Question -
Answer -
We have to find in each figure.
(i) It is given that 
∠AOC+∠COB=180° [Linear pair]∠AOC+∠COB=180° Linear pair
As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Hence 
(ii) As we know that 
= x [Angles in the same segment] line 
is diameter passing through centre, So,
∠BCA= 90° [Angle inscribed in a semicircle is a right angle ]∠BCA= 90° Angle inscribed in a semicircle is a right angle
(iii) It is given that
(iv) 
(Linear pair) And
Hence,
(v) It is given that 
is an isosceles triangle. Therefore 
Hence, 
(vi) It is given that 
And
ΔOCA is an isosceles triangle.∆OCA is an isosceles triangle.
So
Hence, 
(vii)
(Angle in the same segment) In
we haveHence
(viii)
As 
(Radius of circle)
Therefore, 
is an isosceles triangle. So
(Vertically opposite angles)Hence,
(ix) It is given that
…… (1) (Angle in the same segment) ∠ADB=∠ACB=32°∠ADB=∠ACB=32° ......(2) (Angle in the same segment)
Because
and 
are on the same segment of the circle.Now from equation (1) and (2) we have
Hence,
(x) It is given that 
∠BAC=∠BDC=35°∠BAC=∠BDC=35° (Angle in the same segment)
Now in ΔBDC∆BDC we have
∠BDC+∠DCB+∠CBD=180°⇒35°+65°+∠CBD=180°⇒∠CBD=180°−100°=80°∠BDC+∠DCB+∠CBD=180°⇒35°+65°+∠CBD=180°⇒∠CBD=180°-100°=80°
Hence, 
(xi)