Question
In a $\triangle\text{ABC},$ it is given that $\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1$ and $\text{CD}\perp\text{AC}.$ Find $\angle\text{ECD}.$
✓
Answer
In the given $\triangle\text{ABC},$ we have,$\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1$
Let $\angle\text{A}=3\text{x},\angle\text{B}=2\text{x},\angle\text{C}=\text{x}.$ Then,$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow3\text{x}+2\text{x}+\text{x}=180^\circ$
$\Rightarrow6\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$
$\therefore\angle\text{A}=3\text{x}=3\times30^\circ=90^\circ$
$\angle\text{B}=2\text{x}=2\times30^\circ=60^\circ$
and $\angle\text{C}=\text{x}=30^\circ$ Now, in ABC, we have, Ext. $\angle\text{ACE}=\angle\text{A}+\angle\text{B}=90^\circ+60^\circ=150^\circ$$\therefore\angle\text{ACD}+\angle\text{ECD}=150^\circ$
$\Rightarrow\angle\text{ECD}=150^\circ-\angle\text{ACD}$
$\Rightarrow\angle\text{ECD}=150^\circ-90^\circ$ $\big[\text{since}\ \text{AD}\perp\text{CD},\ \angle\text{ACD}=90^\circ\big]$
$\Rightarrow\angle\text{ECD}=60^\circ$
Let $\angle\text{A}=3\text{x},\angle\text{B}=2\text{x},\angle\text{C}=\text{x}.$ Then,$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow3\text{x}+2\text{x}+\text{x}=180^\circ$
$\Rightarrow6\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$
$\therefore\angle\text{A}=3\text{x}=3\times30^\circ=90^\circ$
$\angle\text{B}=2\text{x}=2\times30^\circ=60^\circ$
and $\angle\text{C}=\text{x}=30^\circ$ Now, in ABC, we have, Ext. $\angle\text{ACE}=\angle\text{A}+\angle\text{B}=90^\circ+60^\circ=150^\circ$$\therefore\angle\text{ACD}+\angle\text{ECD}=150^\circ$
$\Rightarrow\angle\text{ECD}=150^\circ-\angle\text{ACD}$
$\Rightarrow\angle\text{ECD}=150^\circ-90^\circ$ $\big[\text{since}\ \text{AD}\perp\text{CD},\ \angle\text{ACD}=90^\circ\big]$
$\Rightarrow\angle\text{ECD}=60^\circ$
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