Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2009 CEMC Pascal Problems/Problem 13

Revision as of 12:08, 7 September 2025 by Anabel.disher (talk | contribs) (Created page with "==Problem== In the diagram, <math>QRS</math> is a straight line. What is the measure of <math>\angle RPS</math>? <math> \text{ (A) }\ 27^{\circ} \qquad\text{ (B) }\ 47^{\circ...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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:

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

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

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

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

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

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

$y = 115^{\circ}$

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

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 $m\angle RPS$.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_CEMC_Pascal_Problems/Problem_13&oldid=259898"