A) \[{{40}^{o}}\]
B) \[{{110}^{o}}\]
C) \[{{45}^{o}}\]
D) \[{{65}^{o}}\]
Correct Answer: C
Solution :
From the figure, we have \[\angle ADB=\angle ABD.\](Since AB = AD.) \[\therefore \]\[\angle ADB={{180}^{o}}-{{110}^{o}}={{70}^{o}}\] But \[\angle ADB=\angle DAC+\angle ACD\] (Exterior angle of \[\Delta \Alpha DC.\]) \[\Rightarrow \]\[{{70}^{o}}={{x}^{o}}+{{25}^{o}}\] \[\Rightarrow \]\[{{x}^{o}}={{70}^{o}}-{{25}^{o}}={{45}^{o}}\]
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