2012PP-II-22
Ans: B
In $\Delta ABE$,
… Read In $\Delta ABE$,
$\because AB=AE$,
$\therefore \ang
$\because AB=AE$,
$\therefore \ang
Since the
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\text{LHS}
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\tan \t
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Add points $E$ and $F$ on $AD$ and $CD$ respectively, s
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\cos 40^\circ
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OA & =
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& BE^2 + BC