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 \[\Delta BCD\] |
\[\angle CBD+\angle BDC+\angle BCD=180{}^\circ \] |
\[\Rightarrow \] \[\theta +\theta +3\theta -180{}^\circ =180{}^\circ \] |
\[\Rightarrow \] \[5\theta =360{}^\circ \]\[\Rightarrow \]\[\theta =72{}^\circ \] |
\[\therefore \] \[\angle BAC=180{}^\circ -2\theta \] |
\[=180{}^\circ -144=36{}^\circ \] |
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