Sunday, December 31, 2006

Happy New Year!

Happy New Year! I hope you all had a great holiday. Seniors, I hope all those essays were finished. Juniors, I hope you all enjoyed not having to write them (this year!). I spent the last week having a great time up in Oregon with my family, and if you thought it was cold here…

You will notice when you log on to our blogs to creat a post that we have switched over to the “new Blogger.” It’s supposed to make things easier and run smoother (we’ll see!) but it will require each of you to switch over as well (and create a Google account) before you can edit or create new posts. Nothing serious, but I wanted to give you a heads up before your turn came around and you panicked!

See you all on the 8th!

Thursday, December 14, 2006

Friday's Test Topics

First, I've posted the solutions to today's vector handouts on our class website. The file is Vector Exercise Solutions.pdf

Here’s a list of topics for Friday’s test:

Precalculus Chapter 6 Test Topics:
Law of Sines given a triangle and directions (use a particular angle/side)
Law of Cosines given a triangle and directions (use a particular angle/side)
Area formulas given a triangle and directions (use a particular angle/side)
Uses of the Law of Sines and the Law of Cosines
Impossible Triangles
Ambiguous Triangles – visual
Sketch a resultant vector given two vectors
Determine horizontal and vertical components (i and j components)
Solve triangles – missing sides, angles, areas
Add vectors – bearing and wind speed
Golf shot problems

That’s it! The format is as expected – ½ calculator, ½ non-calculator. I’ll be in my classroom on Thursday after school and before school on Friday. I’ll also be available online later on Thursday evening. If you have specific questions Thursday night, email me! And don't forget your holiday math songs are due on Monday...

See you in class!

In the spirit of the holidays:
Yes, Virginia, There is a Santa Claus

By Francis P. Church, first published in The New York Sun in 1897. [See The People’s Almanac, pp. 1358–9.]

We take pleasure in answering thus prominently the communication below, expressing at the same time our great gratification that its faithful author is numbered among the friends of The Sun:

Dear Editor—
I am 8 years old. Some of my little friends say there is no Santa Claus. Papa says, “If you see it in The Sun, it’s so.” Please tell me the truth, is there a Santa Claus?
Virginia O’Hanlon

Virginia, your little friends are wrong. They have been affected by the skepticism of a skeptical age. They do not believe except they see. They think that nothing can be which is not comprehensible by their little minds. All minds, Virginia, whether they be men’s or children’s, are little. In this great universe of ours, man is a mere insect, an ant, in his intellect as compared with the boundless world about him, as measured by the intelligence capable of grasping the whole of truth and knowledge.

Yes, Virginia, there is a Santa Claus. He exists as certainly as love and generosity and devotion exist, and you know that they abound and give to your life its highest beauty and joy. Alas! how dreary would be the world if there were no Santa Claus! It would be as dreary as if there were no Virginias. There would be no childlike faith then, no poetry, no romance to make tolerable this existence. We should have no enjoyment, except in sense and sight. The external light with which childhood fills the world would be extinguished.

Not believe in Santa Claus! You might as well not believe in fairies. You might get your papa to hire men to watch in all the chimneys on Christmas eve to catch Santa Claus, but even if you did not see Santa Claus coming down, what would that prove? Nobody sees Santa Claus, but that is no sign that there is no Santa Claus. The most real things in the world are those that neither children nor men can see. Did you ever see fairies dancing on the lawn? Of course not, but that’s no proof that they are not there. Nobody can conceive or imagine all the wonders there are unseen and unseeable in the world.

You tear apart the baby’s rattle and see what makes the noise inside, but there is a veil covering the unseen world which not the strongest man, nor even the united strength of all the strongest men that ever lived could tear apart. Only faith, poetry, love, romance, can push aside that curtain and view and picture the supernal beauty and glory beyond. Is it all real? Ah, Virginia, in all this world there is nothing else real and abiding.

No Santa Claus! Thank God! he lives and lives forever. A thousand years from now, Virginia, nay 10 times 10,000 years from now, he will continue to make glad the heart of childhood.

About the Exchange
Francis P. Church’s editorial, “Yes Virginia, There is a Santa Claus” was an immediate sensation, and went on to became one of the most famous editorials ever written. It first appeared in the The New York Sun in 1897, almost a hundred years ago, and was reprinted annually until 1949 when the paper went out of business.
Thirty-six years after her letter was printed, Virginia O’Hanlon recalled the events that prompted her letter:
“Quite naturally I believed in Santa Claus, for he had never disappointed me. But when less fortunate little boys and girls said there wasn’t any Santa Claus, I was filled with doubts. I asked my father, and he was a little evasive on the subject.
“It was a habit in our family that whenever any doubts came up as to how to pronounce a word or some question of historical fact was in doubt, we wrote to the Question and Answer column in The Sun. Father would always say, ‘If you see it in the The Sun, it’s so,’ and that settled the matter.
“ ‘Well, I’m just going to write The Sun and find out the real truth,’ I said to father.
“He said, ‘Go ahead, Virginia. I’m sure The Sun will give you the right answer, as it always does.’ ”
And so Virginia sat down and wrote her parents’ favorite newspaper.
Her letter found its way into the hands of a veteran editor, Francis P. Church. Son of a Baptist minister, Church had covered the Civil War for The New York Times and had worked on the The New York Sun for 20 years, more recently as an anonymous editorial writer. Church, a sardonic man, had for his personal motto, “Endeavour to clear your mind of cant.” When controversal subjects had to be tackled on the editorial page, especially those dealing with theology, the assignments were usually given to Church.
Now, he had in his hands a little girl’s letter on a most controversial matter, and he was burdened with the responsibility of answering it.
“Is there a Santa Claus?” the childish scrawl in the letter asked. At once, Church knew that there was no avoiding the question. He must answer, and he must answer truthfully. And so he turned to his desk, and he began his reply which was to become one of the most memorable editorials in newspaper history.

Church married shortly after the editorial appeared. He died in April, 1906, leaving no children.
Virginia O’Hanlon went on to graduate from Hunter College with a Bachelor of Arts degree at age 21. The following year she received her Master’s from Columbia, and in 1912 she began teaching in the New York City school system, later becoming a principal. After 47 years, she retired as an educator. Throughout her life she received a steady stream of mail about her Santa Claus letter, and to each reply she attached an attractive printed copy of the Church editorial. Virginia O’Hanlon Douglas died on May 13, 1971, at the age of 81, in a nursing home in Valatie, N.Y.

Wednesday, December 13, 2006

6-6: Vector Addition

"Good evening, I'm Chevy Chase and you're not."

So folks, it's about 1:20 am right about now and I'm consequently not really in a festive mood. Therefore, let's get down to business (TO DEFEAT...THE HUNS!). 6-6 really isn't that bad; it's basically just adding vectors or determining the magnitude and direction of a given vector. But, for the sake of your knowledge, my blog grade, and Sam, let's start at square one....

A vector has both magnitude (length in absolute value) and direction (measured in the angle from the vector to the horizontal axis). Here's the basic "anatomy" of the vector:

The sign for vectors is this:

Okay, so now that we've got the basics down, we can get to vector addition. If you already have two vectors, you can add them (by putting the tip of one to the tail of the other......larry, don't be immature) and get a resultant vector. Sweet italic use, huh? Anyways, here's an example formulated by your resident genius himself, me.

You can see the necessary steps here: First find the magnitude of the resultant vector, then determine the angle of its direction using the given information. This particular vector will have a magnitude of 16.547 ft and a direction of 54.031 degrees. CHYEA.

Alright boys and girls, there's really one more key component of the lesson: deriving the components of a resultant vector from that vector itself. That is, if we have a vector, we can determine the two other vectors that form it when added so:

Use the nifty little formula and it's pretty self-explanatory.


Let's say that on a particular football play, Larry, using his god given athletic ability, first runs 7 yards at an angle of 48 degrees from the horizontal axis and then sharply cuts (as only he can) at an angle of 108 degrees for 6 more yards. Following this brilliant display football prowess, Paul, the starting varsity quarterback (replacing your blogger himself and following in his noble footsteps), reads the Cover 3 defense and decides to hit Larry on the aforementioned intermediate route before the strongside linebacker can cover him (...excuse me, TRY to cover him....I mean, this is Larry the receiving GOD we're talking about here). So then, after that huge digression, determine the resultant vector for the throw Paul needs to make to Larry for the presumably solid play! Ready....BREAK!
(Oh yeah, it looks like this.....hahaha)

Done? Okay, here's the method of solution to this unbearably sweet problem:

We're done.

Well, except for this:


(As you can tell, I was the "N" NOMAR!!!!!!!!.......GARCIAPARRA!!!!)


G'night everybody.

Monday, December 11, 2006

6.5 The Ambiguous Case

SSA (side-side-angle) is called the ambiguous case.
When given the information, side, side, and angle, consecutively, there are two possible triangles that m eet the specifications.

The TWO different answers can be found:

1. swinging in the side without the fixed angle.
A triangle with sides 50, 80, and angle of 26 degrees can be either:

Well, how do you find both measurements?

Then in order to find the corresponding angles, you'd use the LAW OF SINES, and go on from there.

2. mismatched angle and corresponding side
Solve for a.

This is an impossible triangle, because the measure of side 4 can only go far as the corresponding side of a 75 degree angle and a "5" adjacent side.

3. draw a triangle with perpendicular sides (to find how many solutions/triangles are possible with the given SSA information)

*Also, it's helpful to know your basic cos and sin angles (ie. sin30), because Mr. French can always throw in a problem with basic angles in the non-calculator portion.

Practice problem:

***OKAY so these pictures I uploaded look quite funkadelic and you can barely see the equations... so if you click the picture, you should be able to see a clear version of it. Sorry.

Here's a helpful link: yay for ambiguous cases!

Christmas is almost here!!!
Here on Martha Stewart Living, you can do a bunch of stuff... like cook holiday stuff... make crafts haha, etc. The most important thing is the pretty food, though. Have fun.


Thursday, December 07, 2006

6.4 Oblique Triangles: Law of Sines

Here's a visual of any given triangle, showing the relationship between the ANGLE (A) and the SIDE (a)
Here's another visual of any given triangle, but in the setting of an actual circle
Above is the actual formula for the Law of Sines

Note: That you can only use the Law of Sines when you're given at least one set, (consisting of an angle and it's corresponding side length), and the measure of either one other angle or side length. However, if you have the given information, you may use the Law of Sines to find either the missing angle measure OR side length.


when you're solving for sin, you're technically using arcsin and SHOULD (if you do it correctly) come up with TWO values that give you the same BE CAREFUL!!!!!!

1) Set the problem up
2) Cross multiply
3) Get isolate the variable, b, and evaluate
4) You have your answer!

isn't she just the PRETTIEST baby EVER?!!!
if you want to see more....ask me...
because i for sure have
pictures in my

Wednesday, December 06, 2006

Friday's Quiz Topics

Here’s a list of topics for Friday’s quiz:

Precalculus Quiz 6.1-4 Topics
Law of cosines
Law of sines
Area of a triangle (sine formula and Hero’s formula)

Relationship of sides and corresponding angles
Determining the appropriate variation of the law of cosines
Solving triangles: unknown sides or angles (generic triangles or word problems!)
Determing the area of triangles (sine formula and Hero’s formula)
Demonstration of impossible triangles.
Ambiguity of the sine function in determining angles in a triangle.

That’s it. I’ll be in early on Friday and available after school on Thursday until 3:30.. See you in class!

"I shall be telling this with a sigh
Somewhere ages and ages hence:
Two roads diverged in a wood, and I --
I took the one less traveled by,
And that has made all the difference."

- Robert Frost
The Road Not Taken

Tuesday, December 05, 2006

Finding the area of a trianglezor!!!

Section 6-3: Area of a Triangle

This section is pretty simple. It pretty much centers around two equations for the area of a triangle:

- Area = ½ (b)(c)sinA
- Hero’s Formula: Area = sqrt(s(s-a)(s-b)(s-c)), where s is the semiperimeter, ½ (a+b+c)

Area = ½ (b)(c)sinA

Proving this formula is a combination of the area formula ½ (b)(h) and the definition of sine.

Since the height (h) is not given as information, we must find a new way to calculate the height. We are, however, given the angle A.

c x SinA = h/c x c
(c)sinA = h

By substituting the new h formula into the old area formula, we get:

Area = ½ (b)(c)sinA

Here’s an example problem:
Find the area of this triangle.

Because we are not given an angle of the triangle, we cannot plug in the information straight into the triangle area formula. First, we must find an angle using the law of cosines. For the sake of this example, we will use angle A.

15^2 = 17^2 + 18^2 – 2(17)(18)cosA

We are solving for cosA, so the formula boils down to:

cosA = (17^2 + 18^2 – 15^2) / ((2)(17)(18))
cosA = .63398…
cos-1A = 50.655°

Now, we can substitute the given information into the area formula:

Area = ½ (17)(18)sin(50.655°)
Area = 118.32 units2

Hero’s Formula

Now some crazy guy named Hero decided to make our lives easier if we were to find ourselves in the situation described above, that is, if we were given only the sides of the triangle and no angle. Instead of going all the way through the law of cosines to figure out the angle, we can use this formula:

Area = sqrt(s(s-a)(s-b)(s-c)), where s is the semiperimeter, ½ (a+b+c)

Using the same triangle in the sample problem above, the formula would work out to be:

Area = sqrt(25(25-15)(25-17)(25-18))
Area = sqrt(25(10)(8)(7))
Area = 118.32 units2

Voila! The same answer that we got above.

So… that’s about it folks! Here is a link that has more details on finding the area of a triangle. (O, and I’m sorry about the sqrt() signs, mathtype isn’t working on my computer)

Now for my personalization. I thought long and hard about this one, but I decided nothing would make me happier than to prove that girls are evil. So here’s something that proves that girls are evil.

Girls = Time x Money
Time is money, so Time = Money
Girls = Money x Money, or Money2
Money is the root of all evil, so Money = sqrt(evil)
Girls = (sqrt(evil))2
Girls = evil


Monday, December 04, 2006

Section 6.2 Law of cosines

Allright people, we probably have the most challenging lesson ever right here!

or not...

Its just the law of cosines, which we all learned in geometry. This law is used to relate the three sides of a triangle to one angle.

The law of cosines is: c2 = b2 + a2 – 2ab*Cos(C)

You can use this equation to find any side or angle as long as you are given either two sides and an angle or three sides.

You can solve the equation to give you the missing angle too.

C = Cos-1
((c2 -b2 - a2)/-2ab)

xample 1. Given triangle XYZ, if y = 8, z = 6, and X = 172 Degrees. Find x.

x2 = 82 + 62 – 2(8)(6)*Cos(172)
x2 = 100- (-95.066)
x= Squareroot(195.o66)
x = 13.967

Example 2. Given triangle UMP, if u = 12, m = 22, and p = 16. Find M.

M = Cos-1((222 -122 - 162 )/-2(12)(16))
M =
M = 102.636

And that is all you need to know about the law of cosines. For more information see

Leo, you are next man. Do me proud.