Thursday, November 5, 2009
Monday, November 2, 2009
a little help
hey guys, how do u find the base angle when ur given an area and height of a triangle, its from the homework,
i need some help
i need some help
Sunday, November 1, 2009
Reflection 11
Spirit week went by really fast and I think I did good on my test. I learned my trig chart in one night. Im also ready for tomorrow night because the saints play. The one new thing we learned 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
By following the chart you will get what you need. Meaning that for Sin you are trying to find the adjacent leg, Cos the opposite leg, and tan the hypotenuse.
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.
You can't draw triangles on here, so you pretty much just follow the SOHCAHTOA chart. Another chart that I have to learn. Also can not do my homework because I forgot my book in my locker.
SOCAHTOA
Sin=opposite leg/hypotenuse
Cos=adjacent leg/hypotenuse
tan=opposite leg/adjacent leg
By following the chart you will get what you need. Meaning that for Sin you are trying to find the adjacent leg, Cos the opposite leg, and tan the hypotenuse.
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.
You can't draw triangles on here, so you pretty much just follow the SOHCAHTOA chart. Another chart that I have to learn. Also can not do my homework because I forgot my book in my locker.
relection 11
So how bout that spirit week guys.
halloween was fun even though i didn't go to homecoming, me and matt nobile hung at my house. ya digg?!
tulane nothing, LSU 42.
AND played a cabbage ball tournament at which i beat RICKY!!!!!!!!!!! too bad he can't see this right now.
so to the math...
Friday we learned about SOHCAHTOA(so-ka-toe-uh)
sin= opposite over hypotenuse
cos= adjacent over hypotenuse
tan= opposite over adjacent
i don't normally understand much but this week all i need is an easier way to memorize my trig chart
and ideas?
halloween was fun even though i didn't go to homecoming, me and matt nobile hung at my house. ya digg?!
tulane nothing, LSU 42.
AND played a cabbage ball tournament at which i beat RICKY!!!!!!!!!!! too bad he can't see this right now.
so to the math...
Friday we learned about SOHCAHTOA(so-ka-toe-uh)
sin= opposite over hypotenuse
cos= adjacent over hypotenuse
tan= opposite over adjacent
i don't normally understand much but this week all i need is an easier way to memorize my trig chart
and ideas?
Reflection #11!
Mkay, so due to Mrs. Heidi's obvious lapse of sanity, i missed on friday. Fortunately, I do know most of everything from the last test NOW so i'll just try and review that?
Six trig functions:
sin: y/r
cos: x/r
tan: y/x
cot: x/y
csc: r/y
sec: r/x
Unit Circle:
90(pi/2): (0,1)
180(pi): (-1,0)
270(3pi/2): (0,-1)
360(2pi):(1,0)
sin(90): 1/1=1
csc(2pi): 1/0- UNDEFINED.
ACCOMPLISHMENT TIME: i know my entire trig chartttt! :)
-------------------------------
if anyone is willing to give me the notes that i missed friday, please feel freee. :)
thanks!
Six trig functions:
sin: y/r
cos: x/r
tan: y/x
cot: x/y
csc: r/y
sec: r/x
Unit Circle:
90(pi/2): (0,1)
180(pi): (-1,0)
270(3pi/2): (0,-1)
360(2pi):(1,0)
sin(90): 1/1=1
csc(2pi): 1/0- UNDEFINED.
ACCOMPLISHMENT TIME: i know my entire trig chartttt! :)
-------------------------------
if anyone is willing to give me the notes that i missed friday, please feel freee. :)
thanks!
Reflection 11
dsjflkasjdlkfjlkjewqj98432u89u43uj98jf9rhjt9843h98h4tniuhfapoi
ok. So we finally learned SOHCAHTOA which is one of those memory things for
cos=adjacent side/hypotenuse
sin=opposite side/hypotenuse
tan=opposite side/adjacent side
to find a side you use a trig function. to find an angle you use the inverse of a trig function.
triangle ABC(capitals are sides) and abc(lower case is angles.
C =90, A=28 a=40degrees
to find the missing angles you do:
tan 28=40/b
b tan 28=40
b=40/tan 28
b=75.229 degrees
and
sin 28=40/c
c sin 28=40
c=40/sin 28
c=85.202 degrees
to find the missing side you can just use pythagoreans theorem
Finding an inverse...
when you find an inverse you are finding 2 angles.
the only time you will have one angle is when the angle is 0,90,180,270, or 360 degrees. ex. type sine inverse of 1 in your calculator and it will give you 90. which is between quadrants one and two and can not be moved inside of a quadrant. If you move it, it will only fall between two other quadrants.
inverse sec of 2.... ok so sec is dealing with x values, so it is positive in the 1st and 4th quadrant. if our 2 was negative we would be finding angles in the 2nd and 3rd quadrant because that is where secant is negative. sec of 2 on our chart is at 60degrees so now we need to move it into the fourth quadrant by making 60 negative and adding 360 to it. you get 300 degrees. so the inverse sec of 2 is 60 and 300 degrees.
ok. So we finally learned SOHCAHTOA which is one of those memory things for
cos=adjacent side/hypotenuse
sin=opposite side/hypotenuse
tan=opposite side/adjacent side
to find a side you use a trig function. to find an angle you use the inverse of a trig function.
triangle ABC(capitals are sides) and abc(lower case is angles.
C =90, A=28 a=40degrees
to find the missing angles you do:
tan 28=40/b
b tan 28=40
b=40/tan 28
b=75.229 degrees
and
sin 28=40/c
c sin 28=40
c=40/sin 28
c=85.202 degrees
to find the missing side you can just use pythagoreans theorem
Finding an inverse...
when you find an inverse you are finding 2 angles.
the only time you will have one angle is when the angle is 0,90,180,270, or 360 degrees. ex. type sine inverse of 1 in your calculator and it will give you 90. which is between quadrants one and two and can not be moved inside of a quadrant. If you move it, it will only fall between two other quadrants.
inverse sec of 2.... ok so sec is dealing with x values, so it is positive in the 1st and 4th quadrant. if our 2 was negative we would be finding angles in the 2nd and 3rd quadrant because that is where secant is negative. sec of 2 on our chart is at 60degrees so now we need to move it into the fourth quadrant by making 60 negative and adding 360 to it. you get 300 degrees. so the inverse sec of 2 is 60 and 300 degrees.
Reflection #11
alright this week we learned how to use trig functions to solve for a right triangles and how to use trig functions. the thing i understood the most was how to use trig functions to solve for a right triangle. here's the steps:
Step 1: your given a triangle where angle C =90, A=28, & B is the other angle and a=40. find b & c.
draw your triangle and label it with the information from above.
**remember SOHCAHTOA
step 2: pick an angle to help you solve the equation.
i chose angle A. so to solve for b you have to have your side b over something. so what is your given side? oh your right if you chose a.
so you take the TAN of your angle and set it equal to the side your tying to solve for over what side you already have.
tan 28=40/b
b tan 28=40
b=40/tan 28
b=75.229
step 3: solve for the last side.
so you take SIN of your angle and set it equal to the side your tying to solve for over what side you already have.
sin 28=40/c
c sin 28=40
c=40/sin 28
c=85.202
alright for something i didn't understand was how to find the trig inverses.
Step 1: your given a triangle where angle C =90, A=28, & B is the other angle and a=40. find b & c.
draw your triangle and label it with the information from above.
**remember SOHCAHTOA
step 2: pick an angle to help you solve the equation.
i chose angle A. so to solve for b you have to have your side b over something. so what is your given side? oh your right if you chose a.
so you take the TAN of your angle and set it equal to the side your tying to solve for over what side you already have.
tan 28=40/b
b tan 28=40
b=40/tan 28
b=75.229
step 3: solve for the last side.
so you take SIN of your angle and set it equal to the side your tying to solve for over what side you already have.
sin 28=40/c
c sin 28=40
c=40/sin 28
c=85.202
alright for something i didn't understand was how to find the trig inverses.
Reflection 11
This week wasn't too bad. I understood a lot even though there's still some things I still don't understand. I'm don't remember if we learned how to do this during this week but the thing I understood the most was trig inverses. You have to use the trig chart for inverses. To find the trig inverses you first find what degrees the number given to you is then figure out what quadrant it is in. From there you have to move into 2 different quadrants..follow these rules:
I-->IV- make theta -ve and +360, I-->III- add 180, I-->II- make theta -ve and +180,
II-->IV- +180. You use those rules to figure out your two angles.
Example:
csc^-1(2)
-30 degrees + 180 degrees=150 degrees
theta=30 degrees, 150 degrees.
*Csc 2 in your trig chart is 30 degrees.
It would be in quadrant I and you would be only moving to quadrant II.
-------------------------------------------------------
For something I don't understand the homework from friday. On number 5, how do you find c?
I-->IV- make theta -ve and +360, I-->III- add 180, I-->II- make theta -ve and +180,
II-->IV- +180. You use those rules to figure out your two angles.
Example:
csc^-1(2)
-30 degrees + 180 degrees=150 degrees
theta=30 degrees, 150 degrees.
*Csc 2 in your trig chart is 30 degrees.
It would be in quadrant I and you would be only moving to quadrant II.
-------------------------------------------------------
For something I don't understand the homework from friday. On number 5, how do you find c?
Reflection #11
Spirit Week. This week was NICE, because we didn't learn too much new stuff, which I like, cuz I never understand new stuff. I actually think I did great on my test, so I'm excited to get it back. One thing I noticed a lot of people saying they got wrong on the test for sure was the problems with two trig functions. No need to fear, Matt Nobile is here.
cot(sin^-1(1/2)
Okay, here's why this is so easy....you see sin? You are not trying to find it, it is given to you. This problem is just like some of those ACT problems, its made jus to trip you up. Don't flip out when you see two trig functions in one problem.
Sin is just a tool to help you find cot, which is what you really want to find. Okay we know that sin=y/x. So if sin=1/2, that means y=1 and r=2. Now draw your triangle. Hypotenuse=2 and your y value is 1. So use pythagorean theorem to find x, and you get the square root of 3. We know that cot=x/y. So with your triangle drawn just put x over y, which is square root of 3/ 1. So your final answer is cot=square root of 3.
cot(sin^-1(1/2)
Okay, here's why this is so easy....you see sin? You are not trying to find it, it is given to you. This problem is just like some of those ACT problems, its made jus to trip you up. Don't flip out when you see two trig functions in one problem.
Sin is just a tool to help you find cot, which is what you really want to find. Okay we know that sin=y/x. So if sin=1/2, that means y=1 and r=2. Now draw your triangle. Hypotenuse=2 and your y value is 1. So use pythagorean theorem to find x, and you get the square root of 3. We know that cot=x/y. So with your triangle drawn just put x over y, which is square root of 3/ 1. So your final answer is cot=square root of 3.
reflection 11
spirit week was soo fun this week. i love not wearing school uniforms:) we didn't do much this week because of our test oh thursday. we just reviewed at the beginning of the week and then mrs robinson wasen't there wednesday so we had mrs fi fi:) haha. i definately did not do good on my test, well at least i think i didn't. i don't understand anything. but anyway, friday we learned right triangles. we learned SOCAHTOA. which is sin=opp.leg over hypotenuse, cos=adj.leg over hypotenuse, and tan=opp.leg over adj.leg. then you just use this to solve the problems and it will help you out.
reflection 11
ok, this week was alright. It was funny watchin ppl on the days for spirit week. I found out that even the smell of a gas mask all day can give u a major headache >_> so dont ever use those for like a whole day. let's see........we blew out Lusher at the football game. what else........pretty much nothin
so, about that math
i did understand the stuff for sin cos and tan
cos=adjacent side/hypotenuse
sin=opposite side/hypotenuse
tan=opposite side/adjacent side
but i dont understand how u find different sides on a triangle when ur given these :(
so, about that math
i did understand the stuff for sin cos and tan
cos=adjacent side/hypotenuse
sin=opposite side/hypotenuse
tan=opposite side/adjacent side
but i dont understand how u find different sides on a triangle when ur given these :(
Reflection 11
Last week went by ok. Wish i could have done a little better on my test though. I went and mixed up a formula and also forgot my points on the unit circle. On the good note i know my trig chart and i'm pretty sure i did enough on that test to get a passing grade. This week we are still working with trig, which means we are working with triangles. Right triangles are triangle that form right angles. When solving you have to use sin, cos, tan known as SOH CAH TOA. this means sin is oppisite over hypotenuse, cos is adjacent leg over hypotenuse, and tan is opp leg over adjacent leg. Nothing hard about solving for the angles just remember SOH CAH TOA. This can only be used for solving right angles.
REFLECTION #11
So this week went by pretty fast if you ask me. It wasn't as stressful as usual but it was still a little difficult. I mean it's trigonometry so how easy do you want it to be? hah. Well on the bright side, the thing I put on my blog last week that I was confused about I'm not confused with anymore. I actually understand it pretty well although I'm nervous about how I did on that test we took Thursday. I'm not going to lie it was pretty tough. Well anyway, Monday thru Wednesday it was basically a review of all we learned in Chapter 7. We worked on problems in the book and that actually helped me understand things better especially some of the word problems that I didn't get. Then on Thursday we took our test. And on Friday we started Chapter 9 which is right triangles. Now I have to say I'm pretty nervous about this chapter because I'll admit I'm not that good with shapes. hah but I'll try and manage. Oh and speaking of, I still have to do that homework now that I think about it. :P
So anyway something I understood this week is trig inverses. I didn't understand them last week at all but I'm glad I can explain it now because I actually know what I'm talking about!
Here's an example:
cos inverse (-1/2)
*Okay first looking at this problem you notice that 1/2 is negative and it's dealing with cos. So you are looking for where cos is negative. (remember cos relates to the x axis) So you look in the quadrants and check off the quadrants where cos is negative. which are the 2nd and 3rd quadrants. Next you find 1/2 on your trig chart to get your reference angle which is 60 degrees. Now to find your other 2 angles you first have to go from the 1st quadrant to the 2nd. That means you have to make 60 degrees negative and add 180 to it and you get 120 degrees for your 1st angle. Then to find your 2nd angle you have to go from the 1st quadrant to the 3rd quadrant which means you just have to add 180 to 60 and you get 240 degrees as your other angle.
*Now I would give an example of something we learned on friday, but I'm still a little iffy on the right triangle stuff as of right now so I don't want to work a problem incorrectly if you know what I mean :)
So anyway something I understood this week is trig inverses. I didn't understand them last week at all but I'm glad I can explain it now because I actually know what I'm talking about!
Here's an example:
cos inverse (-1/2)
*Okay first looking at this problem you notice that 1/2 is negative and it's dealing with cos. So you are looking for where cos is negative. (remember cos relates to the x axis) So you look in the quadrants and check off the quadrants where cos is negative. which are the 2nd and 3rd quadrants. Next you find 1/2 on your trig chart to get your reference angle which is 60 degrees. Now to find your other 2 angles you first have to go from the 1st quadrant to the 2nd. That means you have to make 60 degrees negative and add 180 to it and you get 120 degrees for your 1st angle. Then to find your 2nd angle you have to go from the 1st quadrant to the 3rd quadrant which means you just have to add 180 to 60 and you get 240 degrees as your other angle.
*Now I would give an example of something we learned on friday, but I'm still a little iffy on the right triangle stuff as of right now so I don't want to work a problem incorrectly if you know what I mean :)
reflection #11
this week was a general review of last week: trigonometry. we worked out conversions, functions, unit circle, and the trig chart.
CONVERSIONS:
convert 15.13 degrees into o mins. secs.
multiply .13 by 60
15 degrees 7.8 mins
multiply .8 by 60
15 degrees 7 mins 48 secs.
to convert back into 15.13, divide secs. by 3600 and mins. by 60
UNIT CIRCLE:
90 degs. = (0,1) pi/2
180 degs. = (-1,0) pi
270 degs. = (0,-1) 3pi/2
360 degs. = (1,0) 2pi
sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y
CONVERSIONS:
convert 15.13 degrees into o mins. secs.
multiply .13 by 60
15 degrees 7.8 mins
multiply .8 by 60
15 degrees 7 mins 48 secs.
to convert back into 15.13, divide secs. by 3600 and mins. by 60
UNIT CIRCLE:
90 degs. = (0,1) pi/2
180 degs. = (-1,0) pi
270 degs. = (0,-1) 3pi/2
360 degs. = (1,0) 2pi
sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y
Reflection 11
okay, so this week was based entirely on trig functions and how to use them. i finally have the chart and unit circle down, and i understand how to do some of the word problems quickly.
SOH CAH TOA was pretty easy to understand as well
S=sin
O=opposite angle
H=hypotenuse
C=cos
A=adjacent angle
H=hypotenuse
T=tan
O=opposite angle
A=adjacent angle
these are used for solving problems related to right triangles
SOH CAH TOA was pretty easy to understand as well
S=sin
O=opposite angle
H=hypotenuse
C=cos
A=adjacent angle
H=hypotenuse
T=tan
O=opposite angle
A=adjacent angle
these are used for solving problems related to right triangles
Reflection #11
Well, to tell you the truth, I have no clue what I'm doing in trig.
This week, one thing I understand:
SOH CAH TOA
S = sin
O = opposite angle
H = hypotenuse
sin = opposite/hypotenuse
C = cos
A = adjacent angle
H = hypotenuse
cos = adjacent/hypotenuse
T = tan
O = opposite angle
A = adjacent angle
tan = opposite/adjacent
**hypotenuse is always the longest side
What I don't understand: everything else
Word problems still get me and I'm slightly unsure about when to use what function. I don't get a lot of the homework on page 334 either.
Ex. #13
An airplane is at an elevation of 35,000 ft when it begins its approach to an airport. Its angle of descent is 6 degrees. a) What is the distance between the airport and the point on the ground directly below the airplane? b) What is the approximate air distance between the plane and the airport? What assumptions did you make in finding this distance?
Any help with this?
This week, one thing I understand:
SOH CAH TOA
S = sin
O = opposite angle
H = hypotenuse
sin = opposite/hypotenuse
C = cos
A = adjacent angle
H = hypotenuse
cos = adjacent/hypotenuse
T = tan
O = opposite angle
A = adjacent angle
tan = opposite/adjacent
**hypotenuse is always the longest side
What I don't understand: everything else
Word problems still get me and I'm slightly unsure about when to use what function. I don't get a lot of the homework on page 334 either.
Ex. #13
An airplane is at an elevation of 35,000 ft when it begins its approach to an airport. Its angle of descent is 6 degrees. a) What is the distance between the airport and the point on the ground directly below the airplane? b) What is the approximate air distance between the plane and the airport? What assumptions did you make in finding this distance?
Any help with this?
Subscribe to:
Posts (Atom)