OPPOSITE ANGLES IN A CYCLIC QUADRILATERAL WORKSHEET

1. In the diagram shown below. ABCD is a cyclic quadrilateral in which BCD = 100° and ABD = 50°. Find ADB.

2. In the diagram shown below, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ∠CBD = 30° and BAC = 50°Find CAD and BCD.

3. In the diagram shown below, O is the center of the 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 BCD, ADC and ABC.

5. In the diagram shown below, PQ is a diameter of a circle with center O. If PQR = 55°SPR = 25° and PQM = 50°. Find QPR, QPM and PRS.

1. Answer :

By Theorem,

BAD + ∠BCD = 180°

Substitute BCD = 100°.

BAD + 100° = 180°

Subtract 100° from each side. 

∠BAD = 80°

In triangle ABD,

∠ABD + ∠BAD + ADB = 180°

50° + 80° + ADB = 180°

130° + ADB = 180°

Subtract 130° from each side.

ADB = 50°

2. Answer :

∠CAD :

The angles at the circumference subtended by the same arc are equal. ∠CAD and ∠CBD at the circumference subtended by the same arc CD.

∠CAD = ∠CBD

∠CAD = 30°

∠BCD :

By Theorem,

∠BAD + ∠BCD = 180°

In the above diagram, ∠BAD = ∠BAC + ∠CAD.

(∠BAC + ∠CAD) ∠BCD = 180°

∠BAC + ∠CAD ∠BCD = 180°

Substitute ∠BAC = 50° and ∠CAD = 30°.

50° + 30° ∠BCD = 180°

80° ∠BCD = 180°

Subtract 80° from each side. 

∠BCD = 100°

3. Answer :

By Theorem,

ABC + ADC = 180°

Substitute ∠ADC = 120°.

ABC + 120° = 180°

Subtract 120° from each side. 

ABC = 60°

∠ACD = 90° (angle in a semi circle is a right angle)

In triangle ABC,

BAC + ACB + ABC = 180°

x° + 90° + 60° = 180°

x + 90 + 60 = 180

x + 150 = 180

Subtract 150 from each side.

x = 30

4. Answer :

∠BCD :

By Theorem,

BAC + ∠BCD = 180°

Substitute BAD = 100°.

100° + BCD = 180°

Subtract 100° from each side.

∠BCD = 80°

∠ADC :

Since AB || DC, AD is transversal.

When two parallel lines intersected by a transversal, same side interior angles are supplementary. 

∠BAD + ∠ADC = 180°

Substitute BAD = 100°.

100° + ∠ADC = 180°

Subtract 100° from each side.

∠ADC = 80°

∠ABC :

By Theorem, 

ABC + ∠ADC = 180°

Substitute ADC = 80°.

∠ABC + 80° = 180°

Subtract 80° from each side. 

∠ABC = 100°

5. Answer :

QPR : 

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

In triangle PRQ,

PRQ + QPR + PQR = 180°

90° + QPR + 55° = 180°

QPR + 145° = 180°

Subtract 145° from each side.

QPR = 35°

QPM : 

In triangle QPM,

QPM + MQP + QMP = 180°

QPM + 50° + 90° = 180°

QPM + 140° = 180°

Subtract 140° from each side. 

QPM = 40°

∠PRS : 

By Theorem,

PQR + PSR = 180°

55° + PSR = 180°

Subtract 55° from each side. 

PSR = 125°

In triangle PSR,

PSR + SPR + PRS = 180°

125° + 25° PRS = 180°

150° + PRS = 180°

Subtract 150° from each side. 

PRS = 30°

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