A) \[45{}^\circ \]
B) \[75{}^\circ \]
C) \[30{}^\circ \]
D) \[85{}^\circ \]
Correct Answer: D
Solution :
\[\angle ABD+\angle BDA+\angle BAD=180{}^\circ [ASP]\] \[30{}^\circ +\angle BDA+\angle 110{}^\circ =180{}^\circ \] \[\angle BDA=180{}^\circ -140{}^\circ =40{}^\circ \] \[\angle BDA+\angle BDC=75{}^\circ \] \[\angle BDA=75{}^\circ -\angle BDA=75{}^\circ -40{}^\circ =35{}^\circ \] In triangle\[DBC.\] \[\angle BDC+\angle DCB+\angle CBD=180{}^\circ \] \[35{}^\circ +60{}^\circ \angle CBD=180{}^\circ \] \[\angle CBD=180{}^\circ -95{}^\circ =85{}^\circ \] \[\angle x=\angle CBD\] {Alternate interior angles} \[x=85{}^\circ \]You need to login to perform this action.
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