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LINES AND ANGLES WORKSHEET – 2 Class 9

1. In figure AB and AC are opposite rays

(i) If ∠𝑩𝑨𝑫 = 𝟖𝟓𝟎, 𝒇𝒊𝒏𝒅 ∠𝑪𝑨𝑫

(ii) If ∠𝑪𝑨𝑫 = 𝟏𝟑𝟓𝟎, 𝒇𝒊𝒏𝒅 ∠𝑩𝑨𝑫

Solution

(i) Since ∠𝐵𝐴𝐷 and ∠𝐷𝐴𝐶 from a linear pair.

∠𝐵𝐴𝐷 + ∠𝐷𝐴𝐶 = 1800

850 + ∠𝐷𝐴𝐶 = 1800

∠𝐷𝐴𝐶 = 1800 − 850 = 950

(ii) ∠𝐵𝐴𝐷 + ∠𝐷𝐴𝐶 = 1800

∠𝐵𝐴𝐷 + 1350 = 1800 ∠𝐵𝐴𝐷 = 1800 − 1350 = 450

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

Solution

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

Solution

In the given figure, ST is a straight line and ray QP stands on it.

∴ ∠PQS + ∠PQR = 180º (Linear Pair)

∠PQR = 180º − ∠PQS (1)

∠PRT + ∠PRQ = 180º (Linear Pair)

∠PRQ = 180º − ∠PRT (2)

It is given that ∠PQR = ∠PRQ.

Equating equations (1) and (2), we obtain

180º − ∠PQS = 180 − ∠PRT

∠PQS = ∠PRT

4. It is given that ∠XYZ = 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

It is given that line YQ bisects ∠PYZ.

Hence, ∠QYP = ∠ZYQ

It can be observed that PX is a line. Rays YQ and YZ stand on it.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180º

⇒ 64º + 2∠QYP = 180º

⇒ 2∠QYP = 180º − 64º = 116º

⇒ ∠QYP = 58º

Also, ∠ZYQ = ∠QYP = 58º

Reflex ∠QYP = 360º − 58º = 302º

∠XYQ = ∠XYZ + ∠ZYQ

= 64º + 58º = 122º

5. If two lines intersect each other, then the vertically opposite angles are equal.

Solution

Given: AB and CD intersects each other at O.

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To Prove: (i) ∠𝐴𝑂𝐷 = ∠𝐶𝑂𝐵

(ii)∠𝐴𝑂𝐶 = ∠𝐵𝑂𝐷

Proof: Since ray OA stands on the line CD at O.

∴∠𝐴𝑂𝐶 + ∠𝐴𝑂𝐷 = 1800 ...(i)

Also, ray OC stands on the line AB at O.

∴∠𝐴𝑂𝐶 + ∠𝐶𝑂𝐵 = 1800 ....(ii)

From (i) and (ii), we have

∠𝐴𝑂𝐶 + ∠𝐴𝑂𝐷= ∠𝐴𝑂𝐶 + ∠𝐶𝑂𝐵

⇒ ∠𝐴𝑂𝐷 = ∠𝐶𝑂𝐵 Similarly, we can prove

∠𝐴𝑂𝐶 = ∠𝐵𝑂𝐷

6. If a transversal intersects two parallel lines. Then each pair of interior angles

supplementary.

Solution

Given: A transversal intersects two parallel lines AB and CD at P and Q

respectively. Making two pairs of consecutive interior angles, ∠1 and

∠2; ∠3 and ∠4. To Prove: ∠1 + ∠2 = 1800

And ∠3 + ∠4 = 1800

Proof: Ray QD stands on line t.

∴∠2 + ∠5 = 1800 (linear pair)

But, ∠5 = ∠1 (corres. ∠𝑠 axiom)

∴ ∠1 + ∠2 = 1800 ... (i)

Now, PQ stands on AB,

∠1 + ∠3 = 1800 ...(ii)

(linear pair axiom)

Also, Q stands on CD

∴ ∠2 + ∠4 = 1800 (iii)

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(linear pair axiom)

Adding (ii) and (iii), we have

(∠1 + ∠2) + (∠3 + ∠4) = 3600

But, ∠1 + ∠2 = 1800

(proved above)

∴ ∠3 + ∠4 = 3600 − 1800

= 1800

Hence, ∠1 + ∠2 = 1800

And ∠3 + ∠4 = 1800.

7. In the given figure, find the values of x and y and then show that AB || CD.

Solution

It can be observed that,

50º + x = 180º (Linear pair)

x = 130º … (1)

Also, y = 130º (Vertically opposite angles)

As x and y are alternate interior angles for lines AB and CD and also measures of these angles are

equal to each other, therefore, line AB || CD.

8. In the given figure, If AB || CD, EF ⊥ CD and ∠GED = 126º, find ∠AGE, ∠GEF and ∠FGE.

Solution

It is given that,

AB || CD

EF ⊥ CD

∠GED = 126º

⇒ ∠GEF + ∠FED = 126º

⇒ ∠GEF + 90º = 126º

⇒ ∠GEF = 36º

∠AGE and ∠GED are alternate interior angles.

⇒ ∠AGE = ∠GED = 126º

However, ∠AGE + ∠FGE = 180º (Linear pair)

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⇒ 126º + ∠FGE = 180º

⇒ ∠FGE = 180º − 126º = 54º

∴ ∠AGE = 126º, ∠GEF = 36º, ∠FGE = 54º

9. In the given figure, if AB || CD, ∠APQ = 50º and ∠PRD = 127º, find x and y.

Solution

∠APR = ∠PRD (Alternate interior angles)

50º + y = 127º

y = 127º − 50º

y = 77º

Also, ∠APQ = ∠PQR (Alternate interior angles)

50º = x

∴ x = 50º and y = 77º

10. In the given figure, ∠X = 62º, ∠XYZ = 54º. If YO and ZO are the bisectors of ∠XYZ and

∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.

Solution

As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,

∠X + ∠XYZ + ∠XZY = 180º

62º + 54º + ∠XZY = 180º

∠XZY = 180º − 116º

∠XZY = 64º

∠OZY = 64

2 = 32º (OZ is the angle bisector of ∠XZY)

Similarly, ∠OYZ =54

2 = 27º

Using angle sum property for ΔOYZ, we obtain

∠OYZ + ∠YOZ + ∠OZY = 180º

27º + ∠YOZ + 32º = 180º

∠YOZ = 180º − 59º

∠YOZ = 121º

11. In the given figure, if AB || DE, ∠BAC = 35º and ∠CDE = 53º, find ∠DCE.

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Solution

AB || DE and AE is a transversal.

∠BAC = ∠CED (Alternate interior angles)

∴ ∠CED = 35º

In ΔCDE,

∠CDE + ∠CED + ∠DCE = 180º (Angle sum property of a triangle)

53º + 35º + ∠DCE = 180º

∠DCE = 180º − 88º

∠DCE = 92º

12. In the given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40º, ∠RPT =

95º and ∠TSQ = 75º, find ∠SQT.

Solution

Using angle sum property for ΔPRT, we obtain

∠PRT + ∠RPT + ∠PTR = 180º

40º + 95º + ∠PTR = 180º

∠PTR = 180º − 135º

∠PTR = 45º

∠STQ = ∠PTR = 45º (Vertically opposite angles)

∠STQ = 45º

By using angle sum property for ΔSTQ, we obtain

∠STQ + ∠SQT + ∠QST = 180º

45º + ∠SQT + 75º = 180º

∠SQT = 180º − 120º

∠SQT = 60º

13. If the bisectors of a pair of corresponding angles formed by transversal with two parallel

lines, prove that the bisectors are parallel.

Solution:

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Given: Two lines AB and CD are parallel and intersected by transversal‘s’ at P & Q respectively.

EP and FQ bisect the corresponding angle ∠𝐴𝑃𝐺 and ∠CQP respectively.

To prove : EP ∥ FQ

Proof: AB ∥ CD (given)

⇒∠APG = ∠CQP (corresponding ∠𝑠) 1

2 (∠APG) =

1

2 (∠CQP)

⇒ ∠𝐸𝑃𝐺 = ∠𝐹𝑄𝑃

But these are corresponding angles

𝐸𝑃 ∥ 𝐹𝑄 (corresponding angle axiom)

14. If two parallel lines are intersected by a transversal, prove that the bisectors of two pairs of

interior angles enclose a rectangle.

Solution:

Given: 𝐴𝐵 ∥ 𝐶𝐷 & a transversal ’t’ intersects them at 𝐸 & 𝐹 respectively. 𝐸𝐺, 𝐹𝐺, 𝐸𝐻 & 𝐹𝐻are

bisectors of the interior angles ∠𝐴𝐸𝐹, ∠𝐶𝐹𝐸, ∠𝐵𝐸𝐹 and ∠𝐸𝐹𝐷 respectively.

To Prove: 𝐸𝐺𝐹𝐻 is a rectangle.

Proof: 𝐴𝐵 ∥ 𝐶𝐷 & transversal ‘t’ intersects them at 𝐸 & 𝐹 recpectively.

⇒ ∠𝐴𝐸𝐹 = ∠ 𝐸𝐹𝐷 (𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒𝑠)

⇒ 1

2 ∠𝐴𝐸𝐹 =

1

2(∠𝐸𝐹𝐷)

⇒ ∠𝐺𝐸𝐹 = ∠𝐸𝐹𝐻

But these are alternate interior angles formed when the transversal 𝐸𝐹 intersects 𝐸𝐺 & 𝐹𝐻. ⇒ 𝐸𝐺 ∥ 𝐹𝐻

Similarly 𝐸𝐻 ∥ 𝐺𝐹. Therefore, 𝐸𝐺 𝐹𝐻 is a ∥ gm

Now ray 𝐸𝐹 stands on line 𝐴𝐵 ∴

∠𝐴𝐸𝐹 + ∠𝐵𝐸𝐹 = 1800

1

2 ∠𝐴𝐸𝐹 +

1

2 ∠𝐵𝐸𝐹 = 90

0

⇒ ∠𝐺𝐸𝐹 + ∠𝐻𝐸𝐹 = 900

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⇒ ∠𝐺𝐸𝐻 = 900

Thus 𝐸𝐺𝐹𝐻, parallelogram has one right angle. ∴

𝐸𝐺𝐹𝐻 is a rectangle.

15. A triangle ABC is right-angled at A L is a point on BC such that AL ⊥ BC. Prove that

∠𝑩𝑨𝑳 = ∠𝑨𝑪𝑩. Solution:

Here, in ∆𝐴𝐵𝐶, ∠𝐴 = 90

0 and 𝐴𝐿 ⊥ 𝐵𝐶,

⇒ ∠𝐶𝐴𝐿 + ∠𝐵𝐴𝐿 = 900 ....(i)

Also, ∠𝐶𝐴𝐿 + ∠𝐴𝐶𝐿 = 900 .....(ii)

From (i) and (ii), we have

∠𝐶𝐴𝐿 + ∠𝐵𝐴𝐿 = ∠𝐶𝐴𝐿 + ∠𝐴𝐶𝐿

⇒ ∠𝐵𝐴𝐿 = ∠𝐴𝐶𝐿

Hence, ∠𝐵𝐴𝐿 = ∠𝐴𝐶𝐵

(∵ ∠𝐴𝐶𝐿 and ∠𝐴𝐶𝐵 are the same angles)

16.

Solution:

Let normal at A and B meet at P.

As mirrors are perpendicular to each other, therefore, BP ∥ QA and AP ∥ OB.

So BP ⊥ PA 𝑖. 𝑒. , ∠BPA = 900

therefore, ∠3 + ∠2 = 900

𝑎𝑛𝑔𝑙𝑒 𝑠𝑢𝑚 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 … (𝑖)

Also, ∠1 = ∠2 and ∠4 = ∠3 .

𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒 = 𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛

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Therefore, ∠1 + ∠4 = 900 𝑓𝑟𝑜𝑚 𝑖 . . (𝑖𝑖)

Adding (i) and (ii), we have

∠1 + ∠2 + ∠3 + ∠4 = 1800

𝑖. 𝑒., ∠CAB + ∠DBA = 1800

Hence, CA ∥ BD

17. The sum of the three angles of a triangle is 𝟏𝟖𝟎𝟎.

Solution:

Given: A triangle ABC.

To Prove: ∠𝐴 + ∠𝐵 + ∠𝐶 = 1800

i.e., ∠1 + ∠2 + ∠3 = 1800

Const, : Through A, draw a line / parallel to BC.

Proof: since 𝑙 ∥ 𝐵𝐶 therefore,

∠2 + ∠4 ....(i)

(alternate interior angles)

And ∠3 = ∠5 ....(ii)


Adding (i) and (ii) , we have

∠2 + ∠3 = ∠4 + ∠5

⇒ ∠1 + ∠2 + ∠3 = ∠1 + ∠4∠5

(adding ∠1 on both sides)

⇒ ∠1 + ∠2 + ∠3 = ∠4 + ∠1 + ∠5

⇒ ∠1 + ∠2 + ∠3 + 1800

∵ sum of angles at a point on a line is 1800

Thus, the sum of the three angles of a triangle is 1800.

Proof : Since l ∥ BC, therefore,

∠2 = ∠4 ...... (i)


And ∠3 = ∠5 ..(ii)


Adding (i) and (ii), we have

∠2 + ∠3 = ∠4 + ∠5

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⇒ ∠1 + ∠2 + ∠3 = ∠1 + ∠4 + ∠5

(adding ∠1 on both sides)

⇒ ∠1 + ∠2 + ∠3 = ∠1 + ∠4 + ∠5

⇒ ∠1 + ∠2 + ∠3 = 1800

(∵ sum of angles at point on a line is 1800)

⇒ ∠𝐴 + ∠𝐵 + ∠𝐶 = 1800

Thus, the sum of the three angles of a triangle is 1800.

18. In Fig., the sides AB and AC of ∆ABC are produced to points E and D respectively. If

bisectors BO and Co of ∠𝐂𝐁𝐄 and ∠BCF respectively meet at point O, then prove that

∠BOC = 𝟗𝟎𝟎 −𝟏

𝟐∠𝐁𝐀𝐂.

Solution:

Ray BO is the bisector of ∠CBE.

Therefore, ∠CBO =1

2∠CBE

= 1

2 (1800 − 𝑦)

= 900 −𝑦

2 (1)

Similarly, ray CO is the bisector of ∠BCD.

Therefore, ∠𝐵𝐶𝑂 =1

2 ∠𝐵𝐶𝐷

= 1

2(1800 − 𝑧)

= 900 −𝑧

2 (2)

In ∆BOC, ∠BOC + ∠BCO + ∠CBO = 1800 (3)

Substituting (1) and (2) in (3), you get

∠𝐵𝑂𝐶 + 900 −𝑧

2+ 900 −

𝑦

2= 1800

So, ∠BOC =𝑧

2+

𝑦

2

Or, ∠𝐵𝑂𝐶 = 1

2(𝑦 + 𝑧) (4)

But, 𝑥 + 𝑦 + 𝑧 = 1800

(Angle sum property of a triangle)

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Therefore, 𝑦 + 𝑧 = 1800 − 𝑥

Therefore, (4) becomes

∠𝐵𝑂𝐶 =1

2 (1800 − 𝑥)

= 900 −𝑥

2

= 900 −1

2∠𝐵𝐴𝐶

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LINES AND ANGLES WORKSHEET – 2 Class 9 - Career ...

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