Reflection #14

Okay so this week was a little rough, but i'm finally glad it's over and it's THANKSGIVING BREAK! This week we learned more about trigonometry in Chapter 8. We learned some formulas. Here they are:

For a line:

m=tan(alpha) where m=slope, alpha=angle of inclination

For a conic:

tan 2(alpha)=B/A-C

For a conic if A=C then alpha=pi/4

Example: 2x + 5y = 15 find the angle of inclination

m=-2/5 tan(alpha)=-2/5

alpha=tan^-1(-2/5)

alpha= 158.199 degrees, 338.199 degrees

To arrive at this answer, it uses the same principle as trigonometric inverses. You use the same procedure to find the inverse and you should get two angles as your answers.

The only problem I'm really gonna have is with the curves of sine and cosine. I'm just gonna have to practice some more and do homework, and we also have to remember those formulas B-Rob said we were gonna have to remember when we get back from the break. So i hope i do good on that test on the Friday we get back.

area of a scalene triangle

ok so this may be a dumb question but how do you find the area of a triangle such as the one on number 1 of the mixed review?

reflection 13

another week goes by, and im looking forward to thanksgiving break. this week was based on triangles.

law of cosines
opp. leg^2= adj. leg^2 + other adj. leg^2-2(adj. leg)(adj. leg) cos(angle b/w).

reflection #13!

Ok, so its reflection 13, and im really tired. hahaha. its 1130 which means its 2 and a half hours after my bed time! i need some sleep! so this is what it is, the math of course.

The law of cosines formula is opp. leg^2= adj. leg^2 + other adj. leg^2-2(adj. leg)(adj. leg) cos(angle b/w).

i just know the formula and how to use it!

------------------------------------------------------
can someone help me with deciding when to use the law of sin or the law of cos...because i have no idea which one to use and when to use it!!!!!!!!!!!!!!

reflection #13!

Okay, so this is REALLY reflection 13!
for all of you that i confused last sunday, SORRRY! :p

anyways, this week was kind of my favorite being that we learned another formula and that's pretty much the only time i understand ANYTHING.
so as for this blog, i'm just going to explain law of cosines:
well for starters, law of cosines is not used with right triangles and is normally used when only given length of sides and the angle in between rather than two angles and a side(law of sines).

FORMULA:

(opposite leg)^2=(adjacent leg)^2+(other opposite leg)^2-2(leg)(leg)cos(angle in between)

for example:
for a triangle with C=36 degrees a=5 b=6

c^2=5^2+6^2-2(5)(6)cos36
c=SQRRT(25+36-2(5)(6)cos36)
c= 3.53
--------------------------------------------

So, on the Ch. 9 quiz and study guide, i was COMPLETELY confused as to how to completely solve the triangle because i was confused if you could switch back and forth from law of sines, to law of cosines? once you've found it one way, and you're looking for a side or angle is it possible to switch to another way? will the answers still be the same?

if anyone can understand that question(sorry i couldn't explain it any better), please help!
thanks!

Reflection 13

Well this week wasn't that bad. It was pretty easy and we learned how to do the law of cosines.

For the law of cosines you use it on a triangle with no right angle. You have to now know one side but have the other two and the angle in the middle of those. The formula is:

(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)

This is how you use it:

Given triangle ABC, angle C is 36 degrees, a = 5, and b = 6, find c.

x^2 = 5^2 + 6^2 - 2(5)(6) cos (36 degrees)

Plug it in to your calculator

x = square root of ((6^2 + 5^2 - 2(5)(6) cos 36 degrees))

x = 3.53

Thats really the hardest thing that we did this week so I dont really have any questions. Some times the word problems confuse me a little bit but past that its all good.

Reflection #13

Okay, so this week was not that bad at all, considering we had a four-day week. It went by pretty fast, and we learned, as you would guess it, more trigonometry this week! I am going to describe how to do the law of cosines.

To use the law of cosines, you need a non-right triangle, an angle measure with an unknown opposite side, and two side lengths on each side of that angle measure.

Here's the formula:

(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)

Example:

Given triangle ABC, angle C is 36 degrees, a = 5, and b = 6, find c.

you would then plug in to the formula:

(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)

x^2 = 5^2 + 6^2 - 2(5)(6) cos (36 degrees)

x = sqrt of ((6^2 + 5^2 - 2(5)(6) cos 36 degrees))

x = 3.53

to get this answer, you typically plug the whole equation into your calculator.

The only thing I don't understand is when you have to solve the whole triangle, when you use law of cosines, after you figure out one side or angle from the law of cosines, can you continue to find the other angles and sides by say, using the law of sines? Can someone please give me an example of one of these problems? Thanks!

Reflection 13

Alright soo...I dislike trig very very much. Something I understood though was law of cosines. You use law of cosines for non-right triangles that don't have pairs. The law of cosines formula is opp. leg^2= adj. leg^2 + other adj. leg^2-2(adj. leg)(adj. leg) cos(angle b/w). There are two different problems you use this for ones when you know the opp leg, and ones where you don't. For the problems where your finding the angle you have to do the inverse.

Example: Triangle ABC, a=8, b=5, angle C=60 degrees.
*In this problem you have an angle inbetween two sides and your trying to find the opp. leg. So you would do x^2=8^2+5^2-2(8)(5)cos60 degrees. Square root both sides and your answer is x=7.

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

I don't understand a lot of the word problems. If someone could explain 17, 18, or 19 to me on page 355 from the homework it would help. :)

reflection 13

This week we are still learning about triangles and how to find there angles and legnths.The new formula that we've learn was LAW OF COSINES. I still dnt understand this completly . Iknow your not suppose to use it wit a right triangle but i dnt know when im suppose to use it...

one thing i do know is area = 1/2(leg) (leg) sin( angle between). this is used for a right triangle...

Example you hav a triangle that has a 60 degree angle with sides of 4 and 6...

first you plug your numbers into your formula and then plug into your calculator...

1/2* 4 * 6= 12
12 *sin 60 =10.392
x=10.4

homework?!

Does anybody know all of the homework that we had to do? I know that she is checking tomorrow, i just don't know what she is checking. Can someone please tell me what pages it is.

reflection 13

this week went by extremely fast since we were off on monday. we learned more stuff about triangles like the law of cosines. and that formula is:
(opposite leg)^2=(adjacent leg)^2+(other adjacent leg)^2-2(adj leg)(adj leg)cos(angle between)

heres an example:
if you have a triangle with the sides of 5, 6, and 7 and your looking for the angle in between 5 and 6, you would set up the problem like so...

7^2=6^2+5^2-2(5)(6)cos a
7^2-6^2-5^2= 2(5)(6)cos a
cos a= 7^2-6^2-5^2 over -2(6)(5)
a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))
a= 78.463 degrees

REFLECTION #13

So I thought this week was a pretty good week overall. Especially since we had that day off on Monday. Actually if it was up to me, we'd have Monday off every week because math is crazy on Mondays. hah. But anyway, it wasn't all that bad this week. We continued learning how to solve angles and sides of triangles using different methods. We also did a bunch of word problems dealing with triangles (which I thought were pretty hard, but I will discuss that after) And this week on Wednesday I think, we learned the Law of Cosines and the formula is >> (opp leg) = (adj leg)^2 + (other adj leg) - 2(adj leg)(adj leg) cos(angle between). Now that looks like a long formula to memorize but once you've done a few problems using it, it's easy to remember. Law of Cosines can be used whenever you're dealing with a non-right triangle and you are only given one angle measure and two side lengths. It can also be used when you are dealing with a non-right triangle and you are looking for an angle and you are only given the three side lengths. So as of this week, I completely understand SOHCAHTOA, the Law of Sines, and the majority of Law of Cosines (there's just a few things I don't get about it)

Here's an example of a problem where you would have to use Law of Sines:

Ex. 1.) Angle B = 30 degrees, Angle A = 135 degrees, and side b = 4. Find the measure of angle C and the side lengths of a and c.

*Alright so the first thing you want to do is find your other angle because that's the easiest thing to do first. So you add 30 degrees and 135 degrees together and get 165, then you subtract that from 180 and get 15 degrees as your other angle. So Angle C is 15 degrees.

*the next thing you need to do is find one of your sides. Since you have pairs (30 degrees and side length 4) you can use that in the law of sinces formula. So your equation is sin 30/4 = sin 135/a >>because you are looking for side a. then you cross multiply and get asin30 = 4sin135 So a is approximately 5.657

*then to find side c you set up the formula like this >> sin 30/4 = sin 15/c then you cross multiply and get csin30 = 4sin15 So c is approximately 2.071
AND CONGRATULATIONS !!! YOU'VE SOLVED YOUR TRIANGLE! :)
..but don't get too excited, there are ones waaaaaaayyyy harder than that. ha

Okay now here's an example where you would have to use Law of Cosines:

Ex. 2.) Angle A = 32 degrees, side c is 5 and side b is 6

Okay for this, you simply just plug the numbers you have into the law of cosines equation. but first you have to draw your picture so you can see where everything is and you can plug the numbers in the correct spots of the formula.

So your equation is >> x^2 = 5^2 + 6^2 - 2(5)(6) cos 32
since you have x^2 you have to take the square root of everything next sooo...
x = squareroot of (5^2 + 6^2 - 2(5)(6) cos 32) >>and you just plug that into your calculator and you get 3.181 as you other side length.


*Now for what I don't understand. Well first of all I don't understand a lot of the word problems from our homework, like I said before. And I'm not sure how to find the area of a non-right triangle that you used law of cosines on (like the example I just gave. I don't know how to find the area of that. I'm confused) Well if anyone would like to help me out before that scary test on Tuesday that would be greatly appreciated :)

reflection 13

this blog starting with what i dont understand and sitting across from connor watching him figure it out makes it even more impossible for me to get it. number 21 on page 356 in mixed trig review, i know we worked one in class but i cant follow my scratch work, after you seperate the shape into two triangles what do you do to split the angle, and do you have to find the hypotenuse. i thought maybe just finding the other angle and working it to find the sides would work, but thats not helping. okay moving on to what i do know.

now moving on to law of sines which i do understand this week,

here is the formula for you to follow:

sinA/a=sinB/b=sinC/c

example: a triangle ABC has

simply follow sin A/ 123= sin B/ 16

sin A= 16sin115º/123

and sense you are finding the angle of of you have to use the inverse of sin

A= sin-1((16sin115º)/123)

giving you the answer A= 6.771º

another thing i figured out this week was for the problems 14 and 15 on page 335

problem 15 with the flag pole your are just finding the length from where the viewer is to the top of the flag, meaning in your answer you must add whatever the height is from the ground to the viewer.

:)

Reflection Thirteen

This week we were still learning about triangles and how to find angles and lengths. One new formula we learned was the Law of Cosines. This is what I do not know how to do though. One thing I do know how to do well is A= 1/2 (leg) (leg) Sin (angle between). This is used for non-right triangles.

For Area say you have a triangle with:

A side length of 4
A side length of 5
And an angle of 30 degrees

Then plug it all in: A=1/2 (4) (5) Sin 30 degrees

A= 10 Sin 30 degrees which aproximately = 5.


For another triangle:

A side length of 3
A side length of 8
And an angle of 60 degrees

Then plug it all in: A= 1/2 (3) (8) Sin 60 degrees

A= 12 Sin 60 degrees which aproximately = 10.392

This formula can only be done when you have two given lengths and at least one angle.


I know that the Law of Cosines is (opp. leg)^2=(adj. leg)^2-2(adj. leg)(adj. leg) Cos (angle btw)
I can plug it in but I get confused at the end.

Reflection Thirteen

reflection 13

Dude! Why do ppl have to be retarded! And why does there have to be so many rounds of the playoffs. Why can't there be like 2 rounds, and then u either win it all, or u dont. It would take up less time for the fans, and they would get to c the trophy faster :D. But still, not alot of ppl want to see the end of the season, unless it brings home a trophy. Lookin at our side of the bracket, we got (..) that much of a chance of gettin past the 3rd round. No doubt we gonna get past next week, as long as we play like we did on saturday, but after that, John Curtis >.> Then, if we win by some miracle, we would have to play SCC. Then, after that, if we win by SOME MIRACLE, we would have to play Evangile (I think that's how u spell it) in the state championship. That is one team that would definitely PWN us. XD Just sayin, no offense to the football team, but it doesnt look like yall have a shot at gettin past the Thanksgiving holidays :/ but still, gonna miss playin music at the football games, and yellin alot at the games. Unfortunately, we wont have our alleged leader Tir on friday night, so its gonna b a little hectic >.> but we'll survive. Can't wait to see the Rebels win on friday, its gonna be soooo awesome. We gonna beat the 9th seed the week after we beat the 8th seed, double upsets :D and that bridge! omg, its gonna be crazy tryin to do that thing! :X o well, at least we get groups instead of doin the whole thing by ourselvs (btw, Brob, if ur readin this, u think u can give me some info on that project for stats, i kinda forgot like everything about it :/ but im not gonna not do it, thats why i need some more info :/) anyways......back to math......

so...one thing i did understand this week was the Law of Cosines

(opposite side)^2=(adjacent leg)^2 + (other adjacent leg)^2 -2(adj leg)(other adj leg)cos(angle between the legs)

u use that when you're dealin with non right triangles and ur tryin to find different parts of the triangle

but one thing i didnt understand was the whole area of a non right triangle thing

i get the formula, but its only for a right triangle, how are u supposed to get the area the whole thing after u cut the whole triangle into 2 right triangles, after u find the area of one of the right triangles :/

can anybody help me out with that?