Question
In $\triangle \text{ABC}, \angle \text{ABC}=100^\circ, \angle \text{BAC}=35^\circ$ and $\text{BD}\bot \text{AC}$ meets side $AC$ in $D$. If $BD = 2\ cm$, find $\angle \text{C},$ and length $DC.$
✓
Answer
We know that the sum of all angles of a triangle is $180^\circ $
Therefore, for the given $\triangle \text{ABC},$
we can say that: $\angle \text{ABC}+\angle \text{BAC}+\angle \text{ACB}=180^\circ$
$100^\circ+35^\circ+\angle \text{ACB}=180^\circ$
$\angle \text{ACB}=180^\circ-135^\circ$
$\angle \text{ACB}=45^\circ$
$\angle \text{C}=45^\circ$ If we apply the above rule on $\triangle \text{BCD},$
we can say that: $\angle \text{BCD}+\angle \text{BDC}+\angle \text{CBD}=180^\circ$
$45^\circ+90^\circ+\angle \text{CBD}=180^\circ$
$(\angle \text{ACD}=\angle \text{BCD}$ and $BD$ parallel to AC$)$
$\angle \text{CBD}=180^\circ-135^\circ$
$\angle \text{CBD}=45^\circ$
We know that the sides opposite to equal angles have equal length. Thus, $BD = DC DC = 2cm$
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