Question 1.
In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.

Solution:
In AFAE,
ext. ∠FEB = ∠A + F
= 90° + 40° = 130°
Since AB || CD
∴ ∠ECD = FEB = 130°
Hence, ∠ECD = 130°.

Question 2.
In the fig., AD and CE are the angle bisectors of ∠A and ∠C respectively. If ∠ABC = 90°, then find ∠AOC.

Solution:
∵ AD and CE are the bisector of ∠A and ∠C

In ∆AOC,
∠AOC + ∠OAC + ∠OCA = 180°
⇒ ∠AOC + 45o = 180°
⇒ ∠AOC = 180° – 45° = 135°.

Question 3.
In the given figure, prove that m || n.

Solution:
In ∆BCD,
ext. ∠BDM = ∠C + ∠B
= 38° + 25° = 63°
Now, ∠LAD = ∠MDB = 63°
But, these are corresponding angles. Hence,
m || n

Question 4.
In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a, b, c.

Solution:
Here, 4b + 75° + b = 180° [a straight angle]
5b = 180° – 75° = 105°
b – 105∘5 = 21°
∴ a = 4b = 4 × 21° = 84° (vertically opp. ∠s]
Again, 2c + a = 180° [a linear pair]
⇒ 2c + 84° = 180°
⇒ 2c = 96°
⇒ c = 96∘2 = 48°
Hence, the values of a, b and c are a = 84°, b = 21° and c = 48°.

Question 5.
In figure, if AB || CD. If ∠ABR = 45° and ∠ROD = 105°, then find ∠ODC.

Solution:

Through O, draw a line ‘l’ parallel to AB.
⇒ line I will also parallel to CD, then
∠1 = 45°[alternate int. angles]
∠1 + ∠2 + 105° = 180° [straight angle]
∠2 = 180° – 105° – 45°
⇒ ∠2 = 30°
Now, ∠ODC = ∠2 [alternate int. angles]
= ∠ODC = 30°

Question 6.
In the figure, ∠X = 72°, ∠XZY = 46°. If YO and ZO are bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OYZ and ∠YOZ.

Solution:
In ∆XYZ, we have
∠X + XY + ∠Z = 180°
⇒ ∠Y + ∠Z = 180° – ∠X
⇒ ∠Y + ∠Z = 180° – 72°
⇒ Y + ∠Z = 108°
⇒ 12 ∠Y + 12∠Z = 12 × 108°
∠OYZ + ∠OZY = 54°
[∵ YO and ZO are the bisector of ∠XYZ and ∠XZY]
⇒ ∠OYZ + 12 × 46° = 54°
∠OYZ + 23° = 54°
⇒ ∠OYZ = 549 – 23° = 31°
In ∆YOZ, we have
∠YOZ = 180° – (∠OYZ + ∠OZY)
= 180° – (31° + 23°) 180° – 54° = 126°

Question 7.
In the given figure, m and n are two plane mirrors perpendicular to each other. Show that incident rays CA is parallel to reflected ray BD.

Solution:
Let normals at A and B meet at P.
As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.
So, BP ⊥ PA i.e., ∠BPA = 90°
Therefore, ∠3 + ∠2 = 90° [angle sum property] …(i)
Also, ∠1 = ∠2 and ∠4 = ∠3 [Angle of incidence = Angle of reflection]
Therefore, ∠1 + ∠4 = 90° [from (i)) …(ii]
Adding (i) and (ii), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
i.e., ∠CAB + ∠DBA = 180°
Hence, CA || BD

Question 8
If in ∆ABC, the bisectors of ∠B and ∠C intersect each other at O. Prove that ∠BOC = 90° + 12∠ A.

Solution:
Let ∠B = 2x and ∠C = 2y
∵OB and OC bisect ∠B and ∠C respectively.
∠OBC = 12∠B = 12 × 2x = x
and ∠OCB = 12∠C = 12 × 2y = y
Now, in ∆BOC, we have
∠BOC + ∠OBC + ∠OCB = 180°
⇒ ∠BOC + x + y = 180°
⇒ ∠BOC = 180° – (x + y)
Now, in ∆ABC, we have
∠A + 2B + C = 180°
⇒ ∠A + 2x + 2y = 180°
⇒ 2(x + y) = 12(180° – ∠A)
⇒ x + y = 90° – 12∠A …..(ii)
From (i) and (ii), we have
∠BOC = 180° – (90° – 12∠A) = 90° + 12 ∠A

Question 9.
In figure, if I || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Solution:
Here, ∠1 and ∠4 are forming a linear pair
∠1 + ∠4 = 180°
(2x + y)° + (x + 2y)° = 180°
3(x + y)° = 180°
x + y = 60
Since I || m and n is a transversal
∠4 = ∠6
(x + 2y)° = (3y + 20)°
x – y = 20
Adding (i) and (ii), we have
2x = 80 = x = 40
From (i), we have
40 + y = 60 ⇒ y = 20
Now, ∠1 = (2 x 40 + 20)° = 100°
∠4 = (40 + 2 x 20)° = 80°
∠8 = ∠4 = 80° [corresponding ∠s]
∠1 = ∠3 = 100° [vertically opp. ∠s]
∠7 = ∠3 = 100° [corresponding ∠s]
Hence, ∠7 = 100° and ∠8 = 80°

Question 10.
In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28o and ∠QRT = 65°. Find the values of x, y and z.

Solution:
Here, PQ || SR .
⇒ ∠PQR = ∠QRT
⇒ x + 28° = 65°
⇒ x = 65° – 28° = 37°
Now, in it. ∆SPQ, ∠P = 90°
∴ ∠P + x + y = 180° [angle sum property]
∴ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Now, ∠SRQ + ∠QRT = 180° [linear pair]
z + 65° = 180°
z = 180° – 65° = 115°

Important Link

Quick Revision Notes : Lines and Angles

NCERT Solution : Lines and Angles

MCQs :Lines and Angles

Click here for Free Video Lectures