o mrs. Saatkamp

take it easy ^^

Miss Saatkamp

this is giovanni

i just can't do number 53 from the homework

i've already tried everything and couldn't do it

thanks Giovanni

Class Sumarry 9A 2/20/09

Today Mrs. Saatkamp began by correcting the homework. The problem that most students struggled with was number 26, in which we had to use the quatratic formula to solve the excercise and the answer was 12.(i don't know how to represent this problem on geogebra, sorry) After this, Saatkamp moved on to introduce us to our lesson for the day, Lesson 8-3: "Trigonometry: Ratios in Right Triangles". She changed all of our calculators to work with degrees instead of radians. She also told us to check if our calculators worked by typing the function first and then the number or the other way around. Ex: sin(30) or (30)sin. We must memorize this so we can easily work with the calculator. Saatkamp then moved on to the actual explanation of our lesson. She began by drawing the following diagram on the board:

After the teacher drew this, she went on to explain what sine, cosine, and tangent actually are:

**********A trick to remember all of these formulas is SOH-CAH-TOA: SOH meaning sin= opposite/hypotenuse. CAH meaning cos= adjacent/hypotenuse. TOA meaning tan= opposite/adjacent.**********

Another reminder is to watch out for equal signs. Because of all of the rounding done in this chapter, you might need to use an "approximately" sign.

Later, Mrs. Saatkamp also said that if we needed to isolate the x in any equation with tan, sin or cos, that we needed to use the inverse functions on our calculators: tan^-1, cos^-1 and tan

The HW is page 416/#17-53 all except 50.

Good Luck and have a happy "Carnaval"

Today, 20 of February, we learned the following in Math Class:

Before we Started the new Subject, We corrected the HW assigned yesterday:

Page 401 # 15-32 all

We started the section of Chapter 8 which talks about Trigonometry:

Ms. Saatkamp first asked us to put our calculators to calculate DEGREES (each calculator has a different way of ajusting it), then if your calculators had the options of operation: SIN / COS / TAN... if you don't, you need to buy a calculator with those operations fast because it's IMPOSIBLE to do Chapter 8 without it!

When using the operations above, you need to find out if you put the number BEFORE the operation or AFTER.

For Example: Sin37 OR 37Sin

Right Triangles Ratios

As said in the name, this ratios ONLY apply in Right Triangles, and more specifically, in their ACUTE angles.

On your left, this image of a trangle representes how to set which side is the Opposite Leg and which is the Adjascent Leg (By the way, the Hypotenuse NEVER changes, he will always be the Hypotenuse). It depends of which ACUTE Angle you are working with.

In the image, I chose to work with angle C. Since side c is opposite from angle C, side c will be the Opposite Leg. And the Adjacent Leg will be side a, since it's adjacent to angle C.

***But remmeber, the Hypotenuse is also adjacent to angle C but it NEVER, I repeat NEVER, will be the Adjacent Leg.

If we choosed angle A, the Opposite Leg would be side a and the Adjacent Leg, side c.

Ms. Saatkamp then gave us formulas to apply in order to calculate the the missing:

The formulas are:

Then she gave us examples

Example 1:

Angle A = 17

Side c = 5

Side b = x

You should apply the formula of COS

The Calculation for Example 1 would be this (Note that after an arrow, it is a new step):

Step 1:

Apply the numbers in the formula COS

Step 2:

Multiply both sides by x

Step 3:

Divide both sides by Cos17°

Step 4:

Use the calculator to find the answer

You should press 5 then press / then Cos17 then press = and you should get 5.2284... but simplified should be 5.23

Example 2:

Angle C = x

Side c = 7

Side a = 11

Yow should apply formula TAN

Step 1:
Apply the numbers in the formula TAN

Step 2:

Pass Tan to the other side (Note that it becomes negative in the other side)

Step 3:

Use the calculator to find the answer.

You should press Shift then Tan then open parenthesis then 7/11 then press = to find 32.4711... but simplify, so the final answer will be 32.5

Note: Some calculators instead of shift is Second and in some calculators the is no need to open parenthesis

There is HW =( for Monday March 2

Its Page 416 #17-53 ALL (except #50)

I will try to explain today's class here but she said she will be after school and

lunch times there for questions.

--OK so let's start today's class with Geometric Mean, who can guess what it is

(so ms saatkamp:)

Geometric mean is for example:

x is the geometric mean Between a and b if:

See this order?! That is what means geometric mean,

if any other number was the geometric mean

it would be in those exact places.

Theorem 8-1:

If: you have a right triangle,

and the altitude from the right angle to the hypotenuse

Then: The 3 triangles are similar to each other

(Becareful with the order- ABC~ADB~BDC(all these are triangles)

Why? Take triangle ABD for example,

it has two angles congruent to the big triangle

So it is similar to the big triangle

Due to the same reason Triangle BDC is also similar to the big triangle

So if ABD~ABC and CDB~ABC you can say that

ABD~CBD by transitive property

Theorem 8-2:

If: you have a right triangle,
and the altitude from the right angle to the hypotenuse

READ THIS to understand the THEN part,

can you see that the altitude broke down the hypotenuse into 2 sides

keep thinking on them because

Then: The altitude is the geometric mean of those sides that

I made you think.

So... as BD is the Altitude the fraction will be like this:

Theorem 8-3:

If: you have a right triangle,

and the altitude from the right angle to the hypotenuse

To understand the THEN part choose one of the legs( I chose AB)

Then: The leg you choose is the geometric mean

So... :

HYP=Hypotenuse

COR=corresponding side(in this case AD)

Theorem 8-4: Pythegorean Theorem

(Same diagram from other theorems)

If: its a right triangle

Then: the HYP2(squared)= one leg2 plus the other leg2

Ms Saatkamp, proved this theorem in class and couldn't

do the proving here, because i'm having issues here with images and fractions

So if you want to know how to prove this ask her( sorry miss ¬¬)

Theorem 8-5: Converse of Pythagorean Theorem

(You can still use other theorem's diagrams)

If: The longest side squared is equal to one side squared plus the other squared

Then: It is a right triangle

The HomeWork is pg 401#15-32 all

Class Summary

Well, this is what happened yesterday,

Firstly, sorry for forgetting to post things yesterday, but here it is (my first post!!!). Well, anyways, yesterday we began our class with Mrs. Saatkamp giving us back our quizzes, and even though most people went really well, there was a little bit of confusion in question #2. The problem is that even though the two given numbers equaled up to equal ratios, we didn't have enough information to prove the triangles were similar. People, remember, FORGET looking for equal ratios because they won't always lead you to a correct answer, what everybody should always keep in mind are the theorems of this chapter, because even though we're not doing any provings, you SHOULD know them to understand the content and if you wanna go well on tests and quizzes, you better start studying the theorems and postulates because they practically summarize the content within a simple sentence. Well, I guess there weren't any other major problems with the quiz, most of the problems people had were different from one person to another, meaning that not everybody had the same problems. Well, after Mrs. Saatkamp collected our quizzes she reminded us of the announcements that were made in the blog, that the due date was extended (still to be defined), that we will have some more time in class to work on it, and most importantly, this will be the only project of the quarter (hurray!). For there not to be any complications with GeoGebra, Mrs. Saatkamp then showed us some useful tools that we could use in it. Well, before I explain what was said about GeoGebra, remember that before you put any information, define what scale you will use, and remember, the scale does need to fit on an A4 paper, so don't do something like 1 to 5 'cause it won't fit. Well, moving on, on GeoGebra, you can easily define the squares to be .25 or .5 big (although the units will always be one), all you have to do is:

1) Right click on the coordinate plane, and go to PROPERTIES
2) Click on GRID
3) Enable the option GRID (you can change the grid's color if you'd like, but really, it's totally not useful).
4)Enable the option DISTANCE
5)By clicking on the white boxes next to X and Y, just define what distance you want your grid to have from one box to another [if you're using decimals, use "." (period), not "," (comma), otherwise, it won't work].

Once you've got that settled, you can move on to graphing our beautiful math class into GeoGebra (=DD). To draw a line, something that could for example, represent a wall, follow these simple steps:

1)Click on the third box at the top coming from the left to the right, you should see a line, with a small segment between two points.
2) Hold down the small upside down triangle, on the bottom right corner of the little square.
3) Select the option Segment with given length from point
4) Click on the place in the coordinate plane where you want your segment to start.
5) Define what length (according to you pre-defined scale) your segment will have.
6) Press "enter" and you've got your segment, to make it go up, click on the point where it ends (the one that you didn't click in the beginning) and guide it up.

If you want to be 100% sure that all of your walls form 90 degrees perfectly, follow these simple steps:

1) Click on the fourth square from the right to the left, where there is a drawing of an angle named "A" (if you don't see this drawing, select the option Angle).
2) Click on the two segments that form your angle or the three vertices that for your angle.
3) An angle should appear, but if it's an reflexive (exterior, outside) angle, right click on the angle, select properties, and deselect the option Allow reflex angle.

Now, if you want to create a segment on top of an existing segment, such as to draw the board or a window, these are the steps you should follow:

1) Click on the same place you did to create the first segment, the third square from the left to the right.
2) This time, select the option Segment between two points.
3)Click on where you want this segment to be, and then drag it to where it should end.
4) If you right click it and go to style, you can make it thicker, to differentiate it from the other segment.

But remember, for the steps listed on top, you can also change step 2 to Segment with given length from point, as before, to make the correct length, by selecting Segment between two points, you are the one drawing the segment, meaning that you can easily get it wrong. Well, I guess that was all, after that, we all went back to our desks and the bell rang. I guess there are no major announcements to make, just for the lazy people who still haven't downloaded GeoGebra at home, DO IT, its quick and free, and you'll need it. Today we had the test, I hope everyone went well, and I am totally sure that tomorrow we will spend a lot of time discussing the last question of the test, and no Javier, I can almost assure you that it was NOT 80 (¬¬). Well, announcements about other classes, people don't forget that tomorrow we have a science AND history test (Oh no!!).

See ya'll tomorrow

class summary of 17/02

In today's math class, Ms. Saatkamp started the class by giving our quizes back. There was one question in the quiz that no one in our class got it right, which was #2 on the first page. The question asked if AB//YZ in the condition where: XB=3
BZ=2

AB=6
YZ=10
First of all, to answer this question, you

had to remember that AB and YZ could only be pararell when;
1)

(theorem 7-5)

2) A and B are midpoints( theorem 7-6).

In this question, most people simply did

which would be equal to each other. However, the information is not enough to determine that AB//YZ. So the answer to this question was sometimes true.

----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- -----
After getting our quizes back, Ms. Saatkamp corrected the hw which was assigned last Friday.

The questions that people asked were #27 and #30.

On #27, you had to find the value of x with the given information;

PQ//DF

QF=8

EQ=3

PE=x+2

DP=12
On this question, you had to compare between EQ:QF and PE:DP.
So the calculation would be:

On #30, the book asked to find the perimeter of triangle PQM when t.STV ~t.PQM. You were to first calculate x by making a relationship between PQ:TS and PM:SV.

8x+36=60x+10

x=o.5
Then, to solve the perimeter of t. PQM, there were two ways:

1)Calculate QM and PM, then add the 3 sides.

2)Use theorem7-7 (which Ms. Saatkamp said it was more efficient)

4P=110

P=27.5
----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- -----

Toward the end of the class, Ms. Saatkamp taught us tips and directions on Geogebra which would be very useful in our math project.

a)First thing she taught us was that you can't modify sclaes on Geogebra, but you have to decide how many centimeters it worth to each unit.

b) It is more convenient if you make o.5 units show on geogebra. You can do this by enlarging the graph.

c) "Segment with given lengh from point"

This is a button in Geogebra that you can choose a point and draw a certain lengh of segment just inserting them. Once you have made the segment, you can move around in any angles with the same lengh. This tool would specially help you in the project.

d) Ms. Saatkamp told us that we only need to put 10 desks in Geogebra. You don't want to waste your time putting 30 desks, so Don't forget this!

Good luck on the test tomorrow!

Mayu

Sorry that this post is late, but my internet was not working last night, so it had to be delayed.
Well, last class we basically spent the whole class correcting the homework from the day before, which was page 373-374 #12-29.






12) Green:

I divided the small triangle by the big oneto discover x. The first equation solves for segment FB,which is 4, then the next one solves for DC which is also 4.


15) Blue:
you can see that DB is the median of EC and AC, and by using logical reasoning one can tell that its also the median of GC, so by knowing GF is 18, you know that FC is also 18.

13) Orange:
you can discover it the same way as 15 or by dividing side by side equals altitude divided by altitude.




19) Theorem 7-11 stats that:




29) to discover CF:

because of theorem 7-11



and to discover BD :



Ms. Saatkamp also let us work on the new homework she assigned as a review for the quiz on Monday. The review is:page 388 #13-30.
#27-30 are what will fall on monday´s quiz.
Good luck studying!

9'A' Friday 13th

Yesterday when the class started mrs. Saatkamp told us that she was going to give us Thursday's class time to work on the project, remeber the project is due this FRIDAY and we need to turn the geogeobra document and the one in word. Afterwards we checked the h.w on pg. 373 q's 12-20.We solve the ones that most of the people had problem with, this were q's 12, 19, 24 and 29.


----------------------------------------------------------------------------------------

Number 12 : We need to find DC.
Given: EA=12, BC=8, EADB, D and B are midpoints of EC and AC and
angle DCF and BCF are congruent, therefore we can know that GC is an angle bisector.

EA and DB are proportional, meaning that DB is 6 , this allows us to find that FB which is 4.

Now the only thing left to do is cross multiply DC =4.

----------------------------------------------------------------------------------------

Number 19: Find the value of x.

We know that because they are similar triangles. We cross multiply and we get 280-14x=11x. And x=11.2

----------------------------------------------------------------------------------------

Number 24: Find SN.
Given: SV bisects angle ASB, AS=7, AR=11 and SV=9. Triangle ABS~triangle RTS.

Since both triangles are ~ and SV is an angle bisector we can figure out that AS is proportional to SN and SR is proportional to SV. Knowing this we can cross multiply which gives 63= 18SN and finally SN= 3.5

----------------------------------------------------------------------------------------

Number 28: Find the value of x.
Given: both triangles are similar. The perimeter of triangle ABC is 28.


We are first going to find the perimeter of triangle RST by adding all sides, so the formula would be P=10+ x-4+ x-1.5

P=2x + 4.5
Now that we found the perimeter of RST we can cross multiply giving a result of 16x + 36=280 and therefore x= 15.25

----------------------------------------------------------------------------------------

Number 29: find CF and BD.
Given: BF bisects angle ABC and AC ED. BA=6, BC=7.5, AC=9 and DE=9.

First we are going to find CF, we already know that FC is similar to ED so CF= 9-x. To find x we cross multiply so it's and we get 7.5x= 54-6x
13.5x=54
x=4

Since we found the value of x we now need to replace it and CF=5 and BD is 13.5

----------------------------------------------------------------------------------------

After we finished checking our h.w, mrs. Saatkamp gaves us new h.w due Tuesday, the h.w is the review for the test on Wednesday in pg. 388 q's 13-30. Questions 27-30 are similar to the ones that are going to appear in Monday's quiz so she suggested for us to do does questions in class and then if we had any doubts we could ask her and it would had been a good way of revicing for the quiz, instead we all whent into the math lab and worked in our Geogebra project.