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A B is a diameter of a circle with center O . Chord C D is equal to radius O C . A C and B D produced intersect at P . What is the measure of ∠ A P B in degrees?
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2 solutions
Join
O
D
.
Now
△
O
C
D
is an equilateral triangle since
O
C
=
C
D
=
D
O
.
So,
∠
C
O
D
=
∠
O
D
C
=
∠
D
C
O
=
6
0
∘
.
Let
∠
O
D
B
=
∠
O
B
D
=
x
.
[
O
D
=
O
B
=
r
a
d
i
u
s
]
.
∠
D
O
B
=
1
8
0
−
2
x
.
∠
A
O
C
=
2
x
−
6
0
.
Since
O
A
=
O
C
=
r
a
d
i
u
s
,
∠
O
A
C
=
∠
O
C
A
=
1
2
0
−
x
.
∠
P
C
D
=
x
.
∠
P
D
C
=
1
2
0
−
x
.
Therefore
∠
A
P
B
=
1
8
0
−
x
−
1
2
0
+
x
=
6
0
∘
.
this is an easy question i think its from NCERT 9th standard