In the figure,if PQ parallel ST,angle PQR = 1 | vedclass

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(D) Given: $PQ \parallel ST$,$\angle PQR = 110^o$,and $\angle RST = 130^o$.Construction: Draw a line $EF$ parallel to $ST$ passing through point $R$.S

Vedclass · Accepted Answer

$60$

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(D) Given: $PQ \parallel ST$,$\angle PQR = 110^o$,and $\angle RST = 130^o$.Construction: Draw a line $EF$ parallel to $ST$ passing through point $R$.Since $PQ \parallel ST$ and $EF \parallel ST$,it follows that $PQ \parallel EF$.Since $PQ \parallel EF$ and $QR$ is a transversal,the sum of interior angles on the same side of the transversal is $180^o$ (or by alternate interior angles if extended). Alternatively,using alternate interior angles: $\angle PQR + \angle QRF = 180^o$ is incorrect here; rather,extend $PQ$ to meet $EF$ or use the property that $\angle PQR + \angle QRF = 180^o$ is not applicable directly. Let's use: $\angle PQR + \angle QRF = 180^o$ (consecutive interior angles) is wrong. Correct approach: $\angle PQR + \angle QRE = 180^o$ (consecutive interior angles) $\Rightarrow 110^o + \angle QRE = 180^o \Rightarrow \angle QRE = 70^o$.Similarly,for $ST \parallel EF$ and transversal $RS$,$\angle RST + \angle SRE = 180^o$ (consecutive interior angles) $\Rightarrow 130^o + \angle SRE = 180^o \Rightarrow \angle SRE = 50^o$.Now,$\angle QRS = \angle QRE + \angle SRE = 70^o + 50^o = 120^o$ is incorrect based on the figure. Let's re-evaluate: $\angle PQR = 110^o$. Draw $EF \parallel PQ$. Then $\angle PQR + \angle QRE = 180^o \Rightarrow \angle QRE = 70^o$. Since $PQ \parallel ST$ and $EF \parallel PQ$,then $EF \parallel ST$. Thus $\angle RST + \angle SRE = 180^o \Rightarrow 130^o + \angle SRE = 180^o \Rightarrow \angle SRE = 50^o$. From the figure,$\angle QRS = 360^o - (\angle QRE + \angle SRE)$ is not correct. Actually,$\angle QRS = 180^o - (70^o + 50^o) = 60^o$.

In the figure,if $PQ \parallel ST$,$\angle PQR = 110^o$ and $\angle RST = 130^o$,find $\angle QRS$. (in $^o$)

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