question_answer

In a\[\Delta ABC,\]\[AB=AC\]and D is a point on AB, such that \[AD=DC=BC.\] Then, \[\angle BAC\] is [SSC (FCI) 2012]

A) \[40{}^\circ \]                          

B) \[45{}^\circ \]

C) \[30{}^\circ \]                          

D) \[36{}^\circ \]

Correct Answer: D

Solution :

Given that, \[AB=AC\]and \[AD=CD=AC\]

Let        \[\angle ABC=\theta \]

Then,    \[\angle ACB=\theta \]                \[[\because AB=AC]\]

\[\Rightarrow \]\[\angle BAC=180{}^\circ -2\theta \]

\[\Rightarrow \]\[\angle ACD=180{}^\circ -2\theta \]                      \[[\because AD=CD]\]

\[\Rightarrow \]\[\angle BCD=\angle ACB-\angle ACD\]

\[\Rightarrow \]\[\angle BCD=\theta -(180{}^\circ -2\theta )=3\theta -180{}^\circ \]

and       \[\angle BDC=\theta \]                \[[\because CD=BC]\]

New, in \[\Delta BCD\]

\[\angle CBD+\angle BDC+\angle BCD=180{}^\circ \]

\[\Rightarrow \]               \[\theta +\theta +3\theta -180{}^\circ =180{}^\circ \]

\[\Rightarrow \]               \[5\theta =360{}^\circ \]\[\Rightarrow \]\[\theta =72{}^\circ \]

\[\therefore \]                              \[\angle BAC=180{}^\circ -2\theta \]

\[=180{}^\circ -144=36{}^\circ \]