NCERT Class 9 Solutions: Line and Angles (Chapter 6) Exercise 6.2 – Part 3

Alternate interior angles x and y

Alternate Interior Angles

Alternate interior angles x and y

Q-5 In the figure, if ABCD,APQ=50° and PRD=127°, find x and y.

Give the AB CDalso angle of x and y, ∠APQ = 50° and ∠PRD = 127°

Give the AB Parallell CD Also Angle of X and Y

Give the AB CDalso angle of x and y, ∠APQ = 50° and ∠PRD = 127°

Solution:

Given,

ABCD,APQ=50° and PRD=127°

Now,

  • x=50° (Alternate interior angles.)

  • PRD+RPB=180° (Angles on the same side of transversal.)

  • 127°+RPB=180°

  • RPB=53°

Thus,

  • y+50°+RPB=180° (AB is a straight line.)

  • y+50°+53°=180°

  • y+103°=180°

  • y=77°

Q-6 In the 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. Prover that ADCD .

Give line of PQ and RS it is a parallel the mirror PQ at B,the reflected ray moves along the path BC and strikes the mirror RSat C and again reflects back along CD

Give Line of PQ and RS It Is a Parallel

Give line of PQ and RS it is a parallel the mirror PQ at B,the reflected ray moves along the path BC and strikes the mirror RSat C and again reflects back along CD

Solution:

Construction: Draw BLPQandCMRS

Give two line PQ and RS at a parallel AND ray BL⊥PQ and CM⊥RS

Give Two Line PQ and RS at a Parallel

Give two line PQ and RS at a parallel AND ray BL⊥PQ and CM⊥RS

Since, BLPQandCMRS and PQRS BLCM . Thus,

  • LBC=MCB ………..equation(1) (Alternate interiror Angle)

  • ABL=LBC ………….equation(2) (Angle of incidence=Angle of reflection

  • MCB=MCD …………..equation (3) (Angle of incidence=Angle of reflection)

From (1),(2) and (3),we get ABL=MCD ………..equation(4)

Adding (1)and(4),we get

  • LBC+ABL=MCB+MCD

  • ABC=BCD (ABC=LBC+ABLandBCD=MCB+MCD)

But these are alternate interior angle and they are equal.

Therefore, ADCD .

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