Math class 9 chapter 6 exercise 6.2 question 5 and 6 solutions

Chapter 6, Exercise 6.2 

Question 5. 

In figure, if AB

CD, ∠ APQ = 50° and ∠ PRD = 127°,

find x and y.

Sol. AB

CD and PQ is transversal.

∴ ∠PQR = ∠APQ ⇒ x = 50°

Again AB

CD and PR is transversal.

∴ ∠APR = ∠PRD [Alternate angles]

⇒ ∠APQ + ∠QPR = ∠PRD.

⇒ 50° + y = 127° ⇒ y = 127° – 50° = 77°.

Question 6. 

In Figure,, 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.

Sol. Construction: Draw BE perpendicular to PQ and CF Perpendicular to RS.

Proof: As BE ⊥ PQ and CF ⊥ RS and PQ

RS.

⇒ BE

CF ...(i)

Also, we know

Angle of incidence

 = angle of reflection.

i.e., ∠ABE = ∠EBC = x. ...(ii)

and ∠BCF = ∠FCD = y. ...(iii)

From (i), BE || CF and BC is transversal.

∴ ∠EBC = ∠BCF [Alternate angles]

⇒ x = y ⇒ 2x = 2y

⇒ ∠ABC = ∠BCD [From (ii) and (iii)]

But these are alternate angles. Hence, AB
||
CD.