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

Solution:

Data given :

Lines AB and CD intersect at O.

∠ AOC + ∠BOE = 70° …………….(1)

and ∠BOD = 40°…………………..(2)

Since ∠AOC = ∠BOD *(vertically opposite angles)*

Therefore, ∠AOC = 40° [FROM 2]

and 40° + ∠BOE = 70° [from 1]

∠BOE = 70° – 40° = 30°

Also, ∠AOC + ∠BOE + ∠COE = 180° *(**⸫** AOB is a straight line)*

⇒70° + ∠COE = 180° *[from (1)]*

⇒ ∠COE = 180° – 70° = 110°

Now, reflex ∠COE = 360° – 110° = 250°

Hence, ∠BOE = 30° and reflex ∠ COE = 250°

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

Solution:

In the figure, lines XY and MN intersect at O and ∠ POY = 90°.

Also, given a : b = 2 : 3

Let a = 2x and b = 3x.

Since, ∠XOM + ∠POM + ∠POY = 180°*(Linear pair axiom)*

⇒ 3x + 2x + 90° = 180°

⇒ 5x = 180° – 90°

⇒ x = ^{90°}/_{5} = 18°

∴ ∠XOM = b = 3x = 3 × 18° = 54°

and ∠POM = a = 2x = 2 × 18° = 36°

Now, ∠XON = c = ∠MOY = ∠POM + ∠POY *(Vertically opposite angles)*

= 36° + 90° = 126°

Hence, c = 126°

**In Fig., ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.**

∠PQS + ∠PQR = 180° …………….(1)*(Linear pair axiom)*

∠PRQ + ∠PRT = 180° …………..(2)*(Linear pair axiom)*

But, ∠PQR = ∠PRQ *(Given)*

∴ From (1) and (2)

∠PQS = ∠PRT

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

Sum of all angles in a circle always 360°

Hence, ∠AOC + ∠BOC + ∠DOB + ∠AOD = 360°

⇒ x + y + w + z = 360°

⇒ x + y + x + y = 360°

Given that x = y = w + z

⇒ 2w + 2z = 360°

⇒ 2(w + z) = 360°

⇒ w + z = 180° *(linear pair)*

ie., ∠BOD + ∠AOD = 180°

If the sum of two adjacent angles is 180° then the non common arms of the angles form a line.

Therefore, AOB is a line.

**In Fig., POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ ROS =**^{1}/_{2}(∠ QOS – ∠ POS).

Solution:

Since, OR⊥PQ

⇒ ∠QOS = 90°

Since, PQ is a line OR stands on it

Therefore, ∠POR+∠QOR = 180 ° (linear pair)

∠POR = 90 °

Since, ∠POS + ∠SOR = 90 °………..(i)

∠QOR = 90 °

∠QOS – ∠ROS = 90 ° ……………….. (ii)

From (i) and (ii)

∠POS + ∠SOR = ∠QOS -∠POS

∠ROS + ∠ROS = ∠QOS – ∠POS

2∠ROS = ∠QOS-∠POS

∠ROS = ^{1}/_{2} (∠QOS – ∠POS)

Hence proved

**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:

Given, ∠XYZ = 64° and QY bisects ∠ZYP

⇒∠XYZ + ∠ZYP = 180° *[LINEAR PAIR]*

⇒64° + ∠ZYP = 180°

⇒∠ZYP = 180° – 64°

⇒∠PYQ + ∠ZYQ = 116°

⇒2∠PYQ = 116°

⇒∠PYQ = 58°

Again, ∠XYQ = ∠XYZ + ∠ZYQ

= 64° + 58°

= 122°

## 2 thoughts on “Lines and Angle – Exercise 6.1 – Class IX”

Comments are closed.