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**Q.1Â Â Â Â In figure sides QP and RQ of PQR are produced to points S and T respectively. If 135Âº and 110Âº, ****find **

We have ,

180Âº [Linear pair]

+ 135Âº = 180Âº

180Âº â€“ 135Âº = 45Âº

Now , [By exterior angle theorem]

110Âº = 45Âº +

= 110Âº â€“ 45Âº = 65Âº

Hence = 65Âº

**Q.2Â Â Â Â Â In figure, 62Âº, 54Âº. If YO and ZO are the bisectors of respectively of XYZ, find **

Consider XYZ,

= 180Âº [Angle - sum property]

62Âº + 54Âº + = 180Âº [Since 62Âº , 54Âº]

Since YO and ZO are bisectors of

Therefore

and ,

In OYZ , we have

180Âº [Angle sum property]

27Âº + 32Âº = 180Âº

= 180Âº â€“ 27Âº â€“ 32Âº

= 180Âº â€“ 59Âº = 121Âº

Hence, 32Âº

and = 121Âº

**Q.3Â Â Â Â Â In figure if AB || DE, = 35Âº and 53Âº, find .**

Since AB || DE and transversal AE intersects them at A and E respectively.

Therefore [Alternate angles]

35Âº [Since 35Âº]

In DEC, we have

180Âº [Angle sum property]

35Âº + 53Âº = 180Âº

180Âº â€“ 35Âº â€“ 53Âº

= 180Âº â€“ 88Âº = 92Âº

Hence , 92Âº

**Q.4Â Â Â Â Â In figure, if lines PQ and RS intersect at point T, such that 40Âº, 95Âº and 75Âº, find .**

In PRT, we have

180Âº [Angle - sum property]

40Âº + 95Âº = 180Âº

180Âº â€“ 40Âº â€“ 95Âº

= 180Âº â€“ 135Âº = 45Âº

[Vertically opp. angles]

45Âº [Since RTP = 45Âº (proved)]

In TQS we have

180Âº [Angle - sum property]

45Âº + 75Âº = 180Âº [SinceÂ (proved) and ]

180Âº â€“ 45Âº â€“ 75Âº

= 180Âº â€“ 120Âº = 60Âº

Hence , 60Âº

**Q.5Â Â Â Â Â In figure if PQ PS, PQ|| SR, = 28Âº and = 65Âº, then find the values of x and y. **

Using exterior angle property in SRQ, we have

65Âº = 28Âº + [Since 65Âº , 28Âº]

QSR = 65Âº â€“ 28Âº = 37Âº

Since PQ|| SR and the transversal PS intersects them at P and S respectively.

Therefore 180Âº [Sum of consecutive interior angles is 180Âº]

90Âº = 180Âº

y + 37Âº + 90Âº = 180Âº

y = 180Âº â€“ 90Âº â€“ 37Âº

= 180Âº â€“ 127Âº = 53Âº

In the right , we have

90Âº

x + 53Âº = 90Âº

x = 90Âº â€“ 53Âº = 37Âº

Hence, x = 37Âº

and y = 53Âº

**Q.6Â Â Â Â In figure , the side QR of PQR is produced to a point S. If the bisectors of meet at point T, then prove ****that .**

In we have ext.

Â Â ( By exterior angle theorem )

... (1)

[Since QT and RT are bisectors of respectively therefore ]

In QRT. we have ext. ... (2)

From (1) and (2) , we get

[ Since ]

Hence ,

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