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Triangles: Equilateral

If $$\triangle ABC$$ is an equilateral triangle, what is the value of $$x$$?
Incorrect. [[snippet]] Carefully check your calculations.
Correct. [[snippet]] Since $$AB = AC = BC$$, >$$\angle ABC = \angle BCA = \angle CAB = \frac{180^\circ}{3} = 60^\circ$$ Based on this, >$$\angle ACD = 180^\circ - 60^\circ - 40^\circ = 80^\circ$$. Given that $$\angle CAB = 60^\circ$$, >$$\angle CAD = 90^\circ - 60^\circ = 30^\circ$$. So, >$$x = 180^\circ - 80^\circ - 30^\circ = 70^\circ$$.
Incorrect. [[snippet]] Carefully check your work.
Incorrect. [[snippet]] You might have gotten this answer if you assumed that $$\angle ACD$$ is a right angle.
Incorrect. [[snippet]] You may have gotten this answer if you tried to use the properties of parallel lines. Lines $$AD$$ and $$CE$$ are _not_ parallel.
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