H Pre Calc Ch. 8

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H Pre Calc Ch. 8

2/18/2011

Happy Blogging!

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,

For number 17, you have to first convert the tan20 to sin20/cos20, and then subtract the sin20/cos20-cos70/cos20. Then you have to convert the sin to cos by taking sin20 and changing it to cos(90-20), and this is basically cos(70). Then you subtract, and you should get 0 for the final answer.

I have a question as well. For the 8.1 homework, in the problems like 21-26, how do you know what the dimensions of the triangle are...? So when it says 1/3, for sinx, then how do you know whether they took the 1/3, or they did 2/6, and then simplified?

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.

For number 50, you know that the height of the Statue of Liberty is 305 feet, and the angle of elevation is 20 degrees, and you are supposed to find the distance from the base of the statue to a ship in the harbor.

So would you basically just solve this by taking tan20=305/x, and then solving for x using a calculator. Basically, this is x=305/tan20...where x is supposed to equal...

837.98 feet. Is this the right approach or is there an easier one?

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
For number 32 from 8.3

Originally, if the pole was on flat ground, with no angle of elevation, it would be perpendicular to the ground. This would create an angle of 90 degrees with the ground.
Since the pole is at a 5 degree incline though, it is not 90.

Looking at the picture in the book, and the triangle on the right side of the pole, the angle is now 90 + 5. Because you are elevated 5 degrees above the horizontal. This gives you a SAS triangle, with side 500, angle 95, and side 100. You simply plug in to the law of cosines.

For the triangle on the left, you must subtract 5 degrees. Thusly, 90 - 5, you get 85 degrees. Again you have a SAS triangle, 500 - 85 - 100.

If you notice, both the cos95 and the cos85 are .08715....... The important thing is that cos95 is negative and cos85 is positive. Thus making you add and subtract differently on the right side of the law of cosines.

Hopefully that makes sense.
Take once daily

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).

James...you would :).

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).
Hope that helps:)

Shivani D.

2/28/2011 11:18:08 am

Hi!
Can someone help me with 29 in 8.3?

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.

Chammy: For 33, the first thing you do is find the remaining angle of the big triangle, which is 105 degrees. Then you use the law of sines to find either of the remaining sides of the big triangle. So if you chose to find the length of the Point A to the airplane you set sin 105 degrees/1000 = sin 35 degrees/x and solve for x. You should end up getting 593.8100222. Then you only look at the smaller triangle that includes point A. From there you use the law of sines again to solve for the height by setting 593.8100222/sin 90 = height/sin 40 and you solve for the height and you should get 381.694 feet.

Ratuja Reddy

2/28/2011 02:35:44 pm

Alex: For number 26a, you know that the angle for Sarasota is 130 degrees because 180-50 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+150^2-2(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,

for this question it asked to find the area of the shaded segment whose radius was 8 ft and the central angle was 70 degrees. It gave in the hint that to find the sector the formula is (1/2)r^2 theta. In order to find the shaded area

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 39.095-30.0701 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,

Ok so for the bearing problems, that you are talking about is always strictly TOLD to you in the problem. So for example number 64 in 8.1, since that strictly tells you it turns 90 degrees towards a certain direction, then that's what you label it as.

It could always say that it turns a random measure to a certain direction, for example: N80degreesW, etc. So that's how you know.

Hope that helped you.

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.

So for the bearing problems...do you notify the solution ALWAYS off of the N/S line? I think this is right, from what I realized on the quiz.

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.

Thanks.

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.

Can someone please help me on number 42 in the review section from tonight's homework?

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:

With this question, you have to remember that each side of the barn is 10 feet long
So when you measure the A2, remember that the radius is only 90.

and for A3, the sides are 10- 90- and 10root2 because the other side includes the diagonal through the house
hope this helps!

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...

Sivani: Yea...you lost me. (@.@)

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."

But other than that...from what it looks like when I compare the wording of this problem to number 28 in 8.4...sector and segment of a circle both mean the same thing.

Anyone else concur?

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.

A SECTOR of a circle includes the edge of the circle; pizza shape.

A SEGMENT of a circle is the triangular shape that you should be looking for (otherwise, what would be the point of learning Heron's formula or anything else? :P)

Ashwin Chakilum

3/2/2011 12:41:23 pm

OH WAIT! Disregard my last comment.

Turns out that the segment that #49 is asking for is actually the PIZZA-SHAPE, including the center of the circle AND the edge.

Calculations for 5/36 of the area of that circle yields the correct solution as portrayed in the answer key.

Chandan Yashraj

3/2/2011 12:44:32 pm

So Ashwin...basically the segment and sector are the same thing in essence??

The way I'm looking at the problems, this is what I am finding.

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?

Ashwin Chakilum link

3/2/2011 12:50:33 pm

#49 is a bit vague.

It's telling us to find the SEGMENT of a circle, which is the moon-shaped part of the circle created when it is cut off by a secant line. [click my name to see what I mean, otherwise the URL is http://mathworld.wolfram.com/CircularSegment.html ]

A SECTOR would be the entire pizza shape.

So there were really no "triangle" calculations in that problem if we are supposed to get ~15.71 for a solution.

Weird...

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 180-the 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.

..and yea im confused on 42 too.

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.
For 42:
1. You have your triangle, and the line from point A to the point that says error discovered is equal to 70 miles (420 mph, 10 minutes).
You are basically left with, as Muhammed said, a SAS triangle, to solve it, you use the Law of Cosines to find the length of "error discovered" to point B. For the third side you get, 230.25 miles. Then you want to find the value of angle "error discovered". Using the law of cosines again, you find it to be 173.46. You subtract this from 180 degrees and you have to turn approximately 6.5 degrees to get back on course.

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....
I wasn't sure how to get the speed

Shivani D.

3/2/2011 02:29:25 pm

Nithya:
For 36, the problem is asking you for the speed of the glider. From the first right angle to the second right angle, a minute passes. The amount of distance covered in this minute is the base of the big triangle minus the base of the small triangle. Using the law of sines, the base of the big triangle is 167.8 while the base of the small triangle is 35.20. You subtract and the distance covered in a minute is 132.6 feet. You multiply this by 60, and then divide by 5280 to get 1.5 mph.
Hope that helps!

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
number 33 in 8.2 please!

Shivani D.

3/2/2011 03:04:44 pm

Hi Conner!
For number 33:
First, take a look at the drawing. It has a triangle with the height of the triangle drawn, from the airplane to the ground.
You can view this line as a separating line, forming two triangles. Now you'll have two right triangles. Then you find all the other degrees. On the left you have a 40-50-90 triangle, and on the right you have a 35-55-90 triangle. After finding this, you can use the law of sines to find the length of one of the sides. basically, you have sin (50+55)/1000 = sin (40)/x
You find x to equal 593.8 ft. You use the law of sines again to find the height of this triangle (now you focus on the triangle to the right).
You have (sin 35)/x = (sin 90)/ 593.8. You find x to equal 340.6 ft. This is your height.
I hope that helps!

Cody Essling

3/2/2011 03:41:50 pm

How would you do problem 36 in the review?

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