## Lines and Angles

**Chapter 6**

**Exercise 6.1**

**EX 6.1 QUESTION 1.**

** In Fig. 6.13, lines AB and CD intersect at O. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE.**

**Solution:**

Given: Lines AB and CD intersect at 0.

∠BOD = 40°

∠AOC + ∠BOE = 70° …..(i)

∠AOC = ∠BOD [vertically opposite angles]

∴ ∠AOC = 40° [∠BOD = 40°]

from equation (i)

40°+ ∠BOE = 70°

⇒∠BOE = 70° – 40° = 30°

∠AOC + ∠BOE + ∠COE = 180° [AOB is a straight line]

⇒ 70°+∠COE = 180° [from equation (i)]

⇒∠COE = 180° – 70° = 110°

and ∠COE =360° – ∠COE= 360°- 110° = 250°

**EX 6.1 QUESTION 2.**

** In Fig. 6.14, lines XY and MN intersect at O. If POY = 90° and a: b = 2 : 3, find c.**

**Solution:**

XOY is a straight line.

∴ b+a+∠POY= 180°

But ∠POY = 90° [Given]

∴ b + a = 180° – 90° = 90° …(i)

Also a : b = 2 : 3 ⇒ b =3a/2 …(ii)

Now from (i) and (ii), we get

3a/2 + A = 90°

⇒ 5a/2= 90°

⇒ a =(90°/5)x2 = 36°

From (ii), we get

b = 3/2 x 36° = 54°

XY and MN interstect at O,

∴ c = [a + ∠POY] [Vertically opposite angles]

or c = 36° + 90° = 126°

**EX 6.1 QUESTION 3.**

**In Fi****g. 6.15, PQR = PRQ, then prove that PQS = PRT.**

**Solution:**

ST is a straight line.

∴ ∠PQR + ∠PQS = 180° …(i) [Linear pair]

Similarly, ∠PRT + ∠PRQ = 180° …(ii) [Linear Pair]

From (i) and (ii), we have

∠PQS + ∠PQR = ∠PRT + ∠PRQ

But ∠PQR = ∠PRQ [Given]

∴ ∠PQS = ∠PRT

**EX 6.1 QUESTION 4.**

**In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.**

**Solution:**

Sum of all the angles at a point = 360°

∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360°

But (x + y) = (⇒ + w) [Given]

∴ (x + y) + (x + y) = 360° or,

2(x + y) = 360°

or, (x + y) = = 180°

∴ AOB is a straight line.

**EX 6.1 QUESTION 5.**

**In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = ½ (QOS – POS).**

**Solution:**

Given: POQ is a straight line.

∴ ∠POS + ∠ROS + ∠ROQ = 180°

But OR ⊥ PQ

∴ ∠ROQ = 90°

⇒ ∠POS + ∠ROS + 90° = 180°

⇒ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90° – ∠POS … (i)

Now, we have ∠ROS + ∠ROQ = ∠QOS

⇒ ∠ROS + 90° = ∠QOS

⇒ ∠ROS = ∠QOS – 90° ……(ii)

Adding (1) and (2), we have

2 ∠ROS = (∠QOS – ∠POS)

∴ ∠ROS = 1/2(∠QOS – ∠POS)

**EX 6.1 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.**

**Solution:**

∠PYZ + ∠XYZ = 180° [Linear pair]

⇒∠PYZ + 64° = 180° [XYZ = 64° ]

⇒∠PYZ = 180° – 64° = 116°

But,

∠PYQ =∠ ZYQ = 1/2 ∠PYZ [∠ZYP is bisected by ray YQ}

∴∠PYQ =∠ ZYQ = 1/2 x116° = 58°

∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 112°

and

∠QYP = 360 – ∠QYP = 360° – 58° = 302°