A) \[60{}^\circ \]
B) \[50{}^\circ \]
C) \[40{}^\circ \]
D) \[70{}^\circ \]
Correct Answer: C
Solution :
Through E draw GEH\\AB\\CD AB\\GE and BE is the transversal. \[\therefore \] \[\angle ABE+\angle GEB=180{}^\circ \] \[\Rightarrow \] \[120{}^\circ +\angle GEB=180{}^\circ \] \[\Rightarrow \] \[\angle GEB=60{}^\circ \] CD\\EH and CE is the transversal. \[\therefore \] \[\angle DCE+\angle CEH=180{}^\circ \] \[\Rightarrow \] \[100{}^\circ +\angle CEH=180{}^\circ \] \[\Rightarrow \] \[\angle CEH=80{}^\circ \] Now, \[\angle GEB+\angle BEC+\angle CEH=180{}^\circ \] \[\Rightarrow \] \[60{}^\circ +x+80{}^\circ =180{}^\circ \] \[\therefore \] \[x=40{}^\circ \]You need to login to perform this action.
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