The document Ex 6.1 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 following figure, lines AB and CD intersect at O. If âˆ AOC + âˆ BOE = 70Âº and âˆ BOD = 40Âº, find âˆ BOE and reflex âˆ COE.**

**Solution:** Since AB is a straight line,

âˆ´ âˆ AOC + âˆ COE + âˆ EOB = 180Â°

or (âˆ AOC + âˆ BOE) + âˆ COE = 180Âº

or 70Âº + âˆ COE = 180Âº [âˆµ âˆ AOC + âˆ BOC = 70Âº (Given)]

or âˆ COE = 180Âº - 70Âº = 110Âº

âˆ´ Reflex âˆ COE = 360Âº - 110Âº = 250Âº

âˆµ AB and CD intersect at O.

âˆ´ âˆ COA = âˆ BOD [Vertically opposite angles]

But âˆ BOD = 40Âº [Given]

âˆ´ âˆ COA = 40Âº

Also âˆ AOC + âˆ BOE = 70Â°

âˆ´ 40Âº + âˆ BOE = 70Âº

or âˆ BOE = 70Âº âˆ 40Âº = 30Âº

Thus, âˆ BOE = 30Âº and reflex âˆ COE = 250.

**Question 2. In the following figure, lines XY and MN intersect at O. If âˆ POY = 90Âº and a : b = 2 : 3, find c.**

**Solution**: âˆµ XOY is a straight line.

âˆ´ âˆ b + âˆ a + âˆ POY = 180Âº

But âˆ OPY = 90Âº [Given]

âˆ´ âˆ b + âˆ a = 180Âº âˆ 90Âº = 90Âº.

Also a : b = 2 : 3

âˆ´

Since XY and MN intersect at O,

âˆ´ âˆ c= [âˆ a + âˆ POY] [Vertically opposite angles]

or âˆ c = 36Âº + 90Âº = 126Âº

Thus, the required measure of â€˜câ€™ = 126Âº.

**Question 3. In the following figure, âˆ PQR = âˆ PRQ, then prove that âˆ PQS = âˆ PRT.**

**Solution:** âˆµ ST is a straight line,

âˆ´ âˆ PQS + âˆ PQR = 180Âº ...(1)

Similarly, âˆ PRT + âˆ PRQ = 180Âº ...(2)

From (1) and (2), we have

âˆ PQS + âˆ PQR = âˆ PRT + âˆ PRQ

But âˆ PQR = âˆ PRQ [Given]

âˆ´ âˆ PQS = âˆ PRT

**Question 4. In the following figure, if x + y = w + z, then prove that AOB is a line.**

**Solution:** âˆµ Sum of all the angles at a point = 360Âº

âˆ´ x + y + z + w = 360Âº

or (x + y) + (z + w) = 360Âº

But (x + y) = (z + w) [Given]

âˆ´ (x + y) + (x + y) = 360Âº

or 2(x + y) = 360Âº

or (x + y) = (360^{o}/2)= 180Âº

âˆ´ AOB is a straight line.

**Question 5. In the adjoining figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that âˆ ROS = (1/2) (âˆ QOS âˆ âˆ POS). Solution:** âˆµ POQ is a straight line. [Given]

âˆ´ âˆ POS + âˆ ROS + âˆ ROQ = 180Âº

But OR âŠ¥ PQ

âˆ´ âˆ ROQ = 90Âº

âˆ´ âˆ POS + âˆ ROS + 90Âº = 180Âº

â‡’ âˆ POS + âˆ ROS = 90Âº â€¦(1)

Now, we have âˆ ROS + âˆ ROQ = âˆ QOS

â‡’ âˆ ROS + 90Âº = âˆ QOS â€¦(2)

From (1) and (2),

we have âˆ ROS + [âˆ POS + âˆ ROS] = âˆ QOS

â‡’ 2âˆ ROS + âˆ POS = âˆ QOS

â‡’ 2âˆ ROS = [âˆ QOS âˆ POS]

âˆ´ âˆ ROS = (1/2)[âˆ QOS âˆ POS]

**Question 6. It is given that âˆ XYZ = 64Âº and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects âˆ ZYP, find âˆ XYQ and reflex âˆ QYP. Solution: **âˆµ XYP is a straight line.

âˆ´ âˆ XYZ + âˆ ZYQ + âˆ QYP = 180Âº

â‡’ 64Âº + âˆ ZYQ + âˆ QYP = 180Âº [âˆµ YQ, bisects âˆ ZYP, so âˆ QYP = âˆ ZYQ]

â‡’ 64Âº + 2âˆ QYP = 180Âº

â‡’ 2âˆ QYP = 180Âº - 64Âº = 116Âº

â‡’ âˆ QYP = (116^{0}/2)= 58Âº

âˆ´ Reflex âˆ QYP = 360Âº - 58Âº = 302Âº

Since âˆ XYQ = âˆ XYZ + âˆ ZYQ

â‡’ âˆ XYQ = 64Âº + âˆ QYP [âˆµ âˆ XYZ = 64Âº (Given) and âˆ ZYQ = âˆ QYP]

â‡’ âˆ XYQ = 64Âº + 58Âº [âˆµ âˆ QYP = 58Âº]

= 122Âº

Thus, âˆ XYQ = 122Âº and reflex âˆ QYP = 302Â°

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