Question

Q- In the Figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.



ɴᴏ sᴘᴀᴍs ᴘʟᴢ!​

Answer 1

$\large \bf \orange {➤ \: \: Given:-}$

In the figure, PQ || ST, <PQR = 110° and <RST = 130°.

$\large \bf \orange {➤ \: \: To \: find:-}$

We have to find <QRS.

$\large \bf \orange {➤ \: \: Construction:-}$

Draw XY || ST and PQ through R.

$\large \bf \orange {➤ \: \: Solution:-}$

Since ST || RV,

<TSR + <SRY = 180° [Interior angles]

130° + <SRY = 180°

<SRY = 50°

Since PQ || RU,

<PQR + <QRX = 180° [Interior angles]

110° + <QRX = 180°

<QRX = 70°

Since XY is a straight line,

<QRX + <QRS + <SRY = 180° [Linear pair]

70° + <QRS + 50° = 180°

120° + <QRS = 180°

<QRS = 60°

➤ Therefore, <QRS = 60°.

_____________________

All done :)

Answer 2

$\huge{\rm{Required \ Answer :)}}$

$\orange{\bold{\underline{\underline{Answer:}}}}$

$\pink{\tt{\therefore{PQR ~is ~60°}}}$

$\blue{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\purple{\underline \bold{Given :}} \\ \tt: \implies {PQ||} {ST}$

$\purple{\underline \bold{ :}} \\ \tt: \implies {PQR=} {110°}$

$\purple{\underline \bold{ :}} \\ \tt: \implies {RST=} {130°}$

$\red{\underline \bold{To \: Find:}} \\ \tt: \implies QRS~=~?$

$\bold{As \: we \: know \: that}$

$\bf\underline{The~angles~on~the~same~side~of~transversal~is~equal~to~180°}$

•$\bf\blue{According~the~question :}$

　　　　PQR + QRX = 180°

　　　　　QRX = 180° - 110°

　　　　　　QRX = 70°

•$\bf\green{In~the~same~way :}$

　　　　RST + SRY = 180°

　　　　SRY = 180° - 130°

　　　　　SRY = 50°

•$\bf\red{The~linear~pairs~on~the~line~XY -}$

　　　QRX + QRS + SRY = 180°

•$\bf\green{Substituting~their~values -}$

　　　　QRS = 180° - 70° - 50°

　　　　　QRS = 60°

$\boxed {\sf {\purple { Therefore~,~QRS ~is~60°.}}}$

$\bf{\red{▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬}}$

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