MCQ
In a $\triangle\text{ABC},$ it is given that $\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1$ and $\angle\text{ACD}=90^\circ.$ If $BC$ produced to $E$ then $\angle\text{ECD}=?$
- ✓$60^\circ$
- B$50^\circ$
- C$40^\circ$
- D$25^\circ$
✓
Answer
Correct option: A.
$60^\circ$
we know that
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ .....(Angle sum property)
$\therefore\angle\text{A}=\Big(180\times\frac{3}{6}\Big)=90^\circ$
$\angle\text{B}=\Big(180\times\frac{2}{6}\Big)=60^\circ\ \text{and}$
$\angle\text{C}=\Big(180\times\frac{1}{6}\Big)=30^\circ$
Now,
$\angle\text{ACE}=\angle\text{A}+\angle\text{B}$ .....(Exterior angle is equal to sum of the remote interior angles)
$=90^\circ+60^\circ$
$=150^\circ$
$\angle\text{ACE}=\angle\text{ECD}+\angle\text{ACD}$
$\therefore150^\circ=\angle\text{ECD}+90^\circ$
$\therefore\angle\text{ECD}=60^\circ$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ .....(Angle sum property)
$\therefore\angle\text{A}=\Big(180\times\frac{3}{6}\Big)=90^\circ$
$\angle\text{B}=\Big(180\times\frac{2}{6}\Big)=60^\circ\ \text{and}$
$\angle\text{C}=\Big(180\times\frac{1}{6}\Big)=30^\circ$
Now,
$\angle\text{ACE}=\angle\text{A}+\angle\text{B}$ .....(Exterior angle is equal to sum of the remote interior angles)
$=90^\circ+60^\circ$
$=150^\circ$
$\angle\text{ACE}=\angle\text{ECD}+\angle\text{ACD}$
$\therefore150^\circ=\angle\text{ECD}+90^\circ$
$\therefore\angle\text{ECD}=60^\circ$
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