## Sunday, June 12, 2011

### Chapter 2 Project

We picked a problem form our textbook and we solved it.I chose a problem about a fish tank and this is how you slove it.

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## Monday, May 9, 2011

### Ch. 2 - Solving Equation in One Variable

Click for clearer view.

## Friday, May 6, 2011

### How to Do Chapter 2 Project

Chose a word problem to write up and solve. Please use illustrations to show what is happening in the real world.

## Thursday, May 5, 2011

### Ch7 Project How to do Make a Budget

Hi my name is Rany and this is my Ch7 Project. I am asked to pick a career and a house that i will be living in. I chose to be a Video Game Programmer and my Salary will be \$58000 a year for my pay and each year, i gain a 10% of my pay each year and so 10% of \$58000 is \$5800 so, i get an amout of \$5800 each year. The college that i am planning to go to is San Jose state as my college and how i am going to pay for that is through my Financial aid with the grants i get for being poor. The financial aid will be paying 90% of my college pay and the rest of the 10%, my pay will come from my job. So, that means i will not have any student loans for me to pay off because its best to not have any students loans then to have to pay for some. The Job i will be having is Called Video Game Programmer which i will be working at EA games. I will have a Salary of \$58000 a year. I will have Benefits like Medical, Health, Vision and even though its not concluded on the picture, i also have Vacation benefits from my job. I will also have a Retire Plan called the 401k plan. The place i will be living in is the Summerville Apartments in San Jose Which has 2-3 bedrooms(depending on which one you want) 2 bathrooms and the cost of the apartment of rent is \$1275, But i will have a roommate so i will have to pay half of my rent and my roommate pay for the other half which is \$635 for each of us. I will go to work biking on Mondays, Wednesdays and Fridays. I will use the bus on Tuesdays and Thursdays, so overall i will have to pay about \$50 bucks for my bike and bus. The car that i will be having is an Lexus IS 350 that has a 306 hp, with a six-speed paddle shift sequential automatic electionically controlled transmission. It goes 18mpg in city driving and 22 mpg in high driving. The payment for my Car i will use are about \$120 for my gas and so overall of that i will use \$170 for Transportation each month. That is the end of my Ch7 Project How to make a budget of my life in the future.

## Wednesday, April 27, 2011

Pg. 272 Chapter 2 Review

a)b)
c) The reason why more money is being spent in hospitals is because of the high demand for doctors and new high-tech technologies being developed.

past tech.new tech,

## Sunday, March 20, 2011

### Chapter 5 Project: Determining the Number of Triangles in the Ambiguous Case (SSA): A Construction Exploration

Question 1:
Sin C is the same in both triangles in the ambiguous case because Sin A is the same as Sin C.

Question 2:
The thing that I discovered about the number of triangles you can make given SSA (two sides and an angle is that if BC is < or =" AB,">

Obtuse Triangle
a. BC > AB makes one triangle
b. BC = AB makes zero triangle
c. BC <>

Acute Triangle
a. AC > BC > h makes two triangles
b. BC > AC > h makes one triangle
c. BC = h makes one right triangle
d. BC <>

## Saturday, March 19, 2011

### Ch. 5 - Determining Number of Triangles

Exploration 1. Case #1
Exploration 1. Case #2, 3

Exploration 2. Case #1, 2

Exploration 2. Case #3

Exploration 3.

Question 1: Explain why sin C is the same in both triangles in the ambiguous case (exploration 3). This is why the Law of Sines is also ambiguous in this case.

SinC is the same in both triangle because SinA remains the same since according to the law of sin, sinA= sinC.

Question 2:
What did you discover about the number of triangles you can make given SSA (two sides and an angle)? Summarize the "rules" for all cases.

If angel A is obtuse and...
• BC > AB....1 triangle
• BC = AB....0 triangle
• BC <>
If angel A is acute and....
• BC <>
• BC = h....1 triangle
• BC > AB > h....1 triangle
• AB > BC > h....2 triangles

## Thursday, March 17, 2011

### # CH 5 How to Do Determining the Number of Triangles in the Ambuguous case (SSA): A Contruction Exploration

In each of the three different cases, I constructed three triangles with a 120 degree angle but different relationships between the lengths of the sides.

In Case 1, where line BC was longer than line AB, I discovered that one triangle could be formed while keeping angle A at 120 degrees.
In Case 2, where line BC was equal to line AB, it was impossible for line segment AC to form since line segment BC was not long enough to connect with the dashed line. Therefore, no triangles were formed.
In Case 3, where line BC was shorter than line AB, the same problem in Case 2 arose. Line segment BC was not long enough to connect with the dashed line. Again, no triangles were formed.

In conclusion, I found out that in order for an obtuse triangle to be created, the following rules apply*:

• If BC > AB, then one triangle is possible.
• If BC = AB, then no triangles are possible.
• If BC < AB, then no triangles are possible.
*Note: The lettering scheme I use may be different than the lettering schemes used byothers. However, the same concepts I mention here still apply.
In my constructions:
• Line segment BC refers to the side opposite of the obtuse angle measuring 120 degrees
• Line segment AB refers to the leg of the angle measuring 120 degrees which touches line segment BC as mentioned above

These three cases here were based on acute triangles, and the relationships between the lengths of the sides and the height of the triangle. In each of the cases, in order for the triangle to still have an angle A measuring 30 degrees, point C of line segment BC would have had to connect with the line segment.

In Case 1, line segment BC was shorter than the triangle’s “minimum” height, so therefore no triangles were constructed. The dashed red circle I drew represents the fact that no matter where point C is placed, line BC will never connect with the dashed line.
In Case 2, line segment BC was equal to the height, so one triangle was constructed.
In Case 3, line segment BC was greater than line segment AB which was greater than the height. One triangle was possible.

In conclusion, in an acute triangle:
• If BC < h, then no triangles are possible.
• If BC = h, then one triangle is possible.
• If BC > AB > h, then one triangle is possible.

In Exploration 3, I drew a triangle where line AB was greater than line BC which was greater than the triangle’s height. I discovered that two triangles were possible while keeping an angle A measuring 60 degrees. The triangle in Exploration 3 is similar to the triangles in Exploration 2. If this triangle were an Exploration 2 triangle, then point C would be able to connect to the dashed line in two places, therefore creating two triangles that we can see here. Basically, there is enough room for line BC to exist in two places.
In conclusion, if in an acute triangle:
• If AB > BC > h, then two triangles are possible.

Questions
1. Explain why sin C is the same in both triangles in the ambiguous case (exploration 3). This is why the Law of Sines is also ambiguous in this case.
Let’s recall the values for the triangle in Exploration 3. Line AB equals 8 cm (in the Law of Sines formula, this would be side c), angle A equals 60 degrees, line BC equals 7 ½ cm (LoS formula: side a), and the height equals 7 cm.
The Law of Sines is ambiguous in this case since two different triangles will share part of the same Law of Sines formula since they have a congruent angle and two sides (otherwise known as the misleading SSA congruency conjecture). In the triangle in Exploration 3:
sin 60 = sin C
7.5           8
The Law of Sines here misleads us into thinking there is only one triangle, when two are possible. If one were to solve this equation they would only get one of the triangles. It would require one to solve the sin B/b part (which is where the ambiguity comes in), to get the full picture.
2. Explain from diagram 4 in Exploration 2 why a unique triangle is determined if BC < AB.
In acute triangles, a triangle can be constructed when BC < AB since point C of line BC has enough length to “connect” to AB. However, two triangles are not possible since unlike in Exploration 3, there is not enough “room” for point C to connect in two places without point C overshooting line AB.
3. What did you discover about the number of triangles you can make given SSA (two sides and an angle)? Summarize the "rules" for all cases.
Given SSA, the case in Exploration 3, two triangles can be made.

All in all:

• If in an obtuse triangle:

• If BC > AB, then one triangle is possible.

• If BC = AB, then no triangles are possible.

• If BC < AB, then no triangles are possible.

• If in an acute triangle:

• If BC < h, then no triangles are possible.

• If BC = h, then one triangle is possible.

• If BC > AB > h, then one triangle is possible.

• If AB > BC > h, then two triangles are possible.

Since other people use different lettering schemes, I will list how my line segments correspond to the Law of Sines formula:
• AB = c
• BC = a
• AC = b

## Wednesday, March 16, 2011

### Anna Algebra

Question 1: SinC remains at the same point because SinC is the vertex of the triangle, it determines the height of the triangle and the length of the sides.
Question 2: In an obtuse triangle, you would always want the sides add up to more than the hypotenuse for a triangle to be able to form. In an acute triangle, the height would always need to be the smallest, like BC>AC>h. The general rule to be able to make a triangle would be AC>BC>h.

### How to do Determining the number of Triangles in the Ambiguous case (SSA): A construction exploration

Exploration 1a:

AC = 120 degrees
AB = 6 cm
BC = 7 cm

1) Draw a 120 degree angle, 7 cm, 6 cm, and 5 cm line (6 and 5 cm line also used for other two diagrams in exploration 1)
2) Copy angle A
3) Fix compass to length 6 cm. Place pointy end on A and pencil end on the intial side. This will be point B.
*Step 1-3 are also used to start the other two diagrams of exploration 1
4) Fix compass to 7 cm length. Place pointy end on B and make an arc that intersects the terminal side (ray AC) The intersection will be point C.
5) construct a perpendicular to form height.
Exploration 1b:

Exploration 1c:

1) Repeat Step 1-3 of first diagram of exploration 1
2) Fix compass to 5 cm length. Place pointy end on B and make an arc that intersects ray AC. The arc did not intersect AC, it intersected AB. Therefore creating a line, not a triangle.
Exploration 2a:
1)Draw 30 degree angle with a protractor
2) Copy 30 degree angle
3) Draw ray AC and AB
4) AC can be any length, in this case, I fixed the compass to length 7 cm
5) Construct a perpendicular from AC to ray AB to form a altitude
*These 5 steps are the same for the other two triangles in exploration 2

6) First Case, let BC be less then height, which eventually, it won't form a triangle, because if BC is less than h then BC won't touch ray AB (point B is nonexistence)
Exploration 2b:

Exploration 2C:

Exploration 3:

1) Draw a 30 degree angle, 7 cm line segment and 6 cm line segment
2) Copy the 30 degree angle
3) Fix compass to 7 cm length. Place pointy end of compass on angle A, then place pencil end somewhere along the terminal side of the angle. That point will be AC.
4) Fix compass to 6 cm length. Place pointy end on C, then place pencil end to the left of C and right of C. Make arcs. These will be the two B points.
5) Construct altitude by constructing a perpendicular bisector.
6) Measure h.

Questions:
1) Because sin A remains the same in the ambiguous case, sin c also remain the same
2) If angle A is obtuse, then side BC (long side) has to be greater than side AB in order to form a triangle. A triangle cannot form if BC is less than or equal to AB.
In a right triangle, side BC has to equal the height of the triangle. BC cannot be less than h.
BC could also be greater than h, but BC also has to be greater than AC as well and angle A has to be acute. Though this is not a right triangle, one obtuse triangle and two acute