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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