2015 CEMC Gauss (Grade 8) Problems/Problem 11

Problem

In the diagram, the value of $x$ is

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$\textbf{(A)}\ 40 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 70 \qquad\textbf{(E)}\ 80$

We can name the points on the triangle and the diagram.

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Using this diagram, we can see that $\angle ACB = \angle ECF = x$ because the angles are vertical angles.

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Next, we can use the fact that the sum of the angles in a triangle is $180^{\circ}$ degrees. This means that we have:

$\angle BAC + \angle ACB + \angle CBA = 180$

$\angle BAC + x + 80 = 180$

$\angle BAC = 180 - 80 - x = 100 - x$

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We now see that $\angle BAC$ and $\angle DAC$ are supplementary because they are in a straight line. Thus, we have:

$\angle BAC + \angle DAC = 180$

$100 - x + 140 = 180$

$240 - x = 180$

$x = \boxed {\textbf {(C) } 60}$

~anabel.disher