Lines and Angles : Exercise 6.2 (Mathematics NCERT Class 9th)

Q.1     In figure, find the values of x and y and then show that AB || CD.

Sol.

Since AB || CD and transversal PQ intersects them at R and S respectively.
Since $\angle ARS = \angle RSD$ [Alternate angles]

$\Rightarrow$ $x = y$
But $\angle RSD = \angle CSQ$ [Vertically opp. angles]
$\Rightarrow$ $y = 130^o$ [Since $\angle CSQ = 130^o$]
Hence , x = y = 130º

Q.2      In figure , if AB || CD, CD|| EF and y : z = 3 : 7 , find x.

Sol.

Since CD|| EF and transversal PQ intersects them at S and T respectively.
Since $\angle CST = \angle STF$ [Alternate Angles ]
$\Rightarrow$ $180^o - y = z$ [Since $\angle y + \angle CST = 180^o$ being linear pair]
$\Rightarrow$ $y + z = 180^o$
Given y : z = 3 : 7
So, the sum of ratios = 3 + 7 = 10

Therefore $y = {3 \over {10}} \times 180^o$
$= 3 \times 18^o = 54^o$
and $z = {7 \over {10}} \times 180^o = 7 \times 18 ^o= 126^o$
Since AB || CD and transversal PQ intersects them at R and S respectively.
Since $\angle ARS + \angle RSC = 180^o$ [Consecutive interior angles are supplementary]
$\Rightarrow$ $x + y = 180^o$
$\Rightarrow$ $x = 180^o - y$
$= 180^o - 54^o = 126^o$ [Since y = 54º]
Hence, x = 126º

Q.3     In figure , if AB || CD, EF $\bot$ CD and $\angle GED$ = 126º, find $\angle AGE,\,\angle GEF\,and\,\angle FGE.$

Sol.

Since AB || CD and transversal GE cuts them at G and E respectively.
Since $\angle AGE\, = \angle GED$ [Alternate angles]
$\Rightarrow$ $\angle AGE\, = 126^o$ [Since $\angle GED\, = 126^o$ (given)]
$\angle GEF\, = \angle GED\, - \,\angle FED = 126^o - 90^o = 36^o$
and , $\angle FGE\, = \angle GEC$ [Alternate angles]
$\Rightarrow$ $\angle FGE = 90^o - \angle GEF$
$= 90^o - 36^o = 54^o$
Hence, $\angle AGE = 126^o,\,\angle GEF = 36^o\,\,and\,\,\angle FGE = 54^o$

Q.4     In figure , if PQ || ST, $\angle PQR = 110^o$ and $\angle RST = 130^o$ , find $\angle QRS$.

Sol.         Produce PQ to intersect SR at point M.

Now, PM || ST and transversal SM intersects them at M and R respectively.
Since $\angle SMQ = \angle TSM$  [Alternate angles]
$\Rightarrow$ $\angle SMQ =$ 130º  [Since $\angle TSM=$ 130º(given)]
$\Rightarrow$ $\angle QMR =$ 180º – 130º = 50º [Since $\angle SMQ + \angle QMR =$ 180º( linear pairs)]
Since ray RQ stands at Q on PM.
Since $\angle PQR + \angle RQM =$ 180º  (Linear pair)
$\Rightarrow$ 110º + $\angle RQM$ = 180º
$\Rightarrow$ $\angle RQM =$ 70º

Since $\angle QRS$ = 180º – (70º + 50º) = 60º  [Since sum of the angles of a triangle is 180º]

Q.5      In figure , if AB || CD, $\angle APQ =$ 50º and $\angle PRD =$ 127º , find x and y.

Sol.

Therefore AB || CD and transversal PQ intersects them at P and Q respectively.
Therefore $\angle PQR = \angle APQ$ [Alternate angles]
$\Rightarrow$ x = 50º [Since $\angle APQ =$ 50º (given)]
Since AB || CD and transversal PR intersects them at P and R respectively.
Therefore $\angle APR = \angle PRD$ [Alternate angles]
$\Rightarrow$ $\angle APQ + \angle QPR$ = 127º [Since $\angle PRD =$ 127º]
$\Rightarrow$ 50º + y = 127º [Since $\angle APQ =$ 50º]
$\Rightarrow$ y = 127º – 50º = 77º
Hence , x = 50º and y = 77º

Q.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.

Two plane mirrors PQ and RS are placed parallel to each other i.e., PQ || RS. An incident ray AB after reflections takes the path BC and CD.

BN and CM are the normals to the plane mirrors PQ and RS respectively.
Since BN $\bot$ PQ, CM $\bot$ RS and PQ || RS
Therefore BN $\bot$ RS
$\Rightarrow$ BN || CM

Thus BN and CM are two parallel lines and a transversal BC cuts them at B and C respectively.
Therefore $\angle 2 = \angle 3$ [Alternative interior angles]
But , $\angle 1 = \angle 2\,\,and\,\,\angle 3 = \angle 4$ [By laws of reflection]
Therefore, $\angle 1 + \angle 2\,\, = \angle 2 + \angle 2\,\,and\,\,\angle 3 + \angle 4 = \angle 3 + \angle 3$
$\Rightarrow$ $\angle 1 + \angle 2\,\, = 2\left( {\angle 2} \right)\,and\,\,\angle 3 + \angle 4 = 2\left( {\angle 3} \right)$
$\Rightarrow$ $\angle 1 + \angle 2\,\, = \,\,\angle 3 + \angle 4$ [Since $\angle 2\, = \,\angle 3 \Rightarrow 2\left( {\angle 2} \right) = 2\left( {\angle 3} \right)$]
$\Rightarrow$ $\angle ABC = \angle BCD$
Thus, lines AB and CD are intersected by transversal BC such that -
$\angle ABC = \angle BCD$
i.e., alternate interior angles are equal.
Therefore, AB || CD.

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