Wednesday, October 29, 2008

Slides Oct 29 - PreTest

Here they are. Warning: while the pre-test was not cumulative, the test itself is. Give the previous units a quick review. Good luck!

Pre Cal30 S Oct29
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Tuesday, October 28, 2008

Cute Math Site

Despite the cutesy graphics, this looks like a useful site for Grade 11 Pre-Cal.

Cheers, Dr. Eviatar

Monday, October 27, 2008

Sunday, October 26, 2008

October 24, 2008


Last friday, we talked about the Discriminant.

To find the discriminant you need (b² -4ac).

For example: -2x + x² - 4

First, reorder them as ax² + bx + c.

x² -2x -4

Second, use the discriminant formula. (b² -4ac)

-2² -(4)(1)(-4)
4 - 16 = -12

Our disciminant is -12, therefore D Abosulute Value.

For example, the absolute value of |1| is 1 and -1.

Next is Rational Exponents.

√x = x½
³√x = x⅓

Friday, October 24, 2008

Monday, October 20, 2008

Quadratic Equation

Hi there,

Today first we started doing some questions.
and we also did some question related to angles.
When ever you are given a question like this, you should change tan to x.

so it would look like this:-

2x-1=0 or x+1=0
2x=1 x=-1
The roots are (1/2,-1)

So the roots are (1/2,-1)
Now the next step is that we have to find the angles of theta.
angles of theta are
26.6, and 153.4 when theta is 1/2
-45, and 135 when theta is -1.

After that we did a little bit about imaginary numbers.
The general definition for imaginary number is that

For example :-

Next scribe :- Julius

Scribe List

Sorry it has taken this long, but here is the Scribe List. People whose names have been struck out have fulfilled their blogging obligation at least once. When everyone has blogged, we will start again. When you come to choose the next blogger, choose someone whose name is not struck out. People who are missing from this list had better get on SOON or they will be missing about 5% of their mark.


Slides October 20

Here they are ....

Pre Cal30 S Oct20
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Friday, October 17, 2008

Thursday, October 16, 2008

hmm, so i guess it looks like the teacher has already put up the slides of the corrections we did in class for the previous test we just had ! Tomorrow, i'm positive that we are continuing with the rest of the corrections ...

Slides October 16th

Here they are ....

Slides October 15th

Here they are ...

Pre Cal30 S Oct15
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Slides October 14th

Here they are ....

Pc30 S Oct14
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Wednesday, October 8, 2008

prepare fo the test


Dear class I am happy to be chosen as the scribe for today’s class.It good that we are back we the good habit of posting that was about to be stopped because it have been almost one week that nobody have posted. Today Dr AVIATAR gave us a pretest which seemed to be hard at the beginning although we were pretending to do it as if it was a test. The pretest took 30 minutes and after we tried to do the question in group. later she tried to explain the question.

What I want to remind everybody that the test is for tomorrow Wednesday 09 October 2008 and we have to get prepared seriously

There are couple of things that i want to remind every body about tomorrow test

1 if you are given a any problem to solve, problem which has already a drawing and in some of the cases that we are dealing with in this unit relate with triangle

  • when given some sides and angles of the triangle
    we use the sin law to solve the equation
sin (A)/a= sin (B)/b = sin (C)/c
 first s
Our Idea of a Sketch
            Find angle B.

Angle B = 180o - (43o + 57o) = 80o

Now, we use the law of sines to
find the other sides lengths.

c a
----- = -----
sin C sin A

a(sin C)
c = --------
sin A

Plug in any known information.

4.56(sin 570)
c = -------------
sin 43o

Use a calculator to find the sines.

c = -----------

c = 5.61

Now solve for b.

b a
----- = -----
sin B sin A

a(sin B)
b = --------
sin A

Plug in any known information.

4.56(sin 80o)
b = -------------

Use a calculator to find the sines.

b = -----------

b = 6.58

There is another case when you can use the law of sines. When you know two sides and an angle opposite one of the sides, the law of sines can be used. However, with this case, you have to be aware that there might not be a solution, or there may be two! One solution is also possible. Example:

2. Problem: In triangle ABC, a = 15,  b = 25, and angle A = 47o.

Solve the triangle.

Start out by looking for the measure of angle B

a b
----- = -----
sin A sin B

b(sin A)
sin B = --------

Plug in any known information.

25(sin 47o)
sin B = -----------

Use a calculator to find the sine.

sin B = ---------

sin B = 1.219

Since an angle cannot have a sine greater than 1, there is no solution for this triangle
  • When asked to verify number of triangle that can be made from different angles and sides given
when this is the case you have to verify the the domain and the domain and the domain and the range. if the result or the answer is not inclusive in the domain then there is no triangle

  • When given a trigonometric equation (Trigonometric Equations are equations that contain expressions such as sin x. You solve a trigonometric equation in the same way as any other equation.) the most important thing is to start by isolating the term then apply the sin law or cosine law
Solve for x: 2sin x = 1.

Solve for sin x by
dividing by 2.

sin x = (1/2)

The circle below can show the range and domain of sin and cosine of and angle

thank you for reading the blog please prepare your test and good luck

Wednesday, October 8th, 2008

Sorry I didn't blog yesterday, I was too sick to even get on the computer.

So here's what we did in class on Tuesday:

We learned how to solve triangles give certain sides and angles using the sine law.
It's pretty simple stuff, so let's get into it...


Let's say you get a question that looks like this:

The first thing you should do is draw it out. All you need a generic triangle to help you visualize the question:

Right away you should see that "Angle B" is the first thing that needs solving, so to do that, we're going to use the sine law:

This is usually the part where you find out what "Angle B" is using that number you just got and your calculator (2nd function sin on graphing calculators), but in this case, you don't need to.

Why? Because we know that the values of sine only go from -1 to 1, so trying to find an angle with the number you just got will just give you an error. This means that there's no triangle.

So that would be your answer, "No triangle".

We'll solve this question next:

First things first, draw it out:
Now we'll use the sine law to find "Angle B":

Now that we have both "Angle A" and "Angle B", we can find "Angle C":

All we have to do now is find "Side C". Since we have a right triangle (one of the angles is 90 degrees), we can find "Side C" using the Pythagorean theorem:


Let's take a look at this question:

First we draw it:
Then we find "Angle B", but this time, "Angle B" has two different angles (two angles with the same sine):
So what do we do with two angles? We do the exact same things to solve the triangle, but twice. Once for each angle. Find "Angle C" for the two angles we got for "Angle B":

Now we find "Side C" for the two angles we got for "Angle B". Since this is not a right triangle, we can't use the Pythagorean theorem. We'll have to use the sine law to find "Side C":

That's about it! Don't forget to study for the test. The next scribe is ale.

Gooood Night!

Slides October 8th

Tuesday, October 7, 2008

Slides October 7th

Here they are ....

More with math with Niwatori-san

Hiya folks!

Hope you been studying there's the test on Wednesday!

Well anyways besides that what we did in class yesterday was learning about ambiguous case. It's a short unit but if you don't really get it I'll just dumb it down to the level where elementary kids can work around. There is no offense to what I just said ^v^

We started out with the word Ambiguity if you don't know what that means it just means having uncertainty so the basis around this is that we're learning about how to figure out something uncertain with the information given when being asked.

This diagram works around the Sine Law

We then moved onto the cosine law which is another way to get the triangle sides which in this case that formula is:

a^2 = b^2+c^2-2bcCosA

and this single formula can be changed so that you can look the specific side you want by plugging instead of the a^2 you could plug in b^2/c^2 and then putting in the angle relating to that side like b^2 with CosB and c^2 with CosC and so forth.

We also learned how to know if the triangle can be drawn or not.

The diagram above shows how to know if the triangle is not possible.

This next diagram will show you how to wrap around getting two triangles!

Well that concludes my part and for next time have fun in math, it's not just here we're using it. We may use it in the future and who knows what we might use it for!

Until next time with Niwatori-san! GABATTE!! people (japanese for work hard or go for it!)

The next scribe is.... EricT

*passes the baton* *Clap clap clap*^0^

Friday, October 3, 2008

Calling all bloggers!

Since Rhaeanne and Louisa have both left this group, we need someone else to help keep the textbook going. If you have not yet had the opportunity to blog for the class, this is your chance to shine!

Slides October 3rd

Here they are ....

Pre-Cal 30S October 3rd, 2008
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Thursday, October 2, 2008

Slides October 1st

Here are Mrs. Karras's slides from yesterday.

Slides October 2nd

Here they are. Mrs. Karras' slides from yesterday to follow.

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