The document Ex 6.3 NCERT Solutions- Lines and Angles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.

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**Question 1. In the adjoining figure, sides QP and RQ of Î”PQR are produced to points S and T respectively. If âˆ SPR = 135Âº and âˆ PQT = 110Âº, find âˆ PRQ. Solution: **âˆµ TQR is a straight line,

âˆ´ âˆ TQP + âˆ PQR = 180Âº [Linear pair]

â‡’ 110Âº + âˆ PQR = 180Âº

â‡’ âˆ PQR = 180Âº - 110Âº = 70Âº

Since, the side QP of **Î”**PQR is produced to S.

âˆ´ Exterior angle so formed is equal to the sum of interior opposite angles.

âˆ´ âˆ PQR + âˆ PRQ = 135Âº

â‡’ 70Âº + âˆ PRQ = 135Âº [âˆµ âˆ PQR = 70Âº]

â‡’ âˆ PRQ = 135 - 70Âº

â‡’ âˆ PRQ = 65Âº

**Question 2. In the adjoining figure, âˆ X = 62Âº, âˆ XYZ = 54Âº. If YO and ZO are the bisectors of âˆ XYZ and âˆ XZY respectively of Î”XYZ, find âˆ OZY and âˆ YOZ. Solution:** In âˆ XYZ, âˆ XYZ + âˆ YZX + âˆ ZXY = 180Âº [âˆµ Sum of angles of a triangle is 180Âº]

But âˆ XYZ = 54Âº and âˆ ZXY = 62Â° [Given]

âˆ´ 54Âº + âˆ YZX + 62Âº = 180Âº

â‡’ âˆ YZX = 180Â° - 54Â° - 62Â° = 64Â°

âˆµ YO and ZO are the bisectors âˆ XYZ and âˆ XZY respectively, [Given]

âˆ OYZ =(1/2)âˆ XYZ =(1/2)(54Âº) = 27Âº

and âˆ OZY =(1/2)âˆ YZX = (1/2) (64Âº) = 32Âº

Now, in **Î”**OYZ, we have:

âˆ YOZ + âˆ OYZ + âˆ OZY = 180Âº [By the angle sum property]

â‡’ âˆ YOZ + 27Âº + 32Âº = 180Âº

â‡’ âˆ YOZ = 180Âº - 27Âº - 32Âº = 121Âº

Thus, âˆ OZY = 32Âº and âˆ YOZ = 121Âº

**Question 3. In the following figure, if AB || DE, âˆ BAC = 35Âº and âˆ CDE = 53Âº, find âˆ DCE.**

**Solution:** âˆµ AB || DE and AE is a transversal. [Given ]

âˆ´ âˆ BAC = âˆ AED [Interior alternate angles]

But âˆ BAC = 35Âº [Given]

âˆ´ âˆ AED = 35Âº

Now, in **Î”**CDE, we have

âˆ CDE + âˆ DEC + âˆ DCE = 180Âº [Using the angle sum property]

âˆ´ 53Âº + 35Âº + âˆ DCE = 180Âº [âˆµ âˆ DEC = âˆ AED = 35Âº and âˆ CDE = 53 (Given)]

â‡’ âˆ DCE = 180Âº - 53Âº - 35Âº = 92Âº Thus, DCE = 92Âº

**Question 4. In the figure, if lines PQ and RS intersect at point T, such that âˆ PRT = 40Âº, âˆ RPT = 95Âº and âˆ TSQ = 75Âº, find âˆ SQT. Solution:** In

â‡’ 95Âº + 40Âº + âˆ PTR = 180Âº

â‡’ âˆ PTR = 180Âº - 95Âº - 40Âº = 45Âº

But PQ and RS intersect at T,

âˆ´ âˆ PTR = âˆ QTS [Vertically opposite angles]

âˆ´ âˆ QTS = 45Âº Now, in **Î”**TQS, we have

âˆ TSQ + âˆ STQ + âˆ SQT = 180Âº [By angle sum property]

âˆ´ 75Âº + 45Âº + âˆ SQT = 180Âº [âˆµ âˆ TSQ = 75Âº and âˆ STQ = 45Âº]

â‡’ âˆ SQT = 180Âº - 75Âº - 45Âº = 60Âº

Thus, âˆ SQT = 60Âº

**Question 5. In the adjoining figure, if PQ âŠ¥ PS, PQ ||SR, âˆ SQR = 28Âº and âˆ QRT = 65Âº, then find the values of x and y. Solution:** In DQRS, the side SR is produced to T.

âˆ´ Exterior âˆ QRT = âˆ RQS + âˆ RSQ

But âˆ RQS = 28Âº and âˆ QRT = 65Âº

âˆ´ From âˆ RQS + âˆ RSQ = âˆ QRT, we have

28Âº + âˆ RSQ = 65Âº

â‡’ âˆ RSQ = 65Âº âˆ 28Âº = 37Âº

Since, PQ || SR and QS is a transversal. [Given ]

âˆ´ âˆ PQS = âˆ RSQ [interior alternate angles]

â‡’ x = 37Âº Again, PQ âŠ¥ PS [Given]

âˆ´ âˆ P = 90Âº Now, in **Î”**PQS, we have

âˆ P + âˆ PQS + âˆ PSQ = 180Âº [By angle sum property]

â‡’ 90Âº + x + y = 180Âº

â‡’ 90Âº + 37Â° + y = 180Âº [âˆµ x = 37Âº]

â‡’ y = 180Âº âˆ 90Âº âˆ 37Âº = 53Âº

Thus, x = 37Âº and y = 53Âº

**Question 6. In the adjoining figure, the side QR of Î”PQR is produced to a point S. If the bisectors of âˆ PQR and âˆ PRS meet at point T, then prove that âˆ QTR = **(1/2)** âˆ QPR.**

**Solution:** In **Î”**PQR, the side QR is produced to S.

âˆ´ Exterior âˆ PRS = Sum of the interior opposite angles

â‡’ âˆ PRS = âˆ P + âˆ Q

Since QT and RT are bisectors of âˆ Q and âˆ PRS respectively,

âˆ´ (1/2)âˆ PRS =(1/2)âˆ P +(1/2)âˆ Q

**â‡’ âˆ TRS =** (1/2) âˆ P + âˆ TQR â€¦(1)

Now, In **Î”**QRT, we have

Exterior âˆ TRS = âˆ TQR + âˆ T ...(2)

From (1) and (2), we have

âˆ TQR + (1/2)âˆ P= âˆ TQR + âˆ T

â‡’ (1/2)âˆ P= âˆ T

i.e. (1/2)** âˆ QPR = âˆ QTR or âˆ QTR = **(1/2)

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