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Question 30
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If ABCD is a square and BCE is an equilateral triangle, what is the measure of the angle DEC?
Solution
$$\angle\ DCB\ =90^0$$ (Since each angle in a square is 90)
$$\angle\ BCE\ =60^0$$ (Since each angle in an equilateral triangle is 60)
$$\angle\ DCE\ =90^0+ 60^0$$ = $$150^0$$
In $$\triangle\ DCE$$, the sides DC and CE are equal, so the angles opposite to them are equal.
$$\angle\ DCE$$ = $$\angle\ CDE$$ = $$\ \frac{\ 180-150}{2}$$
=$$15^0$$
A is the correct answer.
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