Post on 01-Feb-2023
Downloaded from www.careerboosteracademy.com
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
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
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.
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
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)
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
(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)
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
⇒ 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.
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
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:
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
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
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
⇒ ∠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
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
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)
(alternate interior angles)
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)
(alternate interior angles)
And ∠3 = ∠5 ..(ii)
(alternate interior angles)
Adding (i) and (ii), we have
∠2 + ∠3 = ∠4 + ∠5
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539
Downloaded from www.careerboosteracademy.com
⇒ ∠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)
CAREER BOOSTER
Saraf Complex,
Indira Market Road, Wardha
M:9822204977 , 8087341539