MCQ
In a $\triangle A B C$, it is given that $\angle A: \angle B: C=3: 2: 1$ and $\angle A C D=90^{\circ}$. If $B C$ is produced to $E$, then $\angle E C D=$
- A$60^{\circ}$
- B$30^{\circ}$
- C$50^{\circ}$
- D$40^{\circ}$
✓
Answer
A. $60^{\circ}$
We have, $\angle A: \angle B: \angle C=3: 2: 1$
So, let $\angle A=3 x, \angle B=2 x$ and $\angle C=x$.
$\therefore$ $\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow$ $3 x+2 x+x=180^{\circ}$ $\Rightarrow$ $6 x=180^{\circ}$ $\Rightarrow$ $x=30^{\circ}$
$\therefore$ $\angle A=90^{\circ}, \angle B=60^{\circ}$ and $\angle C=30^{\circ}$
Using exterior angle property in $\triangle A C E$, we obtain
$\angle A C E=\angle A+\angle B$ $\Rightarrow$ $\angle A C D+\angle E C D=90^{\circ}+60^{\circ}$$\Rightarrow$$90^{\circ}+\angle E C D=150^{\circ}$$\Rightarrow$$\angle E C D=60^{\circ}$
We have, $\angle A: \angle B: \angle C=3: 2: 1$
So, let $\angle A=3 x, \angle B=2 x$ and $\angle C=x$.
$\therefore$ $\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow$ $3 x+2 x+x=180^{\circ}$ $\Rightarrow$ $6 x=180^{\circ}$ $\Rightarrow$ $x=30^{\circ}$
$\therefore$ $\angle A=90^{\circ}, \angle B=60^{\circ}$ and $\angle C=30^{\circ}$
Using exterior angle property in $\triangle A C E$, we obtain
$\angle A C E=\angle A+\angle B$ $\Rightarrow$ $\angle A C D+\angle E C D=90^{\circ}+60^{\circ}$$\Rightarrow$$90^{\circ}+\angle E C D=150^{\circ}$$\Rightarrow$$\angle E C D=60^{\circ}$
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