## Wednesday, October 29, 2008

### Slides Oct 29 - PreTest

## Tuesday, October 28, 2008

### Cute Math Site

www.mathisfun.com

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

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

### Scribe List

deaarly

Ashley

saaayana

»§ĦΛ§ŦΛ«

## Friday, October 17, 2008

## Thursday, October 16, 2008

### Slides October 16th

## 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

` first s`

` Find `*angle B*.

**Angle B** = 180^{o} - (43^{o} + 57^{o}) = **80**^{o}

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 57*^{0})

*c = -------------*

* sin 43*^{o}

Use a calculator to find the sines.

* 4.56(.8387)*

*c = -----------*

* .6820 *

**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 80*^{o})

*b = -------------*

* .6820 *

Use a calculator to find the sines.

* 4.56(.9848)*

*b = -----------*

* .6820 *

**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: Intriangle ABC,a = 15,b = 25, andangle A = 47.^{o}Solve the triangle.Start out by looking for the measure of angle B

Solution:

a b

----- = -----

sin A sin B

b(sin A)

sin B = --------

a

Plug in any known information.

25(sin 47^{o})

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

15

Use a calculator to find the sine.

25(.7314)

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

15

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 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*.

Solution: 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

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...

EXAMPLE 1

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".

EXAMPLE 2

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:

EXAMPLE 3

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!

-Eric

## Tuesday, October 7, 2008

### Slides October 7th

### More with math with Niwatori-san

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^