Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 6 Lines and Angles Ex 6.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

Question 1.

In the given figure, lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE.

Answer:

Lines AB and CD intersect at O.

∴ ∠ AOC = ∠ BOD (Vertically opposite angles)

Now, ∠ BOD = 40° (Given)

∴ ∠ AOC = 40°

Moreover, ∠ AOC + ∠ BOE = 70° (Given)

∴ 40° + ∠ BOE = 70°

∴ ∠BOE = 70° – 40°

∴ ∠ BOE = 30°

∠ BOD and ∠ BOE are adjacent angles with common arm ray OB.

∴ ∠ DOE = ∠ BOD + ∠ BOE = 40° + 30° = 70°

Reflex ∠ COE = ∠ COD + ∠ DOE

= 180° + 70° (∠ COD is a straight angle as ray OA stands on line CD.)

= 250°

Thus, ∠ BOE = 30° and reflex ∠ COE = 250°.

Question 2.

In the given figure, lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3, find c.

Answer:

Ray OP stands on line XY.

Hence, ∠XOP and ∠POY from a linear pair of angles.

∴ ∠ XOP + ∠ POY = 180°

∴ ∠ XOP + 90° = 180°

∴ ∠ XOP = 90°

∠XOM and ∠MOP are adjacent angles.

∴ ∠ XOM + ∠ MOP = ∠XOP

∴ b + a = 90° …………. (i)

Now, a : b = 2 : 3

If a = 2x, then b = 3x.

∴ 3x + 2x = 90° [From (1)]

∴ 5x = 90°

∴ x = 18°

Then, ∠ XOM = b = 3x = 3 × 18° = 54°

and ∠ MOP = a = 2x = 2 × 18° = 36°

Now, ∠ MOY = ∠ MOP + ∠ POY (Adjacent angles)

∴ ∠ MOY = 36° + 90° = 126°

Lines XY and MN intersect at O.

∴ ∠ XON and ∠ MOY are vertically opposite angles.

∴ ∠ XON = ∠ MOY

∴ c = 126°

Question 3.

In the given figure, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.

Answer:

Ray QP stands on line ST.

∴ ∠ PQR and ∠ PQS form a linear pair of angles.

∴ ∠ PQR + ∠ PQS = 180°

Ray RP stands on line ST.

∴ ∠ PRQ and ∠ PRT form a linear pair of angles.

∴ ∠ PRQ + ∠ PRT = 180°

∴ ∠ PQR + ∠ PRT = 180° (Given : ∠ PQR = ∠ PRQ)

Then, ∠ PQR + ∠ PQS = ∠ PQR + ∠ PRT = 180°

∴ ∠ PQS = ∠ PRT

Question 4.

In the given figure, if x + y = w + z, then prove that AOB is a line.

Answer:

We know that sum of all the angles round any given point is 360°.

∴ x + y + z + w = 360°

∴ x + y + x + y = 360° (Given : x + y = w + z)

∴2x + 2y = 360°

∴ 2 (x + y) = 360°

∴ x + y = 180°

∴ ∠ COB + ∠ COA = 180°

But, ∠ COB and ∠ COA are adjacent angles and their sum is 180°.

∴ ∠ COB and ∠ COA are angles of a linear pair.

Hence, AOB is a line.

Question 5.

In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that.

Answer:

∠ ROS = \(\frac{1}{2}\) (∠ QOS – ∠ POS).

Ray OR is perpendicular to line PQ.

∴ ∠ QOR = ∠ POR = 90°

Now, ∠ QOR and ∠ ROS are adjacent angles with common arm ray OR.

∴ ∠ QOS = ∠ QOR + ∠ ROS

∴ ∠ QOS = 90° + ∠ ROS ……………….. (1)

Similarly, ∠ POS and ∠ ROS are adjacent angles with common arm ray OS.

∴ ∠ POR = ∠ POS + ∠ ROS

∴ 90° = ∠ POS + ∠ ROS

∴ ∠ POS = 90° – ∠ ROS ………………… (2)

Subtracting (2) from (1), we get

∠ QOS – ∠ POS = (90° + ∠ ROS) – (90° – ∠ ROS)

∴ ∠ QOS – ∠ POS = 90° + ∠ ROS – 90° + ∠ ROS

∴ ∠ QOS – ∠ POS = 2∠ ROS

∴ ∠ ROS = \(\frac{1}{2}\) (∠ QOS – ∠ POS)

Question 6.

It is given that ∠ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP.

Answer:

∠ XYZ + ∠ PYZ = 180° [Angles of a linear pair]

∴ 64° + ∠ PYZ= 180° [Given ∠ XYZ = 64°]

∴ ∠ PYZ = 180°-64°

∴ ∠ PYZ = 116°

Ray YQ bisects ∠ PYZ.

∴ ∠ PYQ = ∠ QYZ = \(\frac{1}{2}\) ∠ PYZ = \(\frac{1}{2}\) × 116° = 58°

∴ ∠ XYQ = ∠ XYZ + ∠ QYZ [Adjacent angles]

∴ ∠ XYQ = 64° + 58°

∴ ∠ XYQ = 122°

XY is produced to P.

∴ ∠ XYP is a straight angle.

∴ ∠ XYP = 180°

Reflex ∠ QYP = ∠ XYQ + ∠ XYP

= 122° + 180°

= 302°