A) \[{{72}^{o}}\]
B) \[{{54}^{o}}\]
C) \[{{36}^{o}}\]
D) \[{{60}^{o}}\]
Correct Answer: C
Solution :
In \[\Delta ABC,\] we have,\[AC=BC\] (Given) \[\Rightarrow \] \[\angle CAB=\angle ABC\] Let \[\angle CAB=\angle ABC=\theta \] ...(i) So, \[\angle ACB={{180}^{o}}-2\theta \] ?.(ii) In \[\Delta ADB,\]we have,\[AD=AB\] (Given) \[\Rightarrow \] \[\angle ADB=\angle ABD\] From (i), \[\angle ADB=\angle ABD=\theta \] So, \[\angle DAB={{180}^{o}}-2\theta \] Since. AD bisects \[\angle BAC\] \[\therefore \] \[{{180}^{o}}-2\theta =\frac{\theta }{2}\,\,\,\Rightarrow ={{72}^{o}}\] So, \[\angle ACB={{180}^{o}}-2({{72}^{o}})={{36}^{o}}\][From (ii)]
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