If O is the centre of the circle find the value of x
A
50°
B
C
X
Þ
EB
Answers
Step-by-step explanation:
(i) ∠ACB=∠ADB (Angles in the same segment of a circle)
But ∠ADB=x
o
⇒∠ABC=x
o
Now in ΔABC
∠CAB+∠ABC+∠ACB=180
o
⇒40
o
+90
0
+x
o
=180
o
(AC is the diameter)
⇒130
o
+x
o
=180
o
⇒x
o
=180
0
−130
o
=50
o
(ii) ∠ACD=∠ABD (angles in the same segment)
And, ∠ACD=x
o
Now in triangle OAC,
OA=OC (radii of the same circle)
⇒∠ACO=∠AOC (opposite angles of equal sides)
Therefore, x
o
=62
o
(iii) ∠AOB+∠AOC+∠BOC=360
o
(sum of angles at a point)
⇒∠AOB+80
o
+130
o
=360
o
⇒∠AOB+210
o
=360
o
⇒∠AOB=360
o
−210
o
=150
0
Now arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴∠AOB=2∠ACB
⇒∠ACB=
2
1
∠AOB=
2
1
×150
o
=75
o
(iv) ∠ACB+∠CBD=180
o
⇒∠ABC+75
o
=180
o
⇒∠ABC=180
o
−75
o
=105
o
Now arc AC subtends reflex ∠AOC at the centre and ∠ABC at the remaining part of the circle.
Reflex ∠AOC=2∠ABC
=2×105
o
=210
o
(v) ∠AOC+∠COB=180
o
⇒135
o
+∠COB=180
o
⇒∠COB=180
o
−135
o
=45
o
Now arc BC Subtends reflex ∠COB at the centre and ∠CDB at the remaining part of the circle.
⇒∠COB=2∠CDB
⇒∠CDB=
2
1
∠COB
=
2
1
×45
o
=
2
45
o
=22
2
1
o
(vi) Arc AB subtends ∠AOD at the centre and ∠ACD at the remaining part of the Circle
∠AOD=2∠ACB
⇒∠ACB=
2
1
∠AOD=
2
1
×70
o
=35
o
⇒∠CMO=90
o
⇒∠AMC=90
o
(∠AMC+∠CMO=180
o
)
Now in ΔACM
∠ACM+∠AMC+∠CAM=180
o
⇒35
o
+90
o
+x
o
=180
o
⇒125
o
+x
o
=180
o
⇒x
o
=180−125
o
=55
o