1997 CEMC Pascal Problems/Problem 7

Problem

In the diagram the triangle shown is isosceles with $AB = AC$ .

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The value of $x$ is

$\text{ (A) }\ 40 \qquad\text{ (B) }\ 55 \qquad\text{ (C) }\ 35 \qquad\text{ (D) }\ 50 \qquad\text{ (E) }\ 35$

Since $AB = AC$ , we can see that $m\angle ABC = m\angle ACB$ . We can then let $y$ be the value of these two angles.

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Since $180^{\circ}$ is always the sum of the interior angles in a triangle, we can set up an equation involving $y$ :

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

$y + y + 40 = 180$

$2y + 40 = 180$

$2y = 140$

$y = 70^{\circ}$

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$2x$ and $y$ then form a straight angle, allowing us to set up an equation involving $y$ and $2x$ :

$2x + y = 180$

$2x + 70 = 180$

$2x = 110$

$x = \boxed {\textbf {(B) } 55}$

~anabel.disher

Since $m\angle ACB = 180 - (m\angle ABC + m\angle BAC) = 180 - (y + 40)$ , we can see that $2x = 180 - (180 - (y + 40)) = y + 40$ .

We can use the same process as solution 1 to find that $y = 70^{\circ}$ , so:

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$2x = 70 + 40 = 110$

$x = \boxed {\textbf {(B) } 55}$

~anabel.disher

$35$ is listed twice in the answer choices.