2003 CEMC Pascal Problems/Problem 4

Problem

In the diagram, the value of x is

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$\text{ (A) }\ 40 \qquad\text{ (B) }\ 60 \qquad\text{ (C) }\ 100 \qquad\text{ (D) }\ 120 \qquad\text{ (E) }\ 80$

Solution

We can name the points to make everything easier to follow.

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we can see that angles $\angle DBC$ and $\angle ABC$ form a straight angle, meaning we can set up an equation:

$m\angle DBC + m\angle ABC = 180$

$120 + m\angle ABC = 180$

$m\angle ABC = 60^{\circ}$

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We can now notice that angles $\angle ACB$ and $\angle ECF$ are vertical angles. This means that:

$m\angle ACB = m\angle ECF = 40^{\circ}$

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We now have all of the angles of triangle $ABC$ . The interior angles of a triangle always sum up to $180^{\circ}$ , allowing us to set up an equation:

$x + m\angle ACB + m\angle ABC = 180$

$x + 60 + 40 = 180$

$x + 100 = 180$

$x = 180 - 100 = \boxed {\textbf {(E) } 80}$

~anabel.disher

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2003 CEMC Pascal Problems/Problem 4

Problem

Solution