In the given figure AB=BC and o is the centre of the circle if <AOB=90° determine <AOC
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In figure, A,B and C are three points on the circle with centre O such that ∠AOB=90
o
and ∠AOC=110
o
. Find ∠BAC.
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From the figure we know that
∠AOB+∠AOC+∠BOC=360
o
By substituting the values
90
o
+110
o
+∠BOC=360
o
On further calculation
∠BOC=360
o
−90
o
−110
o
By subtraction
∠BOC=360
o
−200
o
∠BOC=160
o
So we get
We know that
∠BOC=2×∠BAC
It is given that ∠BOC=160
o
∠BAC=
2
160
o
By division
∠BAC=80
o
Therefore, ∠BAC=80
o
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