ANGLES IN A CYCLIC QUADRILATERAL WORKSHEET

1. In the figure given below, PQ is a diameter of a circle with center O. If PQR = 55°SPR = 25° and PQM = 50°. Find :

(i) QPR

(ii) QPM

(iii) PRS. 

2. In the figure shown below, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that DBC = 30° and BAC = 50°Find :

(i) BCD 

(ii) CAD

3. In the figure given below, O is the center of a circle and ADC = 120°. Find the value of x.

4. In the figure given below, ABCD is a cyclic quadrilateral in which AB || DC. If BAD = 100° find

(i) BCD

(ii) ADC

(iii) ABC

5. In the figure given below, ABCD is a cyclic quadrilateral in which BCD = 100° and ABD = 50° find ADB. 

1. Answer :

(i) PRQ = 90° (angles in a semi circle is a right angle)

In triangle PRQ,

PRQ + QPR + PQR = 180°

90° + QPR + 55° = 180°

145° + QPR  = 180°

QPR  =  35°

(ii)  In triangle QPM,

QPM + MQP + QMP = 180°

QPM + 50° + 90° = 180°

QPM + 140° = 180°

QPM = 40°

(iii) PQR + PSR = 180

55 + PSR = 180°

PSR =  125°

In triangle PSR,

PSR + SPR + PRS  =  180°

125° + 25° PRS  =  180°

150° + PRS  =  180°

PRS  =  30°

2. Answer :

Angles in the same segment must be equal

So, ∠CBD and CAD are equal.

∠CAD = CBD = 30°

DAB + DCB = 180

DAC + CAB + ∠BCD = 180°

30° + 50° ∠BCD = 180°

80° + ∠BCD = 180°

∠BCD = 100°

3. Answer :

In the figure given below, O is the center of a circle and ADC = 120°. Find the value of x.

ABCD is a cyclic quadrilateral. we have

ABC + ADC = 180°

ABC + 120° = 180°

ABC = 60°

Also ACB = 90° (angle on a semi circle)

In triangle ABC we have,

BAC + ACB + ABC = 180°

BAC + 90° + 60° = 180°

BAC + 150° = 180°

∠BAC = 30°

Hence the value of x is 30.

4. Answer :

BAD + BCD = 180°

100° BCD = 180°

BCD = 80°

ADC = 80°

ADC + ABC = 180°

80° + ABC = 180°

ABC = 100°

5. Answer :

DAB + DCB = 180°

DAB + 100° = 180°

DAB = 80°

In triangle ADB,

DAB + ABD + BDA = 180°

80° + 50° + ADB = 180°

130° + ADB = 180°

ADB = 50°

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