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 = 2cm, find $\angle \text{C},$ and length DC.
✓
Answer
We know that the sum of all angles of a triangle is 180°
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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