FIND THE VALUE OF X. IN THIS CIRCLE
Answers
Answer:
45 is the answer of x in the circle
Answer:
52
Step-by-step explanation:
Solution
(i) A circle with centre O
∠
A
O
C
=
135
∘
B
u
t
∠
A
O
C
+
∠
C
O
B
=
180
∘
(
L
i
n
e
a
r
p
a
i
r
)
⇒
135
∘
+
∠
C
O
B
=
180
∘
⇒
∠
C
O
B
=
180
∘
−
135
∘
=
45
∘
N
o
w
a
r
e
B
C
s
u
b
t
e
n
d
s
∠
B
O
C
a
t
t
h
e
c
e
n
t
r
e
a
n
d
∠
B
P
C
a
t
t
h
e
r
e
m
a
i
n
i
n
g
p
a
r
t
o
f
t
h
e
c
i
r
c
l
e
∴
∠
B
O
C
=
2
∠
B
P
C
⇒
∠
B
P
C
=
1
2
∠
B
O
C
=
1
2
×
45
∘
=
45
∘
2
∴
∠
B
P
C
=
1
2
∠
B
O
C
=
1
2
∘
o
r
x
=
22
1
2
∘
(ii)
∵
C
D
a
n
d
A
B
a
r
e
t
h
e
d
i
a
m
e
t
e
r
s
o
f
t
h
e
c
i
r
c
l
e
w
i
t
h
c
e
n
t
r
e
O
∠
A
B
C
=
40
∘
B
u
t
i
n
Δ
O
B
C
,
O
B
=
O
C
(
R
a
d
i
i
o
f
t
h
e
c
i
r
c
l
e
)
∴
∠
O
C
B
=
∠
O
B
C
=
40
∘
N
o
w
i
n
Δ
B
C
D
,
∠
O
D
B
+
∠
O
C
B
+
∠
C
B
D
=
180
∘
⇒
x
+
40
∘
+
90
∘
=
180
∘
(
A
n
g
l
e
s
o
f
a
t
r
i
a
n
g
l
e
)
⇒
x
+
130
∘
=
180
∘
⇒
x
=
180
∘
−
130
∘
=
50
∘
(iii)
I
n
c
i
r
c
l
e
w
i
t
h
c
e
n
t
r
e
O
,
∠
A
O
C
=
120
∘
,
A
B
i
s
p
r
o
d
u
c
e
d
t
o
D
∵
∠
A
O
C
=
120
∘
a
n
d
∠
A
O
C
+
c
o
n
v
e
x
∠
A
O
C
=
360
∘
⇒
120
∘
+
c
o
n
v
e
x
∠
A
O
C
=
360
∘
∴
C
o
n
v
e
x
∠
A
O
C
=
360
∘
−
120
∘
=
240
∘
∴
a
r
e
A
P
C
S
u
b
t
e
n
d
s
∠
A
O
C
a
t
t
h
e
c
e
n
t
r
e
a
n
d
∠
A
B
C
a
t
t
h
e
r
e
m
a
i
n
i
n
g
p
a
r
t
o
f
t
h
e
c
i
r
c
l
e
∴
∠
A
B
C
=
1
2
∠
A
O
C
=
1
2
×
240
∘
=
120
∘
B
u
t
∠
A
B
C
a
t
t
h
e
r
e
m
a
i
n
i
g
p
a
r
t
o
f
t
h
e
c
i
r
c
l
e
∴
∠
A
B
C
=
1
2
∠
A
O
C
=
1
2
×
240
∘
=
120
∘
B
u
t
∠
A
B
C
+
∠
C
B
D
=
180
∘
(
L
i
n
e
a
r
P
a
i
r
)
⇒
120
∘
+
x
=
180
∘
⇒
x
=
180
∘
−
120
∘
=
60
∘
∴
x
=
60
∘
(iv) A circle with centre O and
∠
C
B
D
=
65
∘
B
u
t
∠
A
B
C
+
∠
C
B
D
=
180
∘
(
L
i
n
e
a
r
p
a
i
r
)
⇒
∠
A
B
C
+
65
∘
=
180
∘
⇒
∠
A
B
C
=
180
∘
−
65
∘
=
115
∘
Now are AEC subtends
∠
x
at the centre and
∠
A
B
C
at the remaining part of the cirle
∴
∠
A
O
C
=
2
∠
A
B
C
⇒
x
=
2
×
115
∘
=
230
∘
∴
x
=
230
∘
(v) In circle with centre O AB is chord of the cirlce,
∠
O
A
B
=
35
∘
I
n
△
O
A
B
,
(
O
A
=
O
B
∴
O
B
A
=
∠
O
A
B
=
35
∘
But in
△
O
A
B
∠
O
A
B
+
∠
O
B
A
+
∠
A
O
B
=
180
∘
⇒
35
∘
+
35
∘
+
∠
A
O
B
=
180
∘
⇒
70
∘
+
∠
A
O
B
=
180
∘
⇒
∠
A
O
B
=
180
∘
−
70
∘
=
110
∘
∴
C
o
n
v
e
x
∠
A
O
B
=
360
∘
−
110
∘
=
250
∘
But are AB subtends
∠
A
O
B
at the centre and
∠
A
C
B
at the remaining part of the circle
∴
∠
A
C
B
=
1
52
∠
A
O
B
⇒
x
=
1
2
×
250
∘
=
125
∘
⇒
x
=
125
∘
(vi) IN the circle with centr O, BOC is its diameter,
∠
A
O
B
=
60
∘
Are AB subtends
∠
A
O
B
at the centre of the circle and
∠
A
C
B
at the remaining part of he circle
∴
∠
A
C
B
=
1
2
∠
A
O
B
=
1
2
×
60
∘
=
30
∘
B
u
t
i
n
△
O
A
C
,
O
C
=
O
A
∴
∠
O
A
C
=
∠
O
C
A
=
∠
A
C
B
⇒
x
=
30
∘
(vii) IN the circle,
∠
B
A
C
a
n
d
∠
B
D
C
are in the same segment
∠
D
B
C
+
∠
B
C
D
+
∠
B
D
C
=
180
∘
⇒
70
∘
+
x
+
50
∘
=
180
∘
⇒
x
+
120
∘
=
180
∘
⇒
x
=
180
∘
−
120
∘
=
60
∘
∴
x
=
60
∘
(viii) IN circle with centre O,
∠
O
B
D
=
40
∘
AB and CD are diameters of the circle
∠
D
B
A
a
n
d
∠
A
C
D
are in the same segment
∴
∠
A
C
D
=
∠
D
B
A
=
40
∘
i
n
△
O
A
C
,
O
A
=
O
C
∴
∠
O
A
C
=
∠
O
C
A
=
40
∘
a
n
d
∠
O
A
C
+
∠
O
C
A
+
∠
A
O
C
=
180
∘
⇒
40
∘
+
40
∘
+
x
=
180
∘
⇒
x
+
80
∘
=
180
∘
⇒x=180∘−80∘=100∘
∴x=100∘
(ix) In the circle, ABCD is a cyclic quadrilatral ∠ADB=32
,∠DAC=28∘,and∠ABD=50∘,∠ABDand∠ACD
are in the same segment of a circle
∴∠ABD=∠ACD
⇒∠ACD=50∘Similarly,∠ADB=∠ACB
⇒∠ACB=32∘
Now
∠DCB=∠ACD+∠ACB=50∘+32∘=82∘x=82∘
(x) IN a circle,
∠BAC=35
∘
,∠CBD=65circ∠BACand∠BDC
are in the same segment
∴∠BAC=∠BDC=35∘
In△BCD,∠BDC+∠BCD+∠CBD=180∘
⇒35∘+x+65∘=180∘
⇒x+100=180∘
⇒x+100∘=180∘
⇒x−180∘−100∘=80∘
∴x=80∘
(xi) In the circle,
∠ABDand∠ACD
are in the same segment of a circle
∴ABD=∠ACD=40∘
Now in△CPD,∠CPD+∠PCD
+∠PDC=180∘⇒110∘+40∘+x=180∘
⇒x+150∘=180∘
∴x=180∘−150∘=30∘
(xii) In the circle, two diameters AC and BD intersect each other at O
∠BAC=50∘OA=OB
∴∠OBA=∠OAB=52∘
⇒∠ABD=52
But
∠ABDand∠ACDare n the same segment of the circle
∴∠ABD=∠ACD
⇒52∘=x
∴x = 52
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