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2009 CEMC Pascal Problems/Problem 13

Problem

In the diagram, $QRS$ is a straight line. What is the measure of $\angle RPS$?

$\text{ (A) }\ 27^{\circ} \qquad\text{ (B) }\ 47^{\circ} \qquad\text{ (C) }\ 48^{\circ} \qquad\text{ (D) }\ 65^{\circ} \qquad\text{ (E) }\ 67^{\circ}$

Solution 1

Let $x$ be $m\angle PRQ$, $y$ be $m\angle PRS$, and $z$ be $m\angle RPS$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

We can first use the fact that the sum of the interior angles of a triangle is always $180^{\circ}$. Using $\triangle PQR$, this means that we have:

$x + 67^{\circ} + 48^{\circ} = 180^{\circ}$

$x + 115^{\circ} = 180^{\circ}$

$x = 180^{\circ} - 115^{\circ} = 65^{\circ}$

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Since $QRS$ is a straight line, we know that $65^{\circ}$ and $y$ must sum to be $180^{\circ}$. We then have:

$y + 65^{\circ} = 180^{\circ}$

$y = 115^{\circ}$

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We can also now use $\triangle RPS$ to find $z$:

$z + 38^{\circ} + 115^{\circ} = 180^{\circ}$

$z + 153^{\circ} = 180^{\circ}$

$z = \boxed {\textbf {(A) } 27^{\circ}}$

~anabel.disher

Solution 1.5

Let $x$ be $m\angle PRS$ and $y$ be $m\angle RPS$.

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We can see that $m\angle PRQ = 180 - [m\angle QPR + m\angle PQR]$ and $x = 180 - m\angle PRQ$, so $x = 180 - (180 - [m\angle QPR + m\angle PQR])$.

This simplifies to $x = m\angle QPR + m\angle PQR = 67^{\circ} + 48^{\circ} = 115^{\circ}$.

We can then use the same process as solution 1 to get $y = \boxed {\textbf {(A) } 27^{\circ}}$.

~anabel.disher

Solution 2

Let $x$ be $m\angle RPS$ and $y$ be $m\angle SPQ$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Using triangle $PQS$, we can see that $y + 48^{\circ} + 38^{\circ} = 180^{\circ}$.

This gives:

$y + 86^{\circ} = 180^{\circ}$

$y = 94^{\circ}$

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

We can also see that $m\angle QPR + x = y$, giving:

$x + 67^{\circ} = 94^{\circ}$

$x = \boxed {\textbf {(A) } 27^{\circ}}$

~anabel.disher

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