MCQ
In a $\angle\text{ABC},$ it is given that $\angle\text{A}:\angle\text{B}:\angle\text{C} = 3 : 2 : 1$ and $\angle\text{ACD}=90^\circ.$ If $BC$ is produced to $E$ then $\angle\text{ECD} = ?$
- ✓$60^\circ$
- B$50^\circ$
- C$25^\circ$
- D$40^\circ$
✓
Answer
Correct option: A.
$60^\circ$
Let $ \angle\text{A}=(3\text{x})^\circ,\ \angle\text{B}=(2\text{x})^\circ$ and $\angle\text{C}=\text{x}^\circ$
Then,
$3x + 2x + x = 180^\circ $ [Sum of the angles of a triangle]
$\Rightarrow 6x = 180^\circ $
$\Rightarrow x = 30^\circ $
Hence, the angles are
$\angle\text{A}=3\times 30^\circ=90^\circ,\ \angle\text{B}=2\times 30^\circ=60^\circ$ and $\angle\text{C}=30^\circ$
Side $BC$ of triangle $ABC$ is produced to $E.$
$\therefore\ \angle\text{ACE}=\angle\text{A}+\angle\text{B}$
$\Rightarrow \angle\text{ACD}+\angle\text{ECD}=90^\circ+60^\circ$
$\Rightarrow 90^\circ+\angle\text{ECD}=150^\circ$
$\Rightarrow \angle\text{ECD}=60^\circ$
Then,
$3x + 2x + x = 180^\circ $ [Sum of the angles of a triangle]
$\Rightarrow 6x = 180^\circ $
$\Rightarrow x = 30^\circ $
Hence, the angles are
$\angle\text{A}=3\times 30^\circ=90^\circ,\ \angle\text{B}=2\times 30^\circ=60^\circ$ and $\angle\text{C}=30^\circ$
Side $BC$ of triangle $ABC$ is produced to $E.$
$\therefore\ \angle\text{ACE}=\angle\text{A}+\angle\text{B}$
$\Rightarrow \angle\text{ACD}+\angle\text{ECD}=90^\circ+60^\circ$
$\Rightarrow 90^\circ+\angle\text{ECD}=150^\circ$
$\Rightarrow \angle\text{ECD}=60^\circ$
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