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Madeline Zehnal
2/22/2011 01:33:14 am
For number 17 on chapter 8.1 homework how do you figure it out since it has both tan and cos. Do you have to put in similar expressions???
Chandan Yashraj
2/22/2011 02:30:06 am
Hey Maddie,
Chandan Yashraj
2/22/2011 02:35:17 am
On Friday when half of the class was rehearsing for the pep assembly, Mrs. Johnson was going over how to do numbers like 50 and 56 from the homework, and she said that in order to get credit, you have to draw a figure, which I understand, but I just want to check if I am solving the problem right.
Chandan Yashraj
2/22/2011 02:36:29 am
NOTE: The homework sheet for Chapter 8 rounds the answer to 838 feet.
Maddie Strick
2/25/2011 01:00:11 pm
Can someone help me with #32 in section 8.3?
Chandan Yashraj
2/26/2011 01:49:18 am
Can someone please thoroughly explain number 64 in section 8.1 please? I get the approach and the geometry used to find the sides, but can someone just go over it again? Thanks.
Dr. Hunter  Here to solve all your mathematical, and romantical, needs
2/27/2011 09:56:39 am
Maddie
Chandan Yashraj
2/28/2011 08:20:19 am
Ok so for the bearing problems that we did in class and that "will" be on the quiz tomorrow...so do you always draw the angle off of the N/S line...? So its not like the other problems in the past when it says a certain angle, you draw it off of the x axis (W/E).
Shivani D.
2/28/2011 08:49:05 am
Chandan, the problem will tell you where to draw it. So only if it says West, then you would draw it off the x axis (or East).
Shivani D.
2/28/2011 11:18:08 am
Hi!
Shivani D.
2/28/2011 11:23:32 am
Nevermind! I figured it out:)
Chammy
2/28/2011 12:03:17 pm
Wait can someone help me with number 33 in 8.2?
Alex Masiak
2/28/2011 12:20:42 pm
Cab anyone help me with 26 and 32 in 8.3??
Ratuja Reddy
2/28/2011 01:56:37 pm
Ratuja Reddy
2/28/2011 02:04:32 pm
^Sorry, didn't mean to do that.
Ratuja Reddy
2/28/2011 02:35:44 pm
Alex: For number 26a, you know that the angle for Sarasota is 130 degrees because 18050 is 130. Then you use the law of cosines to solve for the distance between Ft. myers and Orlando. So you should get. c^2=100^2+150^22(150)(100)(cos 130). That should simplify down to 32500(19283.62829). When you solve for c, you should get 227.5602 miles. For part b, I honestly don't know what to do. I used both the law of cosines and sines to solve it several times, but got about 30.32 degrees, I believe. D:
Cody Essling
2/28/2011 02:36:03 pm
For the problems where you have to find the bearing of a ship, will the ship always make a 90 degree turn?
Ratuja
2/28/2011 02:47:32 pm
^I think so, if not, you can still use the law of sines, right?
Ayo
3/1/2011 11:26:26 am
Can someone help me with #25 in 8.4
Sohaila
3/1/2011 11:37:28 am
Ayo,
sohaila
3/1/2011 11:41:41 am
sry..it like posted half of it...but i was going to say in order to find the shaded area you would have to find the area of the sector and subtract it by the area of the triangle. to find the area of the sector i did (1/2)(8^2)(70pi/180). the 8^2 came from the radius and the 70pi/180 i got from converting the degrees into radians (i multiplied 70 degrees times pi/180 degrees. for the answer i got 39.095. and to find the area of the triangle, you would use the formula (1/2)(a)(b)sin(c). and when you plug the numbers in  (1/2)(8)(8)sin70, you get 30.07. so after that i subtracted 39.09530.0701 to get 9.03 ft ^2.
Sohaila
3/1/2011 11:43:10 am
Can someone help me with problem 39?
Sohaila
3/1/2011 11:45:56 am
oh nvm..i got it :)
Chandan Yashraj
3/1/2011 02:18:06 pm
Hey Cody,
Anika Gupta
3/2/2011 07:41:01 am
hi guys i was doing the chapter review for tonight and i really needed help on problem 42. could someone please explain it?
Chandan Yashraj
3/2/2011 07:58:48 am
Chandan Yashraj
3/2/2011 08:00:14 am
Sorry about the empty comment above.
Chandan Yashraj
3/2/2011 08:02:40 am
Can someone PLEASE explain "The Cow Problem", which is number 43 in section 8.4 That would be really helpful.
Chandan Yashraj
3/2/2011 08:05:00 am
Ok so for the types of triangles...what is the difference between SSA and SAS triangles, and ASA and AAS triangles? SSA is always ambiguous, but how about SAS...?
Mihir Surati
3/2/2011 10:21:23 am
SAS triangles are not ambiguous. Only SSA triangles are. To tell the difference between SSA and SAS, it's easy to just draw the triangle and visually see what is given. And when dealing with SSA, remember to use the law of sines, and when dealing with SAS, use the law of cosines.
Chandan Yashraj
3/2/2011 11:14:36 am
Thanks, Mihir.
Chandan Yashraj
3/2/2011 11:51:14 am
From what I understand from number 28 in section 8.4...is is that the 100 degree sector is NOT being used in the tent building? So you would just subtract the circle's area minus the sector area to find the "cone's" surface area?
Shivani D.
3/2/2011 11:58:22 am
Chandan, for 28 that's correct, you subtract the surface area of the sector from the total surface area of the circle.
Jacob Hayes
3/2/2011 12:03:00 pm
Chandan, for the cow problem just take the area of the circle (10000pi ft^2)  area of the barn (100 ft^2) you have to assume that the rope can go through the barn rather than around it, however. I dont know how to do this problem if the rope where to acually wrap around the sides of the barn
Sivani Aluru
3/2/2011 12:07:44 pm
I have a hint for the cow problem:
Chandan Yashraj
3/2/2011 12:17:45 pm
Jacob: Yea I'm having the same problem as you...the rope can't in real life wrap around the barn...
Chandan Yashraj
3/2/2011 12:19:03 pm
Sivani: Ok that's not all I have to say...how do the details you said HELP you in finding the maximum grazing area...it would be best if you just told me (us) straight off. ;)
Shivani D.
3/2/2011 12:24:47 pm
I was wondering, are sectors and segments the same? since 49 on the review asks for the area of a segment, but the answer is the area of the sector of the circle...can someone clear this up for me?
Chandan Yashraj
3/2/2011 12:31:19 pm
When it is saying "segment of a circle", then automatically translate this to "part of a circle", or "piece of a circle."
Justin Temple
3/2/2011 12:32:51 pm
Shivani, I'm having a similar problem, except I don't even know where to begin. Could someone help me with #49?
Ashwin Chakilum
3/2/2011 12:33:25 pm
@ Shivani D.
Ashwin Chakilum
3/2/2011 12:41:23 pm
OH WAIT! Disregard my last comment.
Chandan Yashraj
3/2/2011 12:44:32 pm
So Ashwin...basically the segment and sector are the same thing in essence??
Shivani D.
3/2/2011 12:47:58 pm
But on page 542, number 25, it says otherwise...
Shivani D.
3/2/2011 12:49:07 pm
Can someone also explain to me why in the review, number 42, when I use law of sines I immediately get the outside angle, how many degrees the ship has to turn to get back on track, but when use the law of cosines, I get the inside angle, that needs to be subtracted from 180 to get the outside degree? 3/2/2011 12:50:33 pm
#49 is a bit vague.
Muhammed Alikhan
3/2/2011 01:11:31 pm
Shivani, I know I already answered your question haha, but I think some other people were stuck on this problem. So what I would do for part A is first determine the value of the bottom side of the triangle and you would know that since you're given the time (10 min) and rate (420 mph). Now you have SAS, so you would use law of cosines and find the angle opposide side AB. Subracting this angle from 180 gets you the degrees that the pilot needs to turn.
mattinator
3/2/2011 01:24:13 pm
how do you tell how many answers you get for ambiguous case? i didnt really understand the stuff in class
mattinator
3/2/2011 01:29:09 pm
and how do you find the ambiguous angle?
Meghan Irby
3/2/2011 01:30:31 pm
Heyy! For ambiguous case you will get either zero, one, or two answers. To find the first possible angle you just use the law of sines. To test for another one you do 180the first angle you found. As long as this angle added to the angle given in the problem are not over 180, it can be another answer. You will only have no answers if, when plugged into the law of sines, there is a domain error
mattinator
3/2/2011 01:33:09 pm
ayy gurl thanks
mattinator
3/2/2011 01:39:58 pm
jk meant to say, thank you maddam
Sohaila Mali
3/2/2011 01:43:36 pm
not sure if this was already answered or not..but for #49 i started out by drawing a cirlce. and then i drew a 50 degree angle coming out of the center to the ends of the circle. therefore, the sides of the "segment" would be 6 (the radius) and the angle would be 50 degrees. The area of finding a sector or segment is equal to (1/2)(radius^2)(theta). to find theta i did 50 degrees/1 x pi/180 and i got 50pi/180 as the radians. So i plugged the numbers into the formula and got (1/2)(6^2)(50pi/180) and that gave me 15.7079.
Shivani D.
3/2/2011 01:54:53 pm
Hey Sohaila! Well, for 49, you found the sector, which is the answer in the book, but the problem asks for the segment, so i'm confused about that.
Nithya Sridhar
3/2/2011 02:05:25 pm
I was wondering if someone could help me with problem #36 on the first section of review problems....
Shivani D.
3/2/2011 02:29:25 pm
Nithya:
Connor A
3/2/2011 02:42:17 pm
Connor A
3/2/2011 02:48:37 pm
Hi, so I need some help figuring out
Shivani D.
3/2/2011 03:04:44 pm
Hi Conner!
Cody Essling
3/2/2011 03:41:50 pm
How would you do problem 36 in the review? Comments are closed.

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