Lines and Angles
Chapter 6
Exercise 6.2
EX 6.2 QUESTION 1.
In Fig. 6.28, find the values of x and y and then show that AB CD.
Solution:
50° +x = 180° [Linear pair]
⇒ x =180° – 50° = 130°
y = 130° [Vertically opposite angles]
∴ x = y = 130°
alternate interior angles are equal.
∴ AB || CD
EX 6.2 QUESTION 2.
In Fig. 6.29, if AB CD, CD EF and y : z = 3 : 7, find x.
Solution:
Given: AB || CD, and CD || EF
∴ AB || EF
∴ x = z ….(i) [Alternate interior angles]
Again, AB || CD
⇒ x + y = 180° ….(ii) [Co-interior angles]
⇒ z + y = 180° … (ii) [By (i)]
But y : z = 3 : 7
z = 
⇒ 10z = 7 x 180°
⇒ z = 7 x 180° /10 = 126° ….(iii)
From (i) and (iii),
x = 126°.
EX 6.2 QUESTION 3.
In Fig. 6.30, if AB || CD, EF ⊥ CD and GED = 126°, find AGE, GEF and FGE.
Solution:
Given: AB || CD
∠AGE = ∠GED [Alternate interior angles]
∠AGE = 126°
∠GED = ∠GEF +∠FED
⇒126° = ∠GEF – 90°
⇒∠GEF = 126° – 90° = 36°
given that AB || CD
∴∠FGE + 126° = 180° [sum of Co-interior angles]
⇒∠FGE = 180° – 126° = 54°
EX 6.2 QUESTION 4.
In Fig. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS.
[Hint : Draw a line parallel to ST through point R.]
Solution:
The angles on the same side of the transversal are equal to 180°.
∴ ∠PQR+∠QRX = 180°
∠PQR = 110°
∠QRX = 180°-110°
∴ ∠QRX = 70°
Similarly,
∠RST +∠SRY = 180°
∠RST = 130°
Or, ∠SRY = 180°- 130°
∴ ∠SRY = 50°
Linear pairs on the line XY=
∠QRX+∠QRS+∠SRY = 180°
⇒∠QRS = 180° – 70° – 50° = 60°
∠QRS = 60°
EX 6.2 QUESTION 5.
In Fig. 6.32, if AB || CD, APQ = 50° and PRD = 127°, find x and y.
Solution:
AB || CD and PQ is a transversal.
∴ ∠APQ = ∠PQR [Alternate interior angles]
⇒ 50° = x [ ∵ ∠APQ = 50° (given)]
Again, AB || CD and PR is a transversal.
∴ ∠APR = ∠PRD [Alternate interior angles]
⇒ ∠APR = 127° [ ∵ ∠PRD = 127°]
⇒ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127° [ ∵ ∠APQ = 50°]
⇒ y = 127°- 50° = 77°
Thus, x = 50° and y = 77°.
EX 6.2 QUESTION 6.
In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB CD.
Solution:
Draw ray BE ⊥PQ and CF ⊥ RS
∵ PQ || RS ⇒ BE || CF [∵ BE || PQ and CF || RS]
Now, BL || CM and BC is a transversal.
∴ ∠EBC = ∠FCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection
∠ABE = ∠EBC and ∠FCB = ∠FCD
⇒ ∠ABE = ∠FCD …(2) [By (1)]
Adding (1) and (2), we get
∠EBC + ∠ABL = ∠FCB + ∠MCD
⇒ ∠ABC = ∠BCD
i. e., a pair of alternate interior angles are equal.
∴ AB || CD.
