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Question:42
In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ?DBC = 60° and
?BAC = 40°. Find
i ?BCD,
ii ?CAD.
Solution:
i
?BDC = ?BAC = 40° 
Anglesinthesamesegment
     In ?BCD, we have:
Page 2


 
                       
     
              
   
         
              
   
           
  
          
   
        
 
   
                
Question:42
In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ?DBC = 60° and
?BAC = 40°. Find
i ?BCD,
ii ?CAD.
Solution:
i
?BDC = ?BAC = 40° 
Anglesinthesamesegment
     In ?BCD, we have:
    ?BCD + ?DBC + ?BDC = 180° 
Anglesumpropertyofatriangle         
     ? ?BCD + 60° + 40° = 180°
     ? ?BCD = (180° - 100°) = 80°
ii
?CAD = ?CBD 
Anglesinthesamesegment
               = 60°
Question:43
In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If ?PSR = 150°, find ?RPQ.
Solution:
In cyclic quadrilateral PQRS, we have:
?PSR + ?PQR = 180°
? 150° + ?PQR = 180°
? ?PQR = (180° – 150°) = 30°
? ?PQR = 30°                ...i
Also, ?PRQ = 90° Angleinasemicircle
                 ...ii
Now, in ?PRQ, we have:
?PQR + ?PRQ + ?RPQ = 180°
? 30° + 90° + ?RPQ = 180°  
From(i)and(ii)
? ?RPQ = 180° – 120° = 60°
? ?RPQ = 60°
Question:44
In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is
extended to P, find ?PBC.
Solution:
Reflex ?AOC + ?AOC = 360
°
Page 3


 
                       
     
              
   
         
              
   
           
  
          
   
        
 
   
                
Question:42
In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ?DBC = 60° and
?BAC = 40°. Find
i ?BCD,
ii ?CAD.
Solution:
i
?BDC = ?BAC = 40° 
Anglesinthesamesegment
     In ?BCD, we have:
    ?BCD + ?DBC + ?BDC = 180° 
Anglesumpropertyofatriangle         
     ? ?BCD + 60° + 40° = 180°
     ? ?BCD = (180° - 100°) = 80°
ii
?CAD = ?CBD 
Anglesinthesamesegment
               = 60°
Question:43
In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If ?PSR = 150°, find ?RPQ.
Solution:
In cyclic quadrilateral PQRS, we have:
?PSR + ?PQR = 180°
? 150° + ?PQR = 180°
? ?PQR = (180° – 150°) = 30°
? ?PQR = 30°                ...i
Also, ?PRQ = 90° Angleinasemicircle
                 ...ii
Now, in ?PRQ, we have:
?PQR + ?PRQ + ?RPQ = 180°
? 30° + 90° + ?RPQ = 180°  
From(i)and(ii)
? ?RPQ = 180° – 120° = 60°
? ?RPQ = 60°
Question:44
In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is
extended to P, find ?PBC.
Solution:
Reflex ?AOC + ?AOC = 360
°
? Reflex ?AOC + 130
°
 + x = 360
°
? Reflex ?AOC = 360
°
 - 130
°
? Reflex ?AOC = 230
°
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the
remaining part of the circle.
Here, arc AC subtends reflex ?AOC at the centre and ?ABC at B on the circle.
? ?AOC = 2 ?ABC
?
?ABC =
230°
2
= 115°           ...1
Since ABP is a straight line, ?ABC + ?PBC = 180
°
? ?PBC = 180
°
 - 115
°
? ?PBC = 65
°
            ...2
Hence, ?PBC = 65
°
.
Question:45
In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced. If
?ABC = 92° and ?FAE = 20°, find ?BCD.
Solution:
Given: ABCD is a cyclic quadrilateral.
Then ?ABC + ?ADC = 180°
? 92° + ?ADC = 180°
? ?ADC = (180° – 92°) = 88°
Again, AE parallel to CD.
Thus, ?EAD = ?ADC = 88°  Alternateangles
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
? ?BCD = ?DAF
? ?BCD = ?EAD + ?EAF
            =  88° + 20° = 108°
Hence, ?BCD = 108°
Question:46
In the given figure, BD = DC and ?CBD = 30°, find m( ?BAC).
Page 4


 
                       
     
              
   
         
              
   
           
  
          
   
        
 
   
                
Question:42
In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ?DBC = 60° and
?BAC = 40°. Find
i ?BCD,
ii ?CAD.
Solution:
i
?BDC = ?BAC = 40° 
Anglesinthesamesegment
     In ?BCD, we have:
    ?BCD + ?DBC + ?BDC = 180° 
Anglesumpropertyofatriangle         
     ? ?BCD + 60° + 40° = 180°
     ? ?BCD = (180° - 100°) = 80°
ii
?CAD = ?CBD 
Anglesinthesamesegment
               = 60°
Question:43
In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If ?PSR = 150°, find ?RPQ.
Solution:
In cyclic quadrilateral PQRS, we have:
?PSR + ?PQR = 180°
? 150° + ?PQR = 180°
? ?PQR = (180° – 150°) = 30°
? ?PQR = 30°                ...i
Also, ?PRQ = 90° Angleinasemicircle
                 ...ii
Now, in ?PRQ, we have:
?PQR + ?PRQ + ?RPQ = 180°
? 30° + 90° + ?RPQ = 180°  
From(i)and(ii)
? ?RPQ = 180° – 120° = 60°
? ?RPQ = 60°
Question:44
In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is
extended to P, find ?PBC.
Solution:
Reflex ?AOC + ?AOC = 360
°
? Reflex ?AOC + 130
°
 + x = 360
°
? Reflex ?AOC = 360
°
 - 130
°
? Reflex ?AOC = 230
°
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the
remaining part of the circle.
Here, arc AC subtends reflex ?AOC at the centre and ?ABC at B on the circle.
? ?AOC = 2 ?ABC
?
?ABC =
230°
2
= 115°           ...1
Since ABP is a straight line, ?ABC + ?PBC = 180
°
? ?PBC = 180
°
 - 115
°
? ?PBC = 65
°
            ...2
Hence, ?PBC = 65
°
.
Question:45
In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced. If
?ABC = 92° and ?FAE = 20°, find ?BCD.
Solution:
Given: ABCD is a cyclic quadrilateral.
Then ?ABC + ?ADC = 180°
? 92° + ?ADC = 180°
? ?ADC = (180° – 92°) = 88°
Again, AE parallel to CD.
Thus, ?EAD = ?ADC = 88°  Alternateangles
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
? ?BCD = ?DAF
? ?BCD = ?EAD + ?EAF
            =  88° + 20° = 108°
Hence, ?BCD = 108°
Question:46
In the given figure, BD = DC and ?CBD = 30°, find m( ?BAC).
Solution:
BD = DC
? ?BCD = ?CBD = 30°
In ?BCD, we have:
?BCD + ?CBD + ?CDB = 180°  Anglesumpropertyofatriangle
? 30° + 30° + ?CDB = 180°
? ?CDB = (180° – 60°) = 120°
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, ?CDB + ?BAC = 180°
? 120° + ?BAC = 180°
? ?BAC = (180° – 120°) = 60°
? ?BAC = 60°
Question:47
In the given figure, O is the centre of the given circle and measure of arc ABC is 100°. Determine ?ADC and
?ABC.
Solution:
We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the
alternate segment.
Thus, ?AOC = 2 ?ADC
? 100° = 2 ?ADC
? ?ADC = 50°
The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.
Thus, ?ADC + ?ABC = 180°
? 50° + ?ABC = 180°
? ?ABC = (180° – 50°) = 130°
? ?ADC = 50° and ?ABC = 130°
Page 5


 
                       
     
              
   
         
              
   
           
  
          
   
        
 
   
                
Question:42
In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ?DBC = 60° and
?BAC = 40°. Find
i ?BCD,
ii ?CAD.
Solution:
i
?BDC = ?BAC = 40° 
Anglesinthesamesegment
     In ?BCD, we have:
    ?BCD + ?DBC + ?BDC = 180° 
Anglesumpropertyofatriangle         
     ? ?BCD + 60° + 40° = 180°
     ? ?BCD = (180° - 100°) = 80°
ii
?CAD = ?CBD 
Anglesinthesamesegment
               = 60°
Question:43
In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If ?PSR = 150°, find ?RPQ.
Solution:
In cyclic quadrilateral PQRS, we have:
?PSR + ?PQR = 180°
? 150° + ?PQR = 180°
? ?PQR = (180° – 150°) = 30°
? ?PQR = 30°                ...i
Also, ?PRQ = 90° Angleinasemicircle
                 ...ii
Now, in ?PRQ, we have:
?PQR + ?PRQ + ?RPQ = 180°
? 30° + 90° + ?RPQ = 180°  
From(i)and(ii)
? ?RPQ = 180° – 120° = 60°
? ?RPQ = 60°
Question:44
In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is
extended to P, find ?PBC.
Solution:
Reflex ?AOC + ?AOC = 360
°
? Reflex ?AOC + 130
°
 + x = 360
°
? Reflex ?AOC = 360
°
 - 130
°
? Reflex ?AOC = 230
°
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the
remaining part of the circle.
Here, arc AC subtends reflex ?AOC at the centre and ?ABC at B on the circle.
? ?AOC = 2 ?ABC
?
?ABC =
230°
2
= 115°           ...1
Since ABP is a straight line, ?ABC + ?PBC = 180
°
? ?PBC = 180
°
 - 115
°
? ?PBC = 65
°
            ...2
Hence, ?PBC = 65
°
.
Question:45
In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced. If
?ABC = 92° and ?FAE = 20°, find ?BCD.
Solution:
Given: ABCD is a cyclic quadrilateral.
Then ?ABC + ?ADC = 180°
? 92° + ?ADC = 180°
? ?ADC = (180° – 92°) = 88°
Again, AE parallel to CD.
Thus, ?EAD = ?ADC = 88°  Alternateangles
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
? ?BCD = ?DAF
? ?BCD = ?EAD + ?EAF
            =  88° + 20° = 108°
Hence, ?BCD = 108°
Question:46
In the given figure, BD = DC and ?CBD = 30°, find m( ?BAC).
Solution:
BD = DC
? ?BCD = ?CBD = 30°
In ?BCD, we have:
?BCD + ?CBD + ?CDB = 180°  Anglesumpropertyofatriangle
? 30° + 30° + ?CDB = 180°
? ?CDB = (180° – 60°) = 120°
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, ?CDB + ?BAC = 180°
? 120° + ?BAC = 180°
? ?BAC = (180° – 120°) = 60°
? ?BAC = 60°
Question:47
In the given figure, O is the centre of the given circle and measure of arc ABC is 100°. Determine ?ADC and
?ABC.
Solution:
We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the
alternate segment.
Thus, ?AOC = 2 ?ADC
? 100° = 2 ?ADC
? ?ADC = 50°
The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.
Thus, ?ADC + ?ABC = 180°
? 50° + ?ABC = 180°
? ?ABC = (180° – 50°) = 130°
? ?ADC = 50° and ?ABC = 130°
Question:48
In the given figure, ?ABC is equilateral. Find
i ?BDC,
ii ?BEC.
Solution:
i
Given: ?ABC is an equilateral triangle.
i.e., each of its angle = 60°
? ?BAC = ?ABC = ?ACB = 60°
Angles in the same segment of a circle are equal.
i.e., ?BDC = ?BAC = 60°
? ?BDC = 60°
ii
The opposite angles of a cyclic quadrilateral are supplementary.
Then in cyclic quadrilateral ABEC, we have:
?BAC + ?BEC = 180°
? 60° + ?BEC = 180°
? ?BEC = 180°– 60°
= 120°
? ?BDC = 60° and ?BEC = 120°
Question:49
In the adjoining figure, ABCD is a cyclic quadrilateral in which ?BCD = 100° and ?ABD = 50°. Find ?ADB.
Solution:
Given: ABCD is a cyclic quadrilateral.

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