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Find the Missing Angle


Alignments to Content Standards: 8.G.A.5

Task

In the picture below, lines $l$ and $m$ are parallel. The measure of angle $\angle PAX$ is $31^\circ$, and the measure of angle $\angle PBY$ is $54^\circ$. What is the measure of angle $\angle APB$?

Figure_c29d2d9522ecca1b79e131f56e051057

IM Commentary

This task provides us with the opportunity to see how the mathematical ideas embedded in the standards and clusters mature over time. The task "Uses facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure (7.G.5)" except that it requires students to know, in addition, something about parallel lines, which students will not see until 8th grade. As a result, this task is especially good at illustrating the links between related standards across grade levels.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

Students who see the hidden structure of the auxiliary line, and make use of that structure by drawing the line (MP. 7), will be able to solve this problem. They can also use the structure of parallel, perpendicular, angles and triangles from earlier experiences to assist them in efficiently solving this multi-step geometric task.  The teacher could use this as a pre-assessment to learn the extent of the student’s understanding of geometric structures and/or as a post-assessment after multiple opportunities to solve similar problems making sure that some of the problems provide opportunities to use variables. Discussion can then be generated based on these various solutions or the understanding of geometric structures that precipitated these solutions.  Ultimately, the discussion should culminate with identifying efficient ways of solving these types of tasks.

Solution

Extend line AP to intersect line $m$ at point Z.

Since $m$ and $\ell$ are parallel, the alternate interior angles $ZAX$ and $AZB$ have the same measure $31^\circ$.

Angle $PBZ$ (which is the same angle as $PBY$) is given to have measure $54^\circ$.

Since the sum of the angles of a triangle is $180^\circ$ and the sum of $AZB$ and $PBZ$ is $31^\circ + 54^\circ = 85^\circ$, angle $ZPB$ measures $180^\circ - 85^\circ = 95^\circ$.

$ZPB$ and $APB$ together form a straight angle, so their sum is $180^\circ$. So $APB$ measures $180^\circ - 95^\circ = 85^\circ$.

One could also extend $BP$ to meet $\ell$ and proceed similarly.

Ken Mullen says:

over 2 years

How about this for a different kind of solution?

Rotate line ℓ by 31° counterclockwise to go through X. And rotate line m by 31° counterclockwise. The lines are still parallel. The measure of ∠APYʹ is then (54 + 31)°, and that is also the measure of ∠APB because the two angles are alternate interior angles of the parallel lines.

Cam says:

over 2 years

I think I can see the general structure of your argument, but I'm either misunderstanding or you have a few typos. You're rotating line $m$ ... around $B$, right? And rotating $\ell$ so that it goes through P (not X)? And then $Y'$ is the image of $Y$ under this rotation? And you mean $\angle PBY'$?

Regardless, nice idea. My own initial reaction was to draw the line through $P$ parallel to $\ell$, and note that $\angle APB$ was the sum the of two angles formed with that line, each of which were trivial to find by alternate interion angles.

Richard Askey says:

over 3 years

One can also draw a line through P which is perpendicular to either of the two parallel lines, and is then perpendicular to the other parallel line.

Dana says:

about 4 years

removed

Richard Askey says:

almost 5 years

I neglected to mention that the four solutions given each illustrate important ways in which auxiliary lines can be drawn. 1 Extend a line segment. 2 Draw a line through a given point parallel to a given line. 3 Draw a line through a given point perpendicular to a given line. 4 Connect two points. The idea of thinking systematically about how to draw auxiliary lines was mentioned by Durrell and Arnold in a geometry book they wrote almost 100 years ago. Three of these suggestions are in this book as are two others.

Bill says:

almost 5 years

You can read the book Dick mentions above at google books.

Richard Askey says:

almost 5 years

There are other comments which should be made. First, the numerical values of the angles are not needed. They could be "a" and "b", and the same argument shows that the unknown angle is "a" + "b". There are other useful solutions. One is to draw a line through P which is parallel to line l and use alternate interior angles. Another is to draw a line from A to the line m which is perpendicular to m. A fourth is to connect points A and B with a line segment, label the angles in the triangle as x at P, y and z in the other two vertices and use a+y+b+z=180 from the parallel lines cut by a transversal, and x+y+z=180, so x=a+b.

Another question which is related to this question, but it belongs more in high school than in eighth grade, is to state and prove the converse. Instead of assuming that lines l and m are parallel, assume that x=a+b, and show that lines l and m have to be parallel.

Dana says:

about 4 years

Can someone elaborate on the perpendicular line method?

Kristin says:

about 4 years

Basically, when you draw a line $n$ through A that is perpendicular to $m$, it will also be perpendicular to $l$ since $l$ and $m$ are parallel. If $n$ passes through $B$, you form a triangle where two of the angles can be found because they are complementary to $\angle PAX$ and $\angle PBY$, and the third angle in the triangle is $\angle APB$. If $n$ doesn't pass through $B$, you form a quadrilateral where you know or can determine three of the four angles, and the last angle is $\angle PBY$ which you can find because it is the fourth angle in the quadrilateral.

Dana says:

about 4 years

Thank you! This was very helpful. You probably mean that the last angle is ∠APB.

Kristin says:

about 4 years

Yes--thanks for correcting that!

Dana says:

about 4 years

removed