>>Today we started class reviewing some concepts we learned yesterday:

Chord - segment whose endpoints are on the circle. The greatest chord is the diameter.

Central Angle - angle whose vertex is circle's center, and forms major and minor arcs.

Measure of arc - measure of central angle (degrees)

Length of arc - central angle divided by 360 times circumference

Miss Saatkamp also taught us the difference between a point on the circle, a point in the circle and a point outside the circle.

A is center
B is ON the circle

C is IN the circle

D is OUTSIDE the circle

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After that, we corrected homework.

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Today we also learned new theorems and a new concept.

-Theorem 9-1:

if: chord BC is congruent to chord DE (green lines)

then: Arc BC is congruent to DE (pink lines)

*Inscribed Polygons - Polygons whose vertices are ON the circle (ex: Polygon BCDE)

-Theorem 9-2:


if: BC is a diameter; DE is a chord; BC is perpendicular to DE

then: Chord DE and arc DE are bisected

*Extention: With radii the theorem works the same way.

-Theorem 9-3:

if: BC and DE are chords; BC and DE are equidistant from the center (meaning AG is congruent to AF)

then: BC is congruent to DE

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At the end we got our homework, which is only due MONDAY - Page 462 #13-27 all + 32-34

DON'T FORGET!! Projects are due NEXT WEEK!!!

Today's class for 9B

Today we continued the lesson we were learning yesterday, because we didn't finish it. This was about arcs and angles.



We learned some definitions of types of circles and arcs:

Concentric circles: circles (two or more) that have the same center.








Congruent circles: circles with the same radius, and therefore have the same diameter, and so have the same circumference.



Similar circles: All circles are similar, in the same way that all squares are similar, and all equilateral triangles are similar (although circles aren't poligons).



Congruent arcs: Arcs that have the same length (but not the same measure! To find length, you have to use the formula we learned in the previous class).



Then we went to the computer lab to see and try to correct a mistake that someone in the other class had made in the blog post. After noticing and the mistake and correcting it, we went back to the classroom to do the homework assigned.



* Homework is on page 456, #19-51 ALL. *

Today's class

Today we started our class cheking the homework and with the correction of some exercises of the test. After that we moved on with the content. We started Angles and Arcs.

Central Angle: angle whose vertex is the center.

Sum of the central angles: 360

ARCS

Minor Arc: arc AB

Major Arc: arc ACB or major AB

Semicircle:

Tips:

Measure is different than length.

Measure of the arc: central angle.

Length of an arc: l= central angle x C

360

Postulate 9-1: Arc addition postulate.

This postulate is similar to the segment addition postulate. If you add two arcs, it will give you one whole arc.

arc AB + arc BC= arc AC

Concentric circles: Concentric circles are circles that share the same center

Congruent Circles: They have the same r

All circles are similar.

Congruent Arcs: Arcs with the same length.

We don't have any homework for tomorrow.

Lesson 9-3 part 1.

Today in class we recieved are tests back, corrected them, and continued studying chapter 9: circles. 

In the test, no one got number ten right and Mrs. Saatkamp got really mad because no one knew what was the formula to solve quadratic equations. 

Quadratic formula:

Central Angle: an angle whose vertex is the center. 

The sum of all central angles will always be 360º. 

Arcs: a portion of the circumference of a circle. 

- minor arc: smaller arc (represented by ) 

- major arc: bigger arc  (usually represented by )

The MEASURE of an arc is equal to its central angle. 

The LENGTH of an arc is the ratio of central angle:360º times the circumference. 

Semicircle: an arc whose central angle is equivalent to 180º.

Postulate 9-1: Arc Addition postulate. 

two ADJACENT arcs can be added two form a greater one.   

March 18 class

Today we corrected the quiz and started learning about circles and its components. Definitions:

Circle: it is a set of points equidistant to a given center; Name: [a circle with a dot] P

Radius: a segment whose endpoints are the center and a point on the circle. (R)

Chord: a segment whose endpoints are on the circle. (C)

Diameter: a segment whose endpoints are on the circle and crosses the center. (D)

Circumference: the perimeter of a circle.

The Homework for Tuesday is: pg. 449 # 14-38 (all). Don't forget to study for the test on Monday!!!

mrs Saatkamp sorry for the big space white i have created in the blog (i have no idea how i did it)...

Today´s class 03/09/09

Today´s class we started by correcting the homework. It was pag.422 #8-26 all (about angles of elevation and depression). Mrs Saatkamp correct some of them in the board.

After correcting the homework we started a new lesson: 8-5:

Sine rule (Law of sines):

This rule is based on that a triangle (no right triangle), we have to find the legs, angles...we have to use cos, sin and tan, but because is not a right triangle then we cannot use it. So to use sin, cos and tan, we have to draw an altitude, and is then when we get a 90 degrees angle. After that it´s just normal work with sin, cos and tan.

*Mrs Saatkamp, i don´t know what is happening with the images from geogebra but i cannot post any. It says that it´s an internal problem and that i have to try later. So mrs, if you want I can send you the images or whatever you want.

Formula:

Proof:

After that we did 2 exercises to practice, but because I don´t know what is happening with the pictures that I cannot post the images. But the exercises were based on the sine rule, like to find missing information, or a determinate leg...

our homework for tomorrow is pag 429 #10-26 even +27

see ya guys!

*Mrs Saatkamp again, sorry for the images!

09/03/2009 9B

First we corrected the homework from page 422 numbers 8 to 26 except 10 and 23. Mrs. Saatkamp also corrected some of the exercises on the board, and the one the majority of the class needed was 26, which the answer was 13.7.

After that, we started lesson 8-5:
this lesson talks about the Sine Rule (our book will call it: Law of Sines)

Proof:

After learning this law we did 2 exercises to practice:
1)

We had a triangle with the measure of two angles and one side and we had to "solve the triangle" using the Sine Rule, which means to find all the missing information, sides and angles.

2)


We had the measure of two angles, one side, and we had to find the altitude.


Our Homework is P.429 # 10-26 even + 27

Hints: - SOH CAH TOA is only for right triangles
- WE usually will round to the nearest tenth the decimals of this lesson

Today we began our class by starting a new section, section 8–4. We learned 2 new definitions: definitions of the Angle of Elevation and of the Angle of Depression.

Angle of Elevation: angle in which one leg is horizontal and the other leg goes up.

Angle of Depression: angle in which one leg in horizontal and the other leg goes down.

Be careful!

*In order for an angle to be classified as any of the angles above, it needs one horizontal leg!
*In one right triangle, if one acute angle is of elevation, it does not mean the other angle will be of depression!

These definitions will be used in word problems, so we are still going to use the theorems of the last sections.

After that, we talked about the quiz… which wasn’t so good. Don’t forget that in math the best way to learn is by practice. In case you want to improve your grade, there will be a Recuperation Test on Monday, which will be about chapter 7.
Don’t forget the homework! Page 422, # 8 – 26, all except 10 and 23
;D

Today's Class 6/03/09

Mrs. Saatkamp began the class today by giving our quizzes back. The grades were not the best so far so Mrs. Saatkamp tols us that in order to be able to understand our future chapters we need to understand better chapter 8, so I would recomend you to study a little bit more. Then Mrs. Saatkamp answered some questions from the quiz.

6) In this kind of questions you have to use pythagorem theorem so we could use calculator to solve it.




11) Because the triangle is equilateral we can infer that the three angles are congruent, each one measuring 60°. The perimeter of the triangle is 48 so divide it by three and each side should be 16. For this question we had to use the table we learned.



Then we began a new lesson. We learned two new terms:

Angle of Elevation: an angle with a horizontal side and the other side facing up




Angle of Depression: and angle with a horizontal side and the other facing down





So here are some hints so you dont confuse the angles in a right triangle



So in thsi right triangle the
green angle: angle of elevation
pink angle: angle of depression
Be really careful with the blue angle which is NOT a angle of depression. If there was a line from the vertex then the angle formed would be a angle of depression (in this case, the pink angle), this is because the line that was drawn is parallel to the horizontal line.

Todays homework due on Monday is page 422 # 10 to 23 except for 10 and 23

So that was today´s class and dont forget that the recuperation test for CHAPTER 7 is on Monday after school.

Sorry I did not post this yesterday.

Ok so in yesterdays class we began by correcting the homework of the night before (pg 409 11-27 ALL). One of the questions we did was a#23 which gave us the length of the altitude which was 5.2 and we had to find the perimeter of the triangle.

1. Sin 60=5.2/x

2. sqrt(3)/2=5.2/x

3. xsqrt(3)/sqrt(3)=10.4/sqrt(3)

4. 10.4 sqrt(3)/3

5. Move the decimal in 10.4 2 times to the left getting 104 sqrt(3)/3

6. Divide that by 2 getting 52sqrt(3)/3

7.Multiply that by 3 that equals 6

8.6 is the length of one side

9. Multiply 6 by the 3 sides of the triangle concluding

that the perimeter=18

After we finished correcting the homework we used the rest of the class to work on out projects (due March 31)

OH don't forget that we have a quiz tomorrow (march 5,09) on

lesson 8-1: geometric mean

lesson 8-2: special right triangles

lesson 8-3: SOH, CAH, TOA

andy

andy

So in today's class, we started off with checking our homework which was pg. 409 #11-27all. After done checking the problem that the class didn't understand the most was problem 27.

The problem 27 says this:

Triangle PCD is a 30-60-90 with right triangle with right

Now with some logical reasoning, we can count the spaces from point C and D from the y axis and you will get 13 spaces. Now since we have a one angle which we can pick 30 degrees and there is an opposite site and adjacent side we can use TAN to find out the coordinates of P.

Then after you work out the problem and should get:

Also from SOH CAH TOA, we know that so far P=(x, -6). After we have almost everything we just need the x of coordinate P, then we should subtract the answer of PC by -3 since we want to move further left and make sure that we are staying in Quadrant III.

And finally to finished this off the answer for be should be:

At the end make sure to remember the Table that Mrs. Saatkamp taught us because we will probably need in this quiz which is on Thursday. Also the deadline for the math project is on March 31, so make sure that you have everything done before the deadline. Also in tomorrow class, we will just be working on the project so don't stress out!

Kisses,

Flow ;3

Class summary. 020309

Today we began our class with correcting the homework we had over the Carnival.

I think the homework was pretty easy for almost everyone, we went over some questions, but i think things were clear for everone.

One of the ex. we went over, however, was nummber 49.

So you were supposed to find X and Y.

to do that you need to find out how much the two congruent sides are.

you apply the pythagorem theorem, and then when you have solved everything the whole side should be 20,12, which means that each segment is 10,06.

After that you can use TAN to find y;

To find Y you have to do the following;

when you are done with that you do the same thing for X. The answer should be 29.2 and 41.8.

Todays Class!

Special Right Triangles!

we looked at two theorems today in class;

Theorem8-6: In a 45gr-45gr-90gr triangle, the hypotenuse istimes as longas a leg.

Theorem 8-7: In a 30gr-60gr-9gr triangle, the hypotenuse is twice as long as the shorter leg, and the longre leg istimes as long as the shorter leg.

This is a table you easily can use now when you are going to start doing the homework.

Guys We neeeeed to remember this table!! you will not be able to have the table in front of you on tests and Quizes!

So.. now... the Homewooork:) shiihuu!:D

Page 409 #11-27 All!

Good Luck:D See u tomorrow:D:D