at the
circumference subtended by the same arc.
Therefore,
∠ABC=
2
1
∠AOC
=(
2
1
×40
o
)
=20
o
In △BCD
∠DBC+\angle BDC+∠DCB=180[sum of all angle of △]
=>20
o
+100
o
+∠DCB=180
o
=>∠DCB=60
o
=>∠OCB=60
o
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