AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.2

Question 1.

In the given figure three lines \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{CD}}\) and \(\overline{\mathrm{EF}}\) intersecting at ‘O’. Find the values of x, y and z, it is being given that x : y : z = 2 : 3 : 5

Solution:

From the figure

2x + 2y + 2z = 360°

⇒ x + y + z = 180°

x : y : z = 2 : 3 : 5

Sum of the terms of the ratio

= 2 + 3 + 5 = 10

Question 2.

Find the value of x in the following figures.

i)

Solution:

From the figure

3x + 18= 180° – 93° (∵ linear pair )

3x + 18 = 87

3x = 87- 18 = 69

∴ x = \(\frac{69}{3}\) = 23

ii)

Solution:

From the figure

(x – 24)° + 29° + 296° = 360“

(∵ complete angle)

x + 301° = 360°

∴ x = 360° – 301° = 59°

iii)

Solution:

From the figure

2 + 3x = 62

(∵ vertically opposite angles)

3x = 62 – 2

∴ 3x = 60° ⇒ x = \(\frac{60}{3}\)

∴ x = 20°

iv)

Solution:

From the figure

40 + (6x + 2) = 90°

(∵ complementary angles)

6x = 90° – 42°

6x = 48

x = \(\frac{48}{6}\) = 8°

Question 3.

In the given figure lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) intersect at ’O’. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:

Given that ∠AOC +∠BOE = 70°

∠BOD = 40°

∠AOC = 40°

(∵ ∠AOC, ∠BOD are vertically opposite angles)

∴ 40° + ∠BOE = 70°

⇒ ∠BOE = 70° – 40° = 30°

Also ∠AOC + ∠COE +∠BOE = 180°

( ∵ AB is a line)

⇒ 40° + ∠COE + 30° = 180°

⇒ ∠COE = 180° -70° = 110°

∴ Reflex ∠COE = 110°

Question 4.

In the given figure lines \(\overline{\mathrm{XY}}\) and \(\overline{\mathrm{MN}}\) . intersect at O. If ∠POY = 90° and a : b = 2:3, find c.

Solution:

Given that XY and MN are lines.

∠POY = 90°

a : b = 2 : 3

From the figure a + b = 90°

Sum of the terms of the ratio a : b

= 2 + 3 = 5

∴ b = \(\frac{3}{5}\) x 90° = 54°

From the figure b + c = 180°

(∵ linear pair of angles)

54° + c = 180°

c = 180°-54° = 126°

Question 5.

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

Solution:

Given that ∠PQR = ∠PRQ

From the figure

∠PQR + ∠PQS = 180° ………….. (1)

∠PRQ + ∠PRT = 180° …………..(2)

From (1) and (2)

∠PQR + ∠PQS = ∠PRQ + ∠PRT

But ∠PQR = ∠PRQ

So ∠PQS = ∠PRT

Hence proved.

Question 6.

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

Solution:

Given that x + y = w + z = k say

From the figure

x + y + z + w = 360°

(∵ Angle around a point)

Also x + y = z + w

∴ x + y = z + w = \(\frac{360^{\circ}}{2}\)

∴ x + y = z + w = 180°

OR

k + k = 360°

2k = 360°

k = \(\frac{360^{\circ}}{2}\)

(i.e.) (x,y) and (z, w) are pairs of adjacent angles whose sum is 180°.

In other words (x, y) and (z, w) are linear pair of angles ⇒ AOB is a line.

Question 7.

In the given figure \(\overline{\mathrm{PQ}}\) is a line. Ray \(\overline{\mathrm{OR}}\) is perpendicular to line \(\overline{\mathrm{PQ}}\).\(\overline{\mathrm{OS}}\) os is another ray lying between rays \(\overline{\mathrm{OP}}\) and \(\overline{\mathrm{OR}}\) Prove that

Solution:

Given : OR ⊥ PQ ⇒ ∠ROQ = 90°

To prove: ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)

Solution:

Proof: From the figure

∠ROS = ∠QOS – ∠QOR ……………(1)

∠ROS = ∠ROP – ∠POS ……………..(2)

Adding (1) and (2)

∠ROS + ∠ROS = ∠QOS – ∠QOR +∠ROP – ∠POS [ ∵ ∠QOR = ∠ROP = 90° given]

⇒ 2∠ROS = ∠QOS – ∠POS

⇒ ∠ROS = \(\frac{1}{2}\) [∠QOS – ∠POS]

Hence proved.

Question 8.

It is given that ∠XYZ = 64° and XY is produced to point P. A ray \(\overline{\mathrm{YQ}}\) bisects ∠ZYP. Dräw a figure from the given Information. Find ∠XYQ and reflex ∠QYP.

Solution:

∠XYQ = 32°

∠QYP = 32°