Find x.
This is yet another problem where all of the givens appear in an unlabeled diagram. Since this problem involves a circle, we'll label the center O. Five points on the circle are significant, so we can label these in alphabetical order as A, B, C, D, E. Let chords
Clearly this problem involves the circle theorems of Chapter 15 of the U of Chicago text, especially the measures of central (Lesson 15-1) and inscribed (Lesson 15-3) angles. Let's use them both:
Central Angles: Since AOB = 20 is a central angle, Arc AB = 20.
Inscribed Angles: Since CED = 19 is an inscribed angle, Arc CD = 38.
Finally, we must use the Angle-Chord Theorem of Lesson 15-5:
Angle-Chord Theorem:
The measure of an angle formed by two intersecting chords is one-half the sum of the measures of the arcs intercepted by it and its central angle.
Using this theorem, we obtain:
Angle CPD = (Arc AB + Arc CD)/2
x = (20 + 38)/2
x = 58/2
x = 29
Therefore the desired angle is 29 degrees -- and of course, today's date is the 29th. By the way, the Angle-Chord Theorem rarely appears on state tests such as the SBAC. The last major theorem needed for SBAC is the Inscribed Angle Theorem.
This is what I wrote last year about today's lesson:
Lesson 1-2 of the U of Chicago text is called "Locations as Points." (It appears as Lesson 1-1 in the modern edition of the text.) The main focus of the lesson is graphing points on a number line. Indeed, we have another description of a point:
Second Description of a Point:
A point is an exact location.
Yesterday I made a big deal about the first description of a point -- the dot -- since many of our students are interested in pixel-based technology. Locations as points aren't as exciting -- but still, the second description is something we think about every time we find a distance. The definition of distance is highlighted in the text:
Definition:
The distance between two points on a coordinatized line is the absolute value of the difference of their coordinates.
Other than this, the lesson is straightforward. Students learn about zero- through three-dimensional figures, but of course the emphasis is on one dimension. One of the two "exploration questions," which I included as a bonus, is:
- Physicists sometimes speak of space-time. How many dimensions does space-time have?
The answer, of course, is four -- even though there might be as many as ten dimensions in string theory. We ordinarily only include Einstein's four dimensions and don't consider the extra six dimensions of string theory as part of "space-time."
Here's the other bonus question:
- To the nearest 100 miles, how far do you live from each of the following cities?
b. Los Angeles
c. Honolulu
d. Moscow
Well, part b is easy -- I worked in L.A. last year and my daily commute obviously wasn't anywhere near 100 miles, so my distance to L.A. is 0 miles to the nearest 100 miles. The U of Chicago text gives the distance from L.A. to New York as 2451 miles as the crow flies, but 2786 miles by car. I choose to give the air distance in part a, in order to be consistent with parts c and d (for which only air distance is available). We round it up to 2500 miles. My answers are:
a. 2500 miles
b. 0 miles
c. 2600 miles
d. 6100 miles
Hmmm, that's interesting -- I'm only slightly closer to New York than to Honolulu.
Here is the Blaugust prompt for today:
A peek into my classroom - show us your classroom or describe a typical day / hour
Well, yesterday I wrote that I don't have any videos of my teaching. And only once did I take any photos of my classroom (not counting photos submitted to Illinois State). For the sake of this Blaugust post, I'll post those pictures again today.
As for a typical hour, I did write about what I originally wanted a typical 80-minute block to look like, about three weeks before the first day of school:
10 minutes: Warm-Up
10 minutes: Go over homework/previous day's lesson
20 minutes: New lesson (Foldable note taking)
10 minutes: Music break
20 minutes: Guided practice
10 minutes: Closure/Exit Pass
This was set up for a traditional lesson. Of course, soon I learned more about the Illinois State text and its nontraditional lessons. Many parts of this 80-minute plan changed -- but I always kept some form of a Warm-Up, music break, and Exit Pass.
Here are the changes to this plan caused by Illinois State. First of all, the Warm-Up turned into the Illinois State Daily Assessment, which is supposed to take only five minutes, not ten. Going over HW and the previous day's lesson were awkward since there was supposed to be only one traditional lesson per week, and the HW was to be done online. Most of the time, I had the students take notes directly into the Student Journals, which was also where the guided practice was. Thus in the end, the typical 80-minute block became:
5 minutes: Warm-Up (Illinois State Daily Assessment)
10 minutes: Review previous week's lesson (from Illinois State)
20 minutes: New lesson (Illinois State Student Journals)
10 minutes: Music break
25 minutes: Guided practice (Illinois State Student Journals)
10 minutes: Closure/Exit Pass
If I remember correctly, the Illinois State pacing guide assigned one hour to the traditional lesson, and notice that the Illinois State parts of this lesson do add up to one hour. The only non-Illinois State parts of this lesson plan are the music break and Exit Pass.
The traditional lesson, as I wrote earlier, would be one day per week. As I realized much too late, the ideal weekly plan would have been something like this:
Monday: Coding (with coding teacher)
Tuesday: Traditional Lesson
Wednesday: Learning Centers
Thursday: Science
Friday: Weekly Assessment
Once again, there's only repeat posters for Blaugust today. We begin with Benjamin Leis:
http://mymathclub.blogspot.com/
http://mymathclub.blogspot.com/2019/08/15-75-90-alternate-forms.html
It's been a while since I've talked about one of my favorite triangles the 15-75-90. So here's a short post on a new detail that I realized about them the other day.
We think about special right triangles in Lesson 14-1 of the U of Chicago text. In that lesson, we know of two special right triangles -- 45-45-90 and 30-60-90. But as the Exploration question in that lesson implies:
There are many triangles that could be considered special.
And clearly Leis considers the 15-75-90 right triangle to be special. He writes the following problem:
ABCD is a square. Triangle ABF is an equilateral triangle (with F inside the square).
If CD = sqrt(6) + sqrt(2) then EC = ?
This question almost looks like a Pappas problem -- and indeed since the final answer turns out to be 16, this could appear on the sixteenth day of the month. But Pappas never includes 15-75-90 triangles on her calendar -- and Triangle CDE turns out to be 15-75-90. (Hint: Triangle BCF is isosceles.) On the other hand, she includes 45-45-90 and 30-60-90 triangles all the time.
We can think back to the previous Leis post. OK, so maybe we shouldn't really teach the Cubic Formula in Algebra II classes. But can we teach 15-75-90 triangles in Geometry?
Here's an interesting mini-activity to consider when teaching Chapter 14. Just after introducing the trig functions, we ask students to select three acute angles -- say 6, 37, and 86 degrees. Then the teacher writes the following:
sin(6) = ?
cos(37) = ?
tan(86) = ?
Let's enter these on a calculator:
sin(6) = .1045284633
cos(37) = .79863551
tan(86) = 14.30066626
The teacher writes these values on the board, and lets the class "study" them a little. Then without warning, the teacher erases these values and announces a pop quiz. The students must now fill in the missing values:
sin(6) = ?
cos(37) = ?
tan(86) = ?
accurate to ten digits without a calculator. Of course, the students will protest -- it's impossible to have learned all of those digits that fast! Then the teacher "cancels" the pop quiz, and states that the students should have chosen different angles -- 30, 60, and 45 instead of 6, 37, and 86:
sin(30) = .5
cos(60) = .5
tan(45) = 1
Oh, so for some reason, this pop quiz would have been a lot easier with 30, 60, and 45 instead of 6, 37, and 86 degrees. Then this leads to a class discussion -- why is tan(86) such a complicated value but tan(45) so simple? And of course, this leads to special right triangles.
On a Casio calculator that I found in my old classroom, the square root symbol is displayed. This leads to even more simple values for functions:
sin(45) = sqrt(2)/2
cos(30) = sqrt(3)/2
tan(60) = sqrt(3)
The teacher then states that any value that displays an exact value on the screen -- whole number, fraction, or radical -- is fair game for memorization on a pop quiz. Thus it's fair to ask for tan(45) or tan(60), but not tan(86). But then we might stumble upon something:
tan(15) = 2 - sqrt(3)
And students now might wonder, why does tan(15) gives an exact value? Voila -- we just motivated the teaching of 15-75-90 triangles in a Geometry class.
Notice that we can use the addition and subtraction formulas from trig to find tan(15) -- in this case, either tan(60 - 45) or tan(45 - 30) will work. But it's possible to find this values exactly using a purely Euclidean approach, without any use of advanced trig formulas at all.
We begin with Triangle ABC with A = 75, B = 15, C = 90. Let BC = 1 and AC = x. Notice that the tangent of 15 degrees is now AC/BC = x/1, so x = tan 15. Let the hypotenuse AB = y.
Now we reflect this triangle over its leg BC, with D the mirror image of A. Since Angle ABC = 15 -- and its reflection DBC = 15 -- we have Angle ACD = 30. And thus, if we drop a perpendicular from A to
We know one side of this triangle -- the hypotenuse AB = y. Thus its shorter leg is AE = y/2, and thus its longer leg is BE = y sqrt(3)/2. And since DB = y (as DB is the reflection image of AB = y), we can also find DE = y (1 - sqrt(3)/2).
Now ADE turns out to be another 15-75-90 triangle (since there's clearly a right angle at E, and D is the image of A = 75). We can find tan 15 using this triangle:
tan 15 = DE/AE
But we already know that x = tan 15:
x = y(1 - sqrt(3)/2) / (y/2)
x = y(1 - sqrt(3)/2)(2/y)
x = 2(1 - sqrt(3)/2
x = 2 - sqrt(3)
Thus x = tan 15 = 2 - sqrt(3), just as it appears on the calculator.
Notice that y = tan(75), and y is the hypotenuse of a triangle whose legs are now known, and so we can just use the Pythagorean Theorem:
AC^2 + BC^2 = AB^2
1 + (2 - sqrt(3))^2 = y^2
y^2 = 1 + 4 - 4sqrt(3) + 3
y^2 = 8 - 4sqrt(3)
y^2 = 4(2 - sqrt(3))
y = 2sqrt(2 - sqrt(3))
These are the values that appear on the 15-75-90 triangle in the Leis post.
Meanwhile, another Blaugust poster today is Beth Ferguson:
http://algebrasfriend.blogspot.com/
http://algebrasfriend.blogspot.com/2019/08/five-top-posts.html
Just as I do every year, Ferguson decides to link to her five most popular posts -- except it's of all time, not just the past twelve months. As it turns out, her most popular post is from 2016:
http://algebrasfriend.blogspot.com/2016/07/a-collection-of-first-week-activities.html
It's yet another list of activities for the first week to school. Ferguson's list isn't as famous as the ones posted by the two Sara(h)s -- in fact, she acknowledges Sara and Sarah in today's post. In fact, we see that Ferguson posted her list in July 2016, while the Sara(h)s posted the following month. Even though this is her most popular post, her list is still overshadowed by the Sara(h) lists.
But here are those pictures from my classroom three years ago to fulfill first part of the Blaugust prompt, followed by the Lesson 1-2 worksheet.
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