Reflection #13

Okay, for what I understand:
Law of sines and law of cosines

LAW OF SINES:
sin A/a = sin B/b = sin C/c
**used when you have a pair

Example:
Visualize a triangle ABC:
A = 47 degrees
a = 8
B = 92 degrees
Find b.
sin 47/8 = sin 92/b
Cross multiply.
b sin 47 = 8 sin 92
Divide.
b = 8 sin 92 / sin 47
Plug into calculator.
b = 10.932

Visualize a triangle DEF:
D = 82 degrees
d = 12
e = 5
Find E.
sin 82/12 = sin E/5
Cross multiply.
5 sin 82 = 12 sin E
Divide.
sin E = 5 sin 82 / 12
Take the sin inverse of 5 sin 82 / 12.
E = 24.369 degrees

LAW OF COSINES:
(opp. side) = (adj. side)^2 + (other adj. side)^2 - 2(adj. side)(other adj. side) cos (angle between)
**used when you can't use law of sines

Examples:
Visualize a triangle ABC:
A = 78 degrees
b = 8
c = 3
Find a.
a^2 = 8^2 + 3^2 - 2(8)(3) cos 78
Square root both sides and plug into calculator.
a = 7.939

Visualize a triangle DEF:
d = 7
e = 12
f = 9
Find D.
7^2 = 12^2 + 9^2 - 2(12)(9) cos D
Subtract over 12^2 and 9^2.
Divide -2(12)(9).
cos D = 7^2 - 12^2 - 9^2 / -2(12)(9)
Take the cos inverse and plug into calculator.
**Be careful about parenthesizes when you plug into cal.
D = 35.431 degrees



Now, for what I don't understand:
Nothing in general that I don't get, but a few (well actually a lot) problems from homework I'm stuck on.
page 355, #s 7, 8, 11, 17b, 18, & 20-24
page 365 #s 5 & 10
I know it's a lot, but if anyone could explain how to do some of these problems, that you be a big help.

Reflection11, 12, 13 I think

Well for the past week and i half i was in the hospital so i wasnt planning on doing to much math. This morning i went over a few of the notes to try to get back in the rythm of things, and remembered what we did go over b4 my little incident. I did manage to figure out how to solve an area of a non right triangle. It's A=1/2(leg)(leg)sin(angle), and if u arent given an angle measure have to find its inverse. And i was given a few notes on the law of sin, but i'm going to matt house 2morrow so he can teach me how to do them.

reflection 12

this week went by pretty fast i guess. since monday was saints day and all we did was do homework problems and went over stuff most of the week. i thought i did pretty good on my quiz thursday, obviously not because my grade is terrible.. im so bad at math. hah. i actually get sohcahtoa and know how to use it.

this week with the start of non right trianlges and finding the area is very easy. i feel so cool because i actually understand something in advanced math. lol:) you just have to use the formula A= 1/2(leg)(leg)sin(angle between)

heres an example:
i dont have my book, but heres one in my binder from the practice problems..

A-1/2(4)(5)sin30=

60

reflection...

This week wasnt so hard it was kind of a review to me with the SOHCAHTOA functions...
so i really caught on quick.

SohCahToa stands for:
Sin= opposite/hypot.
Cos= adjacent/hypot.
Tan=opposite/adjacent

Areas for triangles
Right triangle= lenth x height

Non-Right triangle=1/2 x (leg) (leg) Sin (angle btw)

Law of Sins
SinA/a=SinB/b=SinC/c

Reflection 12

This week was really hard to me for some reason. I rather chapter 6 then 7. I was pretty pumped about the three days weekend, but it is about over and there is a lot of rain coming from that stupid hurricane/tropical depression. Soo besides that, we learned a lot of triangle formulas and things.

First we learned SOHCAHTOA which stands for:

Sin=opposite/hypotenuse Cos=adjacent/hypotenuse tan=opposite/adjacent which works a right triangle and opposite, hypotenuse, and adjacent stands for the sides of the triangle. Used to find the measure of an angle and length of a side.

Here are the two area formulas:

For a right triangle it is A=height x length

For a non-right triangle A= 1/2 (leg)(leg) Sin (angle btw)

Also learned the Law of Sines

Which is SinA/a = Sin B/b = SinC/c

I do not know how to use the law of sin thingy. Anybody can help?

Reflection #12

Now, for I know from this week...I know the areas of different triangles:

Area of a right triangle: A=(1/2)(base)(height)
Area of a non-right triangle: A=(1/2)(leg)(other leg)sin(angle in between the two legs)

visualize a right triangle: DEF (left to right)
D=90 degrees, d=8, e=6...Find the area.
So, you know your base is e, which is 6.
Now, to find your height, or f, use: a^2 + b^2 = c^2
c is always to hypotenuse, which is d (8).
6^2 + b^2 = 8^2
36 + b^2 = 64
b^2 = 28
b = 5.292
Your height is 5.292
Now, plug into the formula.
A = (1/2)(6)(5.292)
A = 15.875

Now, visualize a non-right triangle: HIJ (left to right)
H = 65 degrees, j = 2, i = 6
Plug into the formula.
A = (1/2)(2)(6)sin(65)
A = 5.438

Now, for what I don't know from this week...I don't know, well, I pretty much understand everything. The only thing I'm a little iffy on is numbers 19 and 20 on page 348.

#19) In triangle ABC, tan A = 3/4, tan B = 1, and a =10. Find b in simplest radical form.

I'm not sure how to start it. Do the trig functions work the same for a non-right triangle? If so, then I can easily work it, but I'm not sure. And if not, how then? I don't know, I'm slightly confused. Any help?

Reflection #12

Okay, this week was pretty easy. We learned alot more about trig and i'm gonna describe different formulas we used this week.

SOHCAHTOA

sin = opposite leg/ hypotenuse

cos = adjacent leg / hypotenuse

tan = opposite leg / adjacent leg

area of a right triangle = 1/2 (b) (h)

area of an isoceles triangle = bh

area of a non-right triangle = 1/2 (leg) (leg) (sin (angle between those 2 legs))

law of sines

sin (angle that is given) / opposite side given = sin (Angle unknown) / opposite side given

or you could say

sin (angle that is given) / opposite side given = sin (angle given) / opposite side unknown

You use SOHCAHTOA when you are dealing with right triangles

You usually divide an isoceles triangle in half and then use SOHCAHTOA on one side and multiply by 2.

You use the law of sines when you are dealing with non-right triangles.

reflection 12!!

Ok so way to go ash! making everybody thinkin this is reflection 13! noob hahaha! Soooo, now to the math. SOHCAHTOA!!! its crazy maaan!!

SOHCAHTOA:
sin (angle)=Opposite leg/Hypotenuse
cos (angle)=Adjacent leg/Hypotenuse
tan (angle)=Opposite leg/Adjacent leg

very simple, right?
memorize this and you should be fine, or atleast for all the right triangles!!!
now for non-right triangles, I'll explain how to find their area!

A=1/2 (leg) (leg) sin(angle between)
A=1/2 (2) (3) sin(35)
A is approximately equal to 1.721!

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soo how do you do the law of sins????????????

them things are crazy!

reflection 13

okay, so I didnt really have the best of weeks but in this class it wasn't that bad.
Finding the area of non right triangles is like the easiest thing we learned all year.

the formula is:
A= 1/2(leg)(leg)sin(angle)

all you have to do is plug in the things that you are given in the triangle that you have and get whatever you need.

you can also find things that you are missing in a triangle like so:

The area of triangle PQR is 30 cm^2, if p=10 and q=20. Find all possible measures of angle R.

30=1/2(10)(20)sinR
30=100sinR
sinR=3/10
inverse sin(3/10)=R

it really is a generally easy thing to do.

the only thing that I really forget to do is find the second angle. for these you always need to get two angles.

for once in my like whole life i dont not understand anything this week. everything that we did was generally easy. I think that this was the easiest and the second easiest was the sohcahtoa stuff.

Reflection

REFLECTION #13

Wow we're only on reflection 13? It feels like I've been doing way more than that. haha. Well anyway, I thought this week was pretty good overall. It went by pretty fast for me, although a few days just dragged on forever. I guess it's because it's math. So this week we continued learning things about triangles. The first thing we learned this week was how to find the area of a triangle that is not a right triangle. We use a formula for that and it is >> A = (leg)(leg)sin(angle between). This formula works to find the area of any type of triangle that is not a right triangle, but you have to have at least two of the leg lengths. The next thing we learned this week was the Law of Sines which is sin A/a = sin B/b = sin C/c. This is a formula that is used when you know the pairs in non-right triangles. And remember, for this you set up a proportion. I kind of got confused with that at first but I understand it now. We also learned how to determine whether you will be drawing a right triangle or a regular triangle by just seeing a few numbers for angles and sides.

Okay so what I think I understood the most this week is finding the area of a non-right triangle. I think it's pretty easy because all you really have to do is plug numbers into a formula and type it into your calculator. So here is an example:

Ex. 1.) The area of triangle ABC is 15. a = 12, b = 5. Find angle C.

*Alright so the first thing you want to do is draw your picture which will be a regular triangle since it does not tell you it is a right triangle. Then you label what you have on you triangle--5 and 12 as legs of the triangle. Since you want to find the measure of angle C you first have to find the area of the triangle so you use the non-right triangle area formula to do that.
So you get: 1/2 (12)(5)sin C = 15
30sinC = 15 <<(after you multiply you get this) sinC = 1/2 C = sin^-1 (1/2) (*remember, to find an angle you have to use inverse) [*for inverses-first you have to figure out your reference angle which in this case is 30 degrees. Then you have to figure out which quadrants your angles will be in. You are looking for where sin is positive so your angles would be located in the 1st and 2nd quadrants. So you know one of your angles is 30 degrees. To find your other angle you have to make 30 degrees negative and add 180 (because you are moving from the first quadrant to the 2nd quadrant) And then you get 150 degrees for your second angle. So your final answers are 30 degrees and 150 degrees. ]

*Now for what I didn't understand this week, which actually wasn't as much as usual. I think I did a pretty good job with understanding things this week. On the other hand, I do get a little confused sometimes on when you can use the law of sines. I think I've used it at the wrong time before. So if anyone would like to help me with that, that would be nice :)

NO SCHOOL TOMORROW :)

reflection #13!

mkay, so this week was a pretty slow one for me.
I definitely picked up on is finding the area of non-right triangles:

A= 1/2(leg)(leg)sin(angle)

A triangle has sides of 7cm and 4cm, the angle between is 73 degrees. Find the area of this triangle:

A=1/2(7)(4)sin(73)
A=14sin(73)
A=13.388cm^2

The area of triangle PQR is 15 cm^2, if p=5 and q=10. Find all possible measures of angle R.

15=1/2(10)(5)sinR
15=25sinR
sinR=3/5
inverse sin(3/5)=R
R=36.870 degrees
143.130 degrees


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as for what i don't know, 9-1? sohcahtoa still gets me, i understand up until when i have to decide whether to use sin, cos, or tan.

HELP?

Reflection

Well to start off this week was rough for me. I guess you can say i am having some trouble with the stuff we are learning...and getting sick this weekend again doesn't help. I am trying to understand this stuff better and its getting there but i got to study harder. This geometry type stuff just isn't my area. I did understand a good bit of the worksheet she gave us on friday though.

EXAMPLE:

C = 90 degrees
b = 7
c = 12

First i found angle A
cos A = 7/12
A = cos^-1(7/12)
= 54.315 degrees

Second i found angle B
sin B = 7/12
B = sin^-1(7/12)
= 35.685 degrees

Next i found length of a
tan 54.315 = a/7
(7) tan 54.315 = a/7 (7)
a = 7 tan 54.315
a = 9.747

Finally i found the area
A = 1/2 bh
= 1/2 (7) (9.747)
= 34.115

Overall, this is becoming easier but if anyone wants to explain this a little better to me...please feel free to do so!

Reflection

okay this week was easy, except that i missed school on...don't even remember what day it was..Everyone in the class said that we learned a lot of stuff on that day. But the stuff that we did learn this week was pretty easy...except for some of the things that i did not catch onto.

First, we learned SOH CAH TOA...which stands for sin: opposite over hypotenuse, cos: adjacent over hypotenuse, and tan: opposite over adjacent. That helps you out so you know what function to use on each triangle.

We also learned the area of a non-right triagle. Which is A=1/2(leg)(leg)sin(angle between).

We learned that the area of a right triangle is A= (1/2)bh.

I missed the day when we learned the law of sines, so if someone can please do a couple of problems and explain, i would appreciate it. I also sometimes get confused when you have a right triangle, i don't know where to start when you have to solve them. What step do you do first.

reflection 12

This week was very, odd, gotta say that. and annoying......anyway, back to the week. meh, not rly much happened this week. so, nothin to rly write on here............except the saints need to pick up their game, 14-3 right now :(

anyway, back to math

one thing i did understand this week, with a little help from sparky.......was the law of sines
i get it how sinA/a = sinB/b


but one thing i didnt understand was the whole, sec and csc and cot on the triangles, that made no sense to me

I actually thought that this week was pretty good. Well, I guess i'm going to post some of the stuff that i learned. soooo...

Area of a right triangle: A=(1/2)(base)(height)
Area of a non-right triangle: A=(1/2)(leg)(other leg)sin(angle in between the two legs)


SOHCAHTOA
sin = opposite leg/ hypotenuse
cos = adjacent leg / hypotenuse
tan = opposite leg / adjacent leg

law of sines
sin (angle that is given) / opposite side given = sin (unknown angle) / opposite side given

Example:Find the area of triangle ABC: A=34 degrees b=3 c=5 you must assume that this is not a right triangle. so 1/2*3*5fsin34 You type that in your calculator and you will get your answer.


BOOM!!!!

Reflection 12

So this week was okay because i'm pretty sure i did great on our quiz. Which is awesome. Trig is turning out to be a little easier than i made it out to be. I think i can handle this stuff.
So i guess i got some splainin' to do.

Here is a word prolem:

A boy stands 25 feet from the base of a flag pole. He looks up at a 32 degree angle to see the top of the flag pole. How tall is the flagpole?

-ok, you have to now draw this triangle, but i cant on a computer. So....draw a right triangle that has a base of 25 feet. Imagine the ground and where the 32 degree angle is, its common sense.
-Now that you've drawn it, use Tan to solve it. Remember SOHCAHTOA? Yea, use it. Tan(angle)=opp/adj. So plug in:
Tan(32)=x/25.

Now solve for x and wahlah, you got the flag poles height.
____________________________________________________
Still challenged with isoceles triangles.

Relfection 12?

Just about the only thing I understood how to do this week was how to find angles of a right triangle.

SOCAHTOA

Sin=opposite leg/hypotenuse Cos=adjacent leg/hypotenuse Tan=opposite leg/adjacent leg

Use what it gives you in the problem to draw out a triangle. Label your sides with the numbers it gives you then decide what side or angle you want to find and use sin,cos, or tan to find it. **Remember when your finding the angle you use inverse.

Hint: If you have an isoceles triangle all you have to do is split it down the middle to make two right triangles. Then you split the length and they have the same angles on both side of the bottom.

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If someone wants to explain something about 9-3 to me that would be great cause the whole thing confuses me.