A) \[{{14}^{o}}\]
B) \[{{16}^{o}}\]
C) \[{{12}^{o}}\]
D) \[{{32}^{o}}\]
Correct Answer: A
Solution :
Since ABC is an equilateral triangle. \[\therefore \] \[\angle CAB=\angle ABC=\angle BCA={{60}^{o}}\] And \[\angle DBA=\angle DAB=({{60}^{o}}-x)\] \[[\because \,\,DA=DB]\] In \[\Delta \,DAB,\] \[\angle DAB\angle DAB+\angle ADB={{180}^{o}}\] \[\Rightarrow \] \[2({{60}^{o}}-x)+{{88}^{o}}={{180}^{o}}\] \[\Rightarrow \] \[2({{60}^{o}}-x)={{92}^{o}}\Rightarrow {{60}^{o}}-x={{46}^{o}}\Rightarrow x={{14}^{o}}\]
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