| In the given figure, \[AB||CD\]and they cut PQ and QR at E, F and G, H, respectively.\[\angle PEB=70{}^\circ \]and \[\angle QHD=140{}^\circ \]and\[\angle PQR=x.\] Find the value of x. |
 |
A) \[20{}^\circ \]
B) \[30{}^\circ \]
C) \[24{}^\circ \]
D) \[32{}^\circ \]
Correct Answer: B
Solution :
| Since, \[AB||CD\]and PQ is the transversal. |
| \[\therefore \]\[\angle PEF=\angle EGH\][corresponding angles] |
| \[\angle EGH=70{}^\circ \] \[[\because \angle PEF=70{}^\circ ]\] |
| \[\angle EGH+\angle HGQ=180{}^\circ \][linear pair] |
| \[\Rightarrow \] \[\angle HGQ=180{}^\circ -70{}^\circ =110{}^\circ \] |
| \[\angle DHQ+\angle GHQ=180{}^\circ \][linear pair] |
| \[\Rightarrow \]\[\angle GHQ=180{}^\circ -140{}^\circ =40{}^\circ \] |
| In \[\Delta GQH,\]\[\angle GQH+\angle GHQ+\angle HGQ=180{}^\circ \] |
| \[\Rightarrow \]\[x+40{}^\circ +110{}^\circ =180{}^\circ \] |
| \[\Rightarrow \]\[x+150{}^\circ =180{}^\circ \]\[\Rightarrow \]\[x=30{}^\circ \] |
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