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If ABCD is a square and BCE is an equilateral triangle, what is the measure of $\angle DEC$?

- $15^{\circ}$
- $30^{\circ}$
- $20^{\circ}$
- $45^{\circ}$

## 1 Answer

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Given that, $ABCD$ is a square and $BCE$ is an equilateral triangle.

Let the side of a square be $x$ cm.

$\angle BCD=90^{\circ}, \angle BCE=60^{\circ}$

Then $\angle DCE=90^{\circ}+60^{\circ}=150^{\circ}$

Let $\angle CDE=\angle DEC=a^{\circ}$ ($\because$ Angle opposite to the equal sides are equal)

$\Rightarrow a+a+150^{\circ}=180^{\circ}$

$\Rightarrow 2a=30^{\circ}$

$\Rightarrow \boxed{a=15^{\circ}}$

$\therefore$ The $\angle DEC=15^{\circ}$

Correct Answer: $\text{A}$

Let the side of a square be $x$ cm.

$\angle BCD=90^{\circ}, \angle BCE=60^{\circ}$

Then $\angle DCE=90^{\circ}+60^{\circ}=150^{\circ}$

Let $\angle CDE=\angle DEC=a^{\circ}$ ($\because$ Angle opposite to the equal sides are equal)

$\Rightarrow a+a+150^{\circ}=180^{\circ}$

$\Rightarrow 2a=30^{\circ}$

$\Rightarrow \boxed{a=15^{\circ}}$

$\therefore$ The $\angle DEC=15^{\circ}$

Correct Answer: $\text{A}$