# OPPOSITE ANGLES OF A CYCLIC QUADRILATERAL EXAMPLES

Opposite angles of a cyclic quadrilateral examples

Here we are going see example problems of finding opposite angles of a cyclic quadrilateral.

Example 1 :

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

Find

(i)  <QPR

(ii)  <QPM and

(iii) <PRS Solution :

(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  =  180 - 145

<QPR  =  35°

(ii)  In triangle QPM,

<QPM + <MQP + <QMP  =  180

<QPM + 50 + 90  =  180

<QPM + 140  =  180

<QPM  =  180 - 140  =  40

<QPM  =  40°

(iii)  <PQR + <PSR  =  180

55  +  <PSR  =  180

<PSR  =  180 - 55

<PSR  =  125

In triangle PSR,

<PSR + <SPR + <PRS  =  180

125 + 25 + <PRS  =  180

150 + <PRS  =  180

<PRS  =  180 - 150

<PRS  =  30

Example 2 :

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

(i) <BCD Angles in the same segment must be equal

<CBD and <CAD will be equal.

<CBD  =  30°

<DAB + <DCB  =  180

<DAC + <CAB + <DCB  =  180°

30 + 50 <DCB  =  180

80 + <DCB  =  180

<DCB  =  180 - 80

<DCB  =  100°

<QPR  =  35°

Example 3 :

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

ABCD is a cyclic quadrilateral. we have

<ABC = 180° - 120°  =  60°

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

In triangle ABC we have,

<BAC + <ACB + <ABC  =  180°

<BAC + 90° + 60°  =  180°

<BAC  =  180° - 150°

=  30°

Hence the value of x is 30°.

Example 4 :

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

(i) <BCD

(iii) <ABC Solution :

100 + <BCD  =  180

<BCD  =  180 - 100

<BCD  =  80°

80 + <ABC  =  180

<ABC  =  180 - 80

<ABC  =  100°

Example 5 :

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

<DAB  +  <DCB  =  180°

<DAB  +  100  =  180°

<DAB  =  180 - 100

<DAB  =  80°

<DAB + <ABD + <BDA  =  180

80 + <ABD + 50  =  180

130 + <ABD  =  180

<ABD  =  180 - 130

<ABD  =  50°

Hence the required angle is 50°. After having gone through the stuff given above, we hope that the students would have understood "Opposite angles of a cyclic quadrilateral examples"

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