Wednesday, November 29, 2006
posted by --thamie-- at 5:57 p.m.
Hiya!!! We didn't do much today in class. We just had a quiz on trigonometry (including sine and cosine laws) and went to the computer lab to find the definitions of the following words:

- complementary angles
- supplementary angles
- consecutive
- polygon
- quadrilateral
- congruent
- diagonal
- bisect

I guess we're doing this to have a heads-up on the geometry section of this unit.

For HOMEWORK ... fill up your Three Point Approach for Words and Concept sheets if you didn't get to finish it in class today.

If you're having any trouble with the definitions, you can go to this >>website<<. (I know the site says "for kids" but I think the activities are pretty cool and helpful.)

Next blogger is Christine !!!
Tuesday, November 28, 2006
posted by Alanna at 4:22 p.m.
Today in class, we started working on SIMILAR FIGURES. We also worked on SURFACE AREA and VOLUME.

*REMINDER: The quiz is moved to tomorrow!*

Class was started off today with Ms. Armstrong drawing a little man on the board...

In the second picture, the length and the width has been doubled. Does that mean that we have doubled the size of the whole picture? Not quite, because we have doubled BOTH the length and the width, the size is actually quadrupled.
Next, instead of doing the rest of the notes, we did a worksheet on area and volume...
We had to find the surface area and volume of 2 cubes. I have tried to upload them, but it is not working, so the first cube was Length=2, width=2, depth=2. The second cube was length=6, width=6 and depth=6.
1) Surface area: 6(LW)
Volume: 2x2x2
2)Surface Area: 6(LW)
Volume: 6x6x6
The next part of the sheet was:
How many times bigger are the sides in the bigger cube? 3
*NOTE: when it says side, it means length of a side, not the whole side.*
How many times bigger is the surface area of the bigger cube compared to the smaller one?
9 times or 32
How many times bigger is the volume of the bigger cube compared to the smaller one?
27 times or 33
HINT: notice that when change change the answer into an exponent, the exponent for surface area is 2, and the exponent for volume is 3.
That's what we did for this sheet. The other side we have for homework.
Okay, now to continue on with some more notes:
To find the "new area or volume"
area= (scale factor)2x(old area)
volume=(scale factor)3x(old volume)
ex. The area of a region is 10cm2
by what factor must each dimension be multiplied to increase the area by 20cm2?
New area is 20+10=30cm2
30 k2(10)
__= ____
10 10
3=k2 square root of 3=k
wow... okay, thats it for what we did in class today. Sorry about the pictures, it wasn't working very well. I hope this helped anyway though.
finish the other side of the worksheet
exercise #24
STUDY! There's a quiz TOMORROW!

Monday, November 27, 2006
posted by Ms. Armstrong at 8:54 p.m.
Hey there Gummy Bears...

Here are some multiple choice quizzes to help you review.
Law of Sines
Law of Cosines
Right angle Triangles
A Challenge

I also came across an interesting question to challenge your brains and stimulate your competitive nature. I would like you to 'discuss' the solution to this question through the comment box. If each person adds their ideas, I would bet that you can get it.

Solve the equation for A and B.

Good luck, have fun, and happy solving :)
Tuesday, November 21, 2006
posted by DestiniLyn at 4:18 p.m.

Tuesday, November 21, 2006;

Today in class, we were taught the Sine & Cosine Laws. We first started out with an activity. Mrs. Armstrong had us create a triangle that had one Obtuse Angle (more than 90 degrees). She then told us to measure each side of the triangle to the exact millimeter, & then find each angle of the triangle. Once we were done, she showed us the Sine Law.

Sin A = Sin B = Sin C
_____ _____ _____
a b c

*You can use the Sine Law whe you are given;

1) SSA ( Side , Side , Angle )
(notice the angle is not inbetween the two sides)


It cannot look like this:

2) AAS ( Angle , Angle , Side )
* Must have one complete ratio.


Here's an example of how to use the formula:

Solve for x and b:

First you can find angle B:

Angle B = 180 - ( 37+54 )

= 180 - 91

= 89 degrees / Angle B = 89 degrees

Now you can plug in the numbers that you have so far :

Sin 37 = Sin 89 = Sin 54

______ ______ ______

41 b x

Then you can find out what side b is equal to:

Sin 37 = Sin 89

_______ _______ } Cross - multiply

41 b

Sin 89 x 41

____________ = b So, side b is equal to 68.

Sin 37

So, now you should have : Sin 37 = Sin 89 = Sin 54

_______ _______ _______

41 68 x

Now you can find out what side x is equal to:

Sin 37 = Sin 54

_____ ______ } Cross - multiply

41 x

Sin 54 x 41

___________ = x So, side x is equal to 55.

Sin 37

The Cosine Law:

* You need at least two sides for the cosine law to apply to a problem.

a2 = b2 + c2 - 2bcCosA

OR: [ b2 = a2 + c2 - 2acCosB ]

OR: [ c2 = a2 + b2 - 2abCosC ]

Here's an example of how to use the fromula:

Solve the triangle WTV.

w2 = v2 + t2 - 2vtCosW

w2 = 7.82 + 92 - 2 ( 7.8 x 9 ) Cos112

w2 = 60.84 + 81 - ( - 52.59 )

w2 = 194.43

The square root of 194.43 is 13.94

w = 13.94

--Now you can the same thing that you did when you were using the Sine Law:

Sin 112 = Sin T

_____ _____ } Cross - mulitply

13.94 9

Sin 112 x 9

_________ = T


Then do Inverse Sine to get the angle ( Depending on what type of calculator you have, the way of doing this may vary... On my calculator, I press the 2nd function button, then Sin -1, then the number I got from the last equation. ) .

T = 36.7 degrees.

To get angle V, all you do is: 180 - (112 + 36.7) = 31.3 V = 31.3 degrees

Our homework was Excersize # 19 & # 25

If you are still having trouble with the Sine & Cosine Laws, here is a website that should help you to better understand it. =)

Just a few reminders:

- There will be a quiz on Monday on what we've covered on Trigonometry .

& if anyone would like to contribute to sending flowers to Chris showing us our support, bring $2.00 to Mrs.Armstrong. She will then go & buy the flowers.

Well, I did my best. Lol. I hope this blog was helpful. The next person to blog is.. Meagan. =)

( I apologise if my images don't show up on the blog. I seem to be having some trouble uploading them. =( )

Monday, November 20, 2006
posted by Jeff R. at 9:25 p.m.
Today's class was pretty easy and hard at the same time. Easy because we only had one lesson to think about. Hard because the lesson that we learned was a little too difficult for me (I don't know about you guys). But I'll try my best to make this post as understandable as possible. Okay, here it goes.

Ms. Armstrong gave us a yellow sheet of paper with 2 questions on it. The first question didn't make sense at all so even Ms. Armstrong had some confusion. So we'll omit that so we can focus on the more important stuff.

The second question was to determine the angles that would produce the following ratios:
a) sin(x) = .707 acute angle = ___ obtuse angle = ___
b) cos(x) = .940 acute angle = ___ obtuse angle = ___
c) tan(x) = 11.43 acute angle = ___ obtuse angle = ___
d) cos(x) = -.970 acute angle = ___ obtuse angle = ___
e) sin(x) = -.5 acute angle = ___ obtuse angle = ___
f) tan(x) = -1.43 acute angle = ___ obtuse angle = ___

Note: It's not as easy at it looks. Here's how you solve each of the above questions but first, keep in mind to round all of your degrees to a whole number.

1) To find the acute angle, punch in the equation in your calculator:
2nd function > SIN/COS/TAN button = SIN/COS/TAN-1 (given ratio).

Let's use letter a) for an example.
So, 2nd funct. > SIN button = SIN-1 (.707) : the answer is 45o.
The degree that comes out of this equation is your acute angle.

2) Now let's find the obtuse angle. To find the obtuse angle, you will need to know which quadrant does the given ratio belongs to.
The positive sine ratio belongs to the the first & second quadrants.
The positive cosine ratio belongs to the first and fourth quadrants
And the positive tangent ratio belongs to the first and third quadrant.
Since the positive sine ratio lands on the 2nd quadrant, you have to subtract your acute angle from 180 degrees since the acute angle is the measurement of how far is your hypoteneuse from the 2nd quadrant's degree (which is 180 degrees).
When finding the positive cosine ratio, subtract the acute angle from 360 degrees since it lands on the 4th quadrant.
Lastly, if you're finding the positive tangent ratio, subtract the acute angle from 270 degrees or add the acute angle to 180 degrees.

And now you're done your positive ratios! Here's how I solved the first three questions:
a) sin(x) = .707 which is 45o.
Now, follow the rules above: 180 - 45 = 135o
acute angle = 45o
obtuse angle = 135o
b) cos(x) = .940 which is 20o
360 - 20 = 340o
acute angle = 20o
obtuse angle = 34o
c) tan(x) = 11.43 which is 85o
acute angle = 85o
obtuse angle = 265o
Now the negatives (questions 4 - 6). The negative ratios are a little bit different.
1) To find the acute angle of a negative ratio, do the same thing as how you would find the acute angle of a positive ratio, here's a refresher:
punch in the equation in your calculator:
2nd function > SIN/COS/TAN button = SIN/COS/TAN -1 (given ratio).
Let's use letter d) for an example.
So, 2nd funct. > SIN button = COS-1 (-.970)
When I used my scientific calculator, what I got was -166 degrees. But that's the not the final answer. Remember, the acute angle should be positive. So the answer will be 166 degrees (yes, just randomly take out the negative sign because I don't think that there is a mathematical format for how to change a negative degree to a positive degree).

2) Then let's find the obtuse angle. Do the same thing as you did to the positive ratios.
d) cos(x) = -.970 which is 166 degrees
acute angle = 166 degrees
obtuse angle = 194 degrees

e) sin(x) = -.5 which is 30 degrees
acute angle = 30 degrees
obtuse angle = 330 degrees

f) tan(x) = -1.43 which is 55 degrees
acute angle = 55 degrees
obtuse angle = 235 degrees

And that's it! Sorry for the delay guys, there was too much in my head, although it's done now (: we also got our unit tests back then we went over some of the questions. Congratulations to the all stars! And oh yeah, I pick Desitini to blog next since the people I know didn't want to since they're "busy", haha. Oh yeah, and reminder to redo question #2 on exercise 18 using the above steps.

Ms. Armstrong, a video for this one is kind of hard to find. If anyone can help me , that'll be great :)

posted by michele h at 12:35 a.m.
On Friday, Ms. Armstrong has us do a little activity to help us understand how the trig functions relate to the y and x axis'. Whether it has a positive or negative value.

Here are some examples, that might help you understand:


(sinØ= opp/adj)

The sine function means that you need to divide the opposite by the adjacent. But instead of using the terms opposite and adjacent, what if there wasnt a degree for us to determine which side would be the opposite? We could use the Cartesian plane.

This is how you would write it:
sinØ =

the "r" in the denominator means radius
and the "y" means the distance in the y axis

In the example below, shows the two quadrants that contain positive sine values:


What factor contributes to the ratio being positive in those 2 quadrants? (What do those 2 quadrants have in common?)

the Y AXIS: as you can see in the diagram above, the two shaded quadrants are the two quadrants where the y value is positive.



(cosØ= adj/hyp)

The cosine functions means you need to divide the adjacent by the hypotenuse. How would that be done on a Cartesian plane?


the "r" in the denominator means radius
and the "x" means the distance in the x axis

In the example below, shows the two quadrants that contain positive cosine values:


What factor contributes to the ratio being positive in those 2 quadrants? (What do those 2 quadrants have in common?)

the X AXIS: as you can see in the diagram above, the two shaded quadrants are the two quadrants where the x value is positive.



(tanØ= opp/adj)

The tangent function means you need to divide the opposite by the adjacent. How would that be done on a Cartesian plane?


which can also be shown like this:


the "x" in the denominator stands for the cosine which is related to the x axis

the "y" in the numerator stands for the sine which is related to the y axis

In the example below, shows the two quadrants that contain positive tangent values:


What factor contributes to the ratio being positive in those 2 quadrants? (What do those 2 quadrants have in common?)

This is a tricky one. Both values have the same sign. If you look at the diagram above you'll see that they're either in both positive, or both negative, so either the the answer will end up being a positive because a postive divided by a positive is positive. And a negative divided by a negative is negative.

HEY! this guy is so old and he's doing the same math as we are, it makes me feel like a genius! HAHA. Here click here to watch this video

Here's another video, it's pretty long and boring but it kind of helped me because it reviews the simple formula's of algebra enjoy ! (it even the pythagorean theorem, the way Ms. Armstrong showed it)

Well that's my short review on Cartesian Plane. Hmm.. who to pick next ?

I think I pick JEFF. Yeah, you are the winkest link. hehe :)

.. For some of the stuff above i used Ms. Armstrong's worksheet questions and materials. And yahoo was my source to find those videos
Thursday, November 16, 2006
posted by S. Tao at 3:56 p.m.
In today's class we fooled around with 30* triangles using SIN ratio. We drew three 30* triangles (1 small, 1 medium, and 1 big) See pictures below...

Small triangle

Mrs. Armstrong told us that when using a2 + b 2 = c2 we should use Leg2 + Leg2 = Hyp2 instead, unless it is labeled that way

Medium triangle

Since these are 30* triangles (using Sin) it will have a pattern in the measurements; in other words the hyp will be twice as big as the opp

Large triangle

The adj is effect by the opp of the triangle as you can see through these examples


Sin : Related to the Y axis

Cos : Related to the X axis

Tan : related to the Y axis divided by the X axis

Here is a chart that will show the relation of the trig ratios

Homework: exercise 18 and the next one to blog is..... Edward
posted by Haiyan at 12:38 p.m.
In yesterday's class we learned how to find a side of two right triangles using trigononetry.

S - sine
O - opposite
H - hypotenuese

C - cosine
A - adjacent
H - hypotenuese

T - tangent
O - opposite
A - adjacent

Find CD ( Hint: Fist find W )

tan 60 =W/2
(2)tan 60=W
W=3.5 cm

tan 60 =X/3.5
(3.5)tan 60=X
X=6.1 cm

Therefore CD=6.1 cm

Angle of Elevation

The angle between the line of sight and the horizontal.

Angle of depression

Homework:Exercise 17

Still need help? here's a site for you.

The next blogger is Shelly
Tuesday, November 14, 2006
posted by kimberley at 6:04 p.m.
Today's class: The first 20 minutes were spent finishing our test from yesterday, "Rational Exponents & Radicals". As a reminder, our units were also due (10% deduction for everyday that it is late). Afterwards we started our new unit, which is going to be a mixture of Trigonometry and Geometry, beginning with Trigonometry.

Trigonometry is basically the computational component of geometry (to get technical), but to keep it within simpiler terms, it has to do a great deal wtih triangles and to start off our unit that is exactly what we created. In our groups, we were asked to make 3 right triangles with angles of 17degrees, 39 degrees and 54 degrees.

How do we create triangles?
let us use the example of c
reating a right triangle with a 17 degree angle.
STEP 1: On your piece of paper, use a ruler to make a straight horizontal line. (view image for further detail)

(To view image, hold shift and click on box)
STEP 2: Using a protractor, place the 90 degree point on one of the ends of your horizontal line.
Make a mark at the top and draw a straight vertical line going down from your mark to
the end point of your horizontal line. (vi
ew image for further detail)

(To view image, hold shift and click on box)
STEP 3: Place the protractor at the 90 degree point on the other end of the horizontal line. Make a
mark at the degree indicated in the instructions (in this case 17 degrees). Then draw a
line from the mark you've made to the end of the horizontal line.
(view image for further

(To view image, hold shift and click on box)
STEP 4: Label the vertex with its degree (which happens to be 17 degrees in this case). (view
image for further details).

(To view image, hold shift and click on box)

After we were done making our triangles, we labelled the sides with hypotenuse, adjacent and opposite.

REVIEW: Hypotenuse - this is always the longest side, opposite of the right angle.
Adjacent - the side next to the angle
you are refering to.
Opposite - the side opposite of the angle you are refering to .
* (depending on which angle you are refering to, the adjacent and opposite may change)

View diagram:

(To view image, hold shift and click on box)

The last thing we did was divide the sides as follows:
1. opposite/
2. adjacent/
3. opposite/
(round to 3 decimal places)

Still unsure about how to label a triangle? Then maybe this site will help, take a look.

Oh and btw, tomorrows blogger is Haiyan.

Monday, November 06, 2006
posted by JamieC at 4:30 p.m.

In today's class, we somehow fit two lessons in one day simply because I had to remind her that we didn't have math class tomorrow. Urgh... **groan** The mistake is mine, but the bad is not mine. the first part of class, we learned about multiplying and dividing radicals. When multiplying radicals, here is one rule, or example to follow.

Ex. - √a × √b = √ab which also happens to equal (ab)1/2.

That's easy enough to do though isn't it? Well of course, that example was just a review of what we've been doing in the previous days. It gets a little more complicated than that. Today, we also learned how to multiply radical binomials. Now this is pretty similar to the things we did in the first unit we did this year on polynomials and factoring them. In the following I will explain, [or try to...] the concept in a step-by-step process in the following example:

Simplify: 6√2 (√6 – 2√4)

1.) In order to simplify this, we must distribute the both the radical and the....oh...the other number, I don't quite know what the exact term is, but in this case it's the 6 right before the square root of 2. You multiply that with the same type of number of both sides of the binomial and then multiplying the radicals with the radicals resulting in...

6√2 (√6 – 2√4)
= 6√18 – 12√12

2.) After getting that answer, simplify the expression even more throughly, breaking it up bit-by-bit into more parts so that we are able to multiply it later on. [SiMpLiFy UnTiL yOu CaN nO lOnGeR sImPlIfY!!!]

6√2 (√6 – 2√4)
= 6√18 – 12√12
= 6√9√2 – 12√4√3
= 6 × 3 √2 -12 × 2 √3 = 18√2 – 24√3
Secondly, we were taught about radicals w/irrational denominators and what they had to do with dividing radicals-- which was pretty much everything...kind of. It was after all, in the form of a fraction, and a fraction is another expression of division. And this is what this might look like:

[click on box, because it's all screwed up.]

...where you divide the like terms [radicals with radicals] and then further simplify.

But when you are stuck in the situation where you have a radical binomial in the denominator, then it goes something like...

It is okay to have a radical in the numerator, but only then are numbers in their simplest form is when they do not have both a negative exponent and a radical in the denominator.

However, if there is a binomial in the denominator, the "difference of squares" must be multiplied. [REVIEW of DIFFERENCE OF SQUARES: Ex. (2x-3)(2x+3) = 4x2 - 9]

An example of this might be...

[click on the box again. I don't know why the other diagram is goofing up.]

...where the terms with difference of squares cancel each other out because of the positive and the negative.
And yes...finally, it's the end for me. At least I think it is...but if you wish to tweak your "rationalizing" skills, I found a website that can give you some online practice where you could answer some questions by filling in the blanks and then check your answer.
I apologize if you didn't get some of the stuff, even though the post was incredibly lengthy. I tried though. And just a quick reminder just in case you missed it. The homework for the rest of the week is Exercise 37 [omit 10/11/16/17/18] and Exercise 38 [omit 11/14/16/18].
The next person to blog is Kim Tran.
Saturday, November 04, 2006
posted by Ms. Armstrong at 5:04 p.m.
Hello Gummy Bears,
Here is the code that Chris used to create square root symbols. If you need to incorporate it into your post, my suggestion would be to do a copy and paste. The only thing that you will need to do is remove the *'s that I have added in.
x<*big>√<*/big><*span style="BORDER-TOP: white 1px solid">y<*/span>, x=Any number before the root sign and y=What is under the root sign.

I am also putting in some links for reviewing material.

roots and radicals (videos)
rational exponents (tutorial with practice questions)
simplifying radical expressions (m/c quiz)
operations with radicals (m/c quiz)

Hope this helps and have fun!!!
Ms Armstrong
posted by Chris at 2:17 p.m.
Today in Friday's class, we learned about: Adding and Subtracting Radicals.

You can only add and subtract like terms.

Example: 4x+3x=7x
22 + 32 = 52
These two examples can be added together because
their bases are the same.

42 + 33 = 42 + 33
These two examples cannot be added together since their bases are not the same.

In order to add/subtract radicals you must first simplify.

Example: 212 - 348 (Starting problem)
24·3 - 316·3 (Break the problem up so there's a perfect square under the root)
43 - 123 (Solve the perfect square and multiply it with the number at the front, if there's no number in front, its considered to be a 1)
-83 (Combine like terms)

Example: 12 + 28 - 375 + 2(Starting problem)
4·3 - 24·2 - 325·3 + 2 (Break the problem up with the highest perfect square that can be multiplyed into the original number)
23 - 42 - 153 + 2 (Solve the perfect square and multiply it with the number at the front, if there's no number in front, its considered to be a 1)
-133 + 52 (Combine like terms)

Note: The next example has the cube root represented by an upper 3.
Example: 324 + 381 + 212 - 348 (Starting problem)
38·3 + 327·3 + 24·3 - 316·3 (Break up the problem with the highest perfect cube that can be multiplyed into the original number)
233 + 333 + 43 - 123 (Solve the perfect cube and multiply it with the number at the front, if there's no number in front, it's considered to be a 1)
33 - 83(Combine like terms, remember to keep square, cube etc. roots seperated.)

Example: 6xx - 7x3 (Starting problem)
6xx - 7xx (Since this problem is rather easy, we can just look at it and take out x2 from x3 which leaves us with x)
-xx (Combine like terms)

Ms.Armstrong also gave us a Radical Puzzle(Square Root) which was to be put correctly into the shape of an equilateral triangle.

Homework was Exercise #34 omitting 10,16,18. If there's any questions feel free to comment. If you need any additional help, click here to view a page about radicals.

Originally posted by Jamie:
"The next person who will post will be "tall guy" Chris. Yes you. Haha. Mwahahahhaha.... =D"
The next person who will post will be the "small person" Jamie. Yes you. Haha. Mwahahahhaha....=D

Thursday, November 02, 2006
posted by Edward at 11:03 p.m.
I finally found some information about pi. No one really knows who specifically figured out pi, but what we do know is that pi existed for over thousands of years and still today it is being used. In about 2000 BC, the babylonians used 3 1/4 as an approximation for pi and the egyptians used 256/81. Pi was used specifically for the measurements of a circle the old mathematicians used to measure really large circles to come up with the decimal numbers.
posted by dinh at 6:39 p.m.
Today in class we got are quiz back from Ms.Armstrong and we got to see what are mark we have overall in precal...

After that she taught us about radicals and how to simplify them.

A radical is an expression under the square root, cubed root, or to the forth root... Our goal is to simplify radicals to their simplest form. Simplifying means to find another expression with the same value.


Rules for simplifying Radicals
1) Break up Whatever is under the root sign into a product of factors. Find the LARGEST PERFECT SQUARE.
2) Split up and simplify.

For homework exercise # 33 omit #5/12/17
For more help with radicals visit Radicals for you.

I pick Chris to scribe next.