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ans pls Questions. Z InFe. 3. O is of the circle, the x ISA: O is the 5. Fig. 15.134. Fig 15122 fig. 15123 o in each the C o

Dear Student,
We have to find in each figure.

(i) It is given that

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.

Now ,x=12COB=2212°

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 

  

 

CAB+ABC+BCA=180°    Angle sum propertyx+40°+90°=180°x=50°

(iii) It is given that



ABC=12Reflex AOC

 

So

And

Then

Hence

(iv)

  (Linear pair)

12236919

   

And
x =


 


 
Hence,

(v) It is given that

is an isosceles triangle.
  

Therefore

And,

In AOB,AOB+OBA+BAO=180°70°+BAO=180°BAO=110°

AOB=2ReflexACB

                     

Hence,

(vi) It is given that

 

And

COA+AOB=180°COA=180°-60°COA=120°

OCA is an isosceles triangle.

So

Hence, 12236919

(vii)                    (Angle in the same segment)

 

In we have

Hence

(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°  ......(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°                (Angle in the same segment)

 Now in BDC we have

BDC+DCB+CBD=180°35°+65°+CBD=180°CBD=180°-100°=80°

Hence,

(xi)

 

                    (Angle in the same segment)

In we have

Hence

(xii)
 

          (Angle in the same segment)

is an isosceles triangle

So,                   (Radius of the same circle)

Then

Hence

Regards

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