In the given figure, BD = DC and =30^ , find . - Vidyadip

Question

In the given figure, $BD = DC$ and $\angle\text{CBD}=30^\circ,$ find $\angle\text{BAC}.$

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Answer

$BD = DC$
$\Rightarrow\ \angle\text{BCD}=\angle\text{CBD}=30^\circ$ In $\triangle\text{BCD},$
we have: $\angle\text{BCD}+\angle\text{CBD}+\angle\text{CDB}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ 30^\circ+30^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\ \angle\text{CDB}=(180^\circ-60^\circ)=120^\circ$
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, $\angle\text{CDB}+\angle\text{BAC}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{BAC}=180^\circ$
$\Rightarrow\ \angle\text{BAC}=(180^\circ-120^\circ)=60^\circ$
$\therefore\ \angle\text{BAC}=60^\circ$

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