Lines and Angles: 9 Math

NCERT Exercise 6.1: 9th math

Lines and Angles Class ninth math NCERT Exercise 6.1 Question (1) In given figure lines AB and CD intersects at O. If ∠AOC + ∠BOE = 70^o and ∠BOD = 40^o, find ∠BOE and reflex ∠COE.

Solution

Given, ∠BOD = 40^o

And, ∠AOC + ∠BOE = 70^o

Then, ∠BOE = ?

And, ∠COE = ?

As given, ∠AOC + ∠BOE = 70^o

⇒ ∠BOD + ∠BOE = 70^o

[Because ∠BOD and ∠AOC are vertically opposite angles and hence are equal.]

⇒ 40^o + ∠BOE = 70^o

⇒ ∠BOE = 70^o – 40^o

⇒ ∠BOE = 30^o - - - - (i)

Now, ∠EOD + ∠COE = 180^o

[Because ∠EOD and ∠COE are formed linear pair of angles together.]

⇒ ∠BOE + ∠BOD + ∠COE = 180^o

⇒ 30^o + 40^o + ∠COE = 180^o

[From equation (i) ∠BOE = 30^o and from question ∠BOD = 40^o]

⇒ 70 + ∠COE = 180^o

⇒ ∠COE = 180^o – 70^o

⇒ ∠COE = 110^o

Thus, ∠BOE = 30^o and ∠COE = 110^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.1 Question (2) In figure lines XY and MN intersects at O. If ∠POY = 90^o and a : b = 2 : 3, find c.

Solution

Given, ∠POY = 90^o

And a : b = 2 : 3

Then, c = ?

∠POY + ∠XOP = 180^o

[Because ∠POY and ∠XOP is a linear pair of angles.]

⇒ 90^o +∠XOP = 180^o

⇒ ∠XOP = 180^o – 90^o

⇒ ∠XOP = 90^o

Now, as given, a : b = 2 : 3

Let a = 2x and b = 3x

Now, a + b = ∠XOP

⇒ a + b = 90^o

⇒ 2x + 3x = 90^o

⇒ 5x = 90^o

`:. x = 90^o/5 `

⇒ x = 18^o

Now, since a = 2x

Thus, after substituting the value of x = 18^o, we get

a = 2 × 18^o

⇒ a = 36^o

And, b = 3x

Thus, after substituting the value of x = 18^o, we get

b = 3 × 18^o

⇒ b = 54^o - - - - - (i)

Now, ∠XOM + ∠XON = 180^o

[Since ∠XOM and ∠XON are linear pair of angles. ]

⇒ b + c = 180^o

[Because angle XOM = b and angle XON = c]

⇒ 54^o + c = 180^o

[Because angle b = 54^o from equation (i)]

⇒ c = 180^o – 54^o

⇒ c = 126^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.1 Question (3) In figure ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution

Given, ∠PQR = ∠PRQ

Then prove that ∠PQS = ∠PRT.

Proof

∠PQR + ∠PQS = 180^o - - - - - (i)

[Because ∠PQR and ∠PQS are linear pair of angles.]

And, ∠PRQ + ∠PRT = 180^o - - - - - (ii)

[Because ∠PRQ and ∠PRT are linear pair of angles]

Now, from equation (i) and equation (ii)

∠PQR + ∠PQS = ∠PRQ + ∠PRT

Now, as given in question ∠PQR = ∠PRQ after substituting ∠PRQ at the place of ∠PQR in above expression, we get

∠PRQ + ∠PQS = ∠PRQ + ∠PRT

⇒ ∠PRQ + ∠PQS – ∠PRQ = ∠PRT

⇒ ∠PQS = ∠PRT Proved

Lines and Angles Class ninth math NCERT Exercise 6.1 Question (4) In figure if x + y = w + z, then prove that AOB is a line.

Solution

Given, x + y = w + z

To prove: AOB is a line

Proof :

∠x + ∠y + ∠w + ∠z = 360^o

[Because sum of angles around a point = 360^o]

⇒ (∠x + ∠y) + (∠w + ∠z) = 360^o

⇒ (∠x + ∠y) + (∠x + ∠y) = 360^o

[Because as given in question x + y = w + z]

⇒ 2(∠x + ∠y) = 360^o

`:. => /_x + /_y = 360^o/2`

⇒ ∠x + ∠y = 180^o

⇒ ∠w + ∠z = 180^o

[Because as given in question x + y = w + z]

Thus, according to Linear Pair Axiom which says that if sum of two adjacent angles s 180^o, then non-common arms of the angles form a line.

Thus, AOB is a line because ∠x + ∠y = 180^o and ∠w + ∠z = 180^o. Proved

Lines and Angles Class ninth math NCERT Exercise 6.1 Question (5) In figure, 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

Given, Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.

Then to Prove ∠ROS = 1/2 (∠QOS – ∠POS)

Proof

Since, OR is perpendicular to PQ

Thus, ∠POR = 90^o

⇒ ∠POS + ∠SOR = 90^o - - - - - (i)

[Because ∠POR = ∠POS + ∠SOR]

And, ∠POS + ∠QOS = 180^o

[Because ∠POS and ∠QOS are linear pair of angles.]

⇒ ∠POS + ∠QOS = 2(∠POS + ∠SOR)

[Since from equation (i) ∠POS + ∠SOR = 90^o, thus 2(∠POS + ∠SOR) = 180^o]

⇒ ∠POS + ∠QOS = 2 ∠POS + 2 ∠SOR

After transposing ∠POS to RHS

⇒ ∠QOS = 2 ∠POS + 2 ∠SOR – ∠POS

⇒ ∠QOS = 2 ∠POS – ∠POS + 2 ∠SOR

⇒ ∠QOS = ∠POS + 2 ∠SOR

⇒ ∠QOS – ∠POS = 2 ∠SOR

⇒ 2 ∠SOR = ∠QOS – ∠POS

⇒ ∠SOR = 1/2(∠QOS – ∠POS) Proved

Lines and Angles Class ninth math NCERT Exercise 6.1 Question (6) It is given that ∠XYZ = 64^o 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^o

And, XY is produced out to point P.

And, A ray YQ bisects ∠ZYP

Thus, from given information figure can be drawn as follows:

Thus, ∠XYQ and reflex ∠QYP = ?

Now, since PX is a straight line

Thus, ∠ZYP + ∠XYZ = 180^o

⇒ ∠ZYP + 64^o = 180^o

⇒ ∠ZYP = 180^o – 64^o

⇒ ∠ZYP = 116^o

Now, as given QY bisects ∠ZYP

Thus, ∠ZYQ = 1/2 ∠ZYP

⇒ ∠ZYQ = 1/2 × 116^o

⇒ ∠ZYQ = 58^o

Now, ∠XYQ = ∠XYZ + ∠ZYQ

⇒ ∠XYQ = 64^o + 58^o

⇒ ∠XYQ = 122^o

Now, Reflex ∠QYP = ∠XYQ + ∠XYP

⇒ ∠QYP = 122^o + 180^o

[Because ∠XYP is a straight angle which is equal to 180^o]

⇒ ∠QYP = 302^o

Thus, ∠XYQ = 122^o and Reflex ∠QYP = 302^o Answer

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