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.

Tuesday, December 12, 2006

6.6 Vector Addition

6.6 Vector Addition

Vector: A line with a direction and a magnitude
-usually described as an arrow

"Vector x" can be described as an x with a line over it

If you only want to refer to the magnitude, put the absolute value bars around the symbol.

Resultant-found by adding the two vectors together
-to add them together, place them head to tail, so

Would Become

-create a third line, line z, to make a triangle.

Once you have a triangle formed, use the law of sines/cosines to find side z. Side z is the resultant.

Another way you can find the resultant is by using a right triangle. This works if you are given two vectors and a direction angle.

Once you have moved the vectors head to tail, use the component formula to find the resulant motion.

Component Form:

Q = A cosx i + A sinx j
Where Q is a vector where
x=Degree of Direction Angle

In other words, the component form is the
Length x cosA (for the x value) + Length x sinA (for the y value)

Note: If a problem ever talks about a bearing, the degrees will move clockwise from the direct north position.

Sample Problem

Shelly is walking in the park at a bearing of 90 degrees. After walking 200 yards, she turns to walk another 65 yards at a bearing of 42 degreed. How far is Shelly from her original point?

Set up a diagram with the resultant vector.

Convert the two given vectors to component for. One vector is 200yd at 90 degrees, and the other is 65yd at 43 degrees. Set up the two vectors' equations to look like

200Cos(90) + 200Sin(90)


65Cos(43) +65Sin(43).

Once you have converted the equations, add the two together.

[200Cos(90) + 200Sin(90)]i + [65Cos(43) + 65in(43)]

Answer: 291.868yds

Here's a website for additional help.

Once again, I'm going to personalize with something bass related. I found this guy on YouTube who built an upright bass out of a cardboard box and a piece of plywood. It actually sounds really good!

Huzzah! It's Allison's Turn!

Monday, December 11, 2006

6.5: The Ambiguous Case

Hey everybody, this section is the one called The Ambiguous Case, which sounds scary, but isn't. It means that there are two options for the length of the side opposite an angle on a triangle where you have SSA.

Here is an example, where the red lines are the two options that complete the triangle, assuming that you know the measure of Angle B:

In this situation, the things that are ambiguous are the length of side a and the measure of angle A. The length of side b does not change, but it moves to a different position.

The goal is to find the value of the third side (in this case, A) depending on the measure of angle A.

In order to solve these types of problems, you need to determine the number of solutions that are possible. You can do this by using...

( sin B = b / a ) or ( a sin B = b)

If the answer to that eqation is 0 (zero), then there is 1 (one) answer.

If b > a sin B , then there are 2 answers.

If b is less than a sin B, then there are no real answers.

Example Problems:

1. Find out how many solutions there will be for the missing angle and what they are, if any.

Angle A is 20 degrees, side a is 38 ft and side b is 45 ft.

First: 38 = 45 sin 20 , which becomes 38 = 15.39, so a > b, and there are two solutions.

Second: Use the Law of sine to find the angle measures: sin 20 / 38 = sin B / 45.

The answers you get from that is 23.89, and then you subtract that from 180 to get the other, 156.11.

2. Find the length of the missing side.

In triangel EFG, angle F = 33 degrees, side f = 6 miles, and side e = 5 miles. Find side g.

First: Set up the equation with the law of cosines: 6^2 = 5^2 + g^2 -2(5)(g)cos 33.

This leaves you with 0 = g^2 -8.39g - 11.

Second: solve this using the quadratic formula (under PROGRAMS on the calculator).

On the calculator, A = 1^2, B = -8.39(1) and C = -11.

Since one of the answers is negative, the only answer for side g is 9.54 miles.

If there is still something you don't understand, ask Mr. French or check out this site:

Reminder: Henry is next!!! YAY

And finally, if you want to further study math, check out this's funny.

Yep, so that's it. Have fun and i'll see you guys on the flip side.

Thursday, December 07, 2006

Lesson 6-4, Law of Sines

Hey everyone
In class today, we learned the concept of Law of Sines.
Law of Sines

a/SinA= b/SinB= c/SinC

Verbally: With any triangle, the ratio of the sine of an angle to the length of the sides opposite that angle is constant.

Dealing with law of sines, some geometry terms should comeback.

Terminology= SAS= Side, Angle, Side

Objective- We use our geometry terminology and law of sines to figure out all missing angles and sides. Always becareful is all the information is provided.

Law of Sines should only be used to come up with the leg of the triangle. You can use the law of sines to find an angle of a triangle, but one gets two values of the inverse sine relation between 0 and 180, either could be the answer. If one uses the law of cosine, one will achieve one answer to the angle.

Example #1

Angle B= 64 Angle C= 38 Side b= 9 ft
Figure out the missing info. Case= Angle, Angle, Side

Angle A= 180 -(38+64)
Angle A= 78

Sin 64/9= Sin38/c
Sin 64(c)= Sin 38(9)
c= Sin38(9)/Sin64

Sin 64/9= Sin 78/a
(9) sin (78)= Sin(64)(a)
(9)Sin 38/ Sin 64=a

Example #2
In ABC triangle, A= 52 degrees, B= 31 degrees, a= 8 cm
Find side b and side c.

Angle C=97 degrees

Sin (97) /c=Sin(52)/8


Internet Source


Claire you are up next have fun.

Personal Touch

I am a huge sports guy, I like basketball, football, soccer, tennis, and etc. So here is a picture of my favorite college and professional basketball teams.

Lakers UCLA

Wednesday, December 06, 2006

Friday's Test 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

Monday, December 04, 2006

6-3 Area of a Triangle

Hey guys! So we officially only have wednesday, thursday, friday, another week after that and then four days after that till christmas break!!!! That's only 12 days- we can do it!...

Anyway lets learn about the Area of a Triangle...

Given the measures of two sides and the included angle, find the area of the triangle.

Property: Area of a Triangle:

As we already know the area of a triangle = (1/2)(base)(height)

We can alter this formula to equal:

sin A= h/c

h= c sin A which can then be replaced in the original A= (1/2)(b)(h) to equal:

Area = (1/2)bc sin A

The area of a triangle equals half the product of two of its sides and the sine of the included angle.

Example Problem 1:

Find the area of triangle ABC given:

a=12 in.

b=10 in.

c= 62 degrees We can plug this into the formula Area = (1/2)bc sin A to get

A= (1/2)(10)(12) (sin62) = 52.9768 in. squared

Property: Hero's Formula:

In triangle ABC, the area is given by

where s is the semiperimeter (half the perimeter) found by taking 1/2 (a+b+c)

Example Problem 2:

Use Hero's Formula to calculate the area of this triangle.

a= 4


c= 8

The perimeter (4+11+8) = 23 so s= (1/2)(23)= 11.5


For Extra Help go to: or

Personalization: I had a ballet performance of the
Nutcracker on saturday and sunday and here's a picture from Snow which comes right before the Land of Sweets!

Reminder: Arash you are next!

Sunday, December 03, 2006

6.2 Oblique Triangles: Law of Cosines

Is everyone excited for the law of cosines?? I know I am!!! Especially since it's all festive in Christmas colors!

Oblique Triangles: Law of Cosines

If you are given three triangles with given sides 3 and 4 and included angles of 60º, 90º, and 120º:
In a right triangle, you can find the third side by using the Pythagorean Theorem:
a squared + b squared = c squared

However, in the 60º triangle, the value of c is less than a + b . For the 120º triangle, the value of c is greater than a + b .

The equation you will use to find the value of the third side from measures of two sides and an included angle is the Law of Cosines (since it involves the cosine of an angle).


  • Given two sides and the included angle of a triangle, use the law of cosines to find the third side
  • Given three sides of a triangle, find the angle

PROPERTY: The Law of Cosines

In tirangle ABC with sides a, b, and c,

a squared = b squared + c squared - 2bc cos A


  • if the angle is 90 degrees, the law of cosines reduces to the Pythagorean Theorem, because cos 90 degrees is zero.
  • if A is obtuse, cos A is negative. so you are subtracting a negative number from b squared + c squared, giving the larger value for a squared.

Applications of the Law of Cosines

you can use the Law of Cosines to calculate either a side or an angle. In each case, there are different parts of a triangle given.


Find side b in triangle ABC, if a = 7 cm, c = 10 cm, and angle B = 71 degrees


b squared = 7 squared + 10 squared - 2(7)(10) cos 71 deg

b squared = 49 + 100 - (140 cos 71 deg)

b squared = 149 - (140 cos 71 deg)

b squared = 103. 4204...

*take the square root of both sides and then you get

b = 10.17 cm


Find angle B if a = 7 ft, b = 14 ft, and c = 11 ft


14 squared = 7 squared + 11 squared - 2(7)(11) cos B

196 = 49 + 121 - 154 cos B

196 = 170 - 154 cos B

*subtract 170 from 196

26 = - 154 cos B

*then divide -154 into 26

-13/77 = cos B

*then take the inverse cos of that fraction to get the value of angle B

cos -1 (-13/77) = B

Angle B = 99.72 deg

Want additonal help?

go here:


Becasue I was dressed so colorfully one day last week, that inspired my friend Taylor to draw me a super awesome picture:

Anna, you're up next!! (we got the best chapter ever. woot!)