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Answer :

Construction : Draw a line parallel to ST through R.

As it is given that,

PQ || ST,\(\angle{PQR}\) = \(110^\circ\) and \(\angle{RST}\) = \(130^\circ\)

We can also say that, AB || PQ || ST

\(\because \) \(\angle{PQR}\) + \(\angle{QRA}\) = \(180^\circ\)

....(interior angles on the same side of
transversal)

\(\therefore \) \(110^\circ\) + \(\angle{QRA}\) = \(180^\circ\)

\(\Rightarrow \) \(\angle{QRA}\) = \(70^\circ\)

\(\because \) \(\angle{ARS}\) = \(130^\circ\) ....(Alternate Interior angle)

As, \(\angle{RST}\) = \(130^\circ\)

Now, so as to find \(\angle{QRS}\), We have,

\(\angle{ARS}\) = \(\angle{ARQ}\) + \(\angle{QRS}\)

\(\Rightarrow \) \(130^\circ\) = \(70^\circ\) + \(\angle{QRS}\)

\(\Rightarrow \) \(\angle{QRS}\) = \(60^\circ\)

- In figure, find the values of x and y and then show that AB ||CD.
- In figure, if AB || CD, CD || EF and y : z = 3:7 , find x.
- In figure, if AB || CD, EF is perpendicular to CD and \(\angle{GED}\) = \(126^\circ\) , find \(\angle{AGE}\),\(\angle{GEF}\) and \(\angle{FGE}\)
- In figure, if AB || CD, \(\angle{APQ}\) = \(50^\circ\) and \(\angle{PRD}\) = \(127^\circ\), find x and y.
- 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.

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