A) \[{{180}^{o}}\]
B) \[{{70}^{o}}\]
C) \[{{190}^{o}}\]
D) \[{{80}^{o}}\]
Correct Answer: C
Solution :
From the figure, \[150-x+70-x+x={{180}^{o}}\] \[\Rightarrow \] \[x={{220}^{o}}-{{180}^{o}}={{40}^{o}}\] Since \[AE||BD,\] \[AE||BD,\] \[y=x\] as they are alternate angles. In \[\Delta BCD,\] \[\angle BDC=x\] (Alternate angles) \[70-x+x+z={{180}^{o}}\] \[\Rightarrow \] \[z={{110}^{o}}\] \[\therefore \]The required sum \[=x+y+z={{40}^{o}}+{{40}^{o}}+{{110}^{o}}={{190}^{o}}\]You need to login to perform this action.
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