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TheSingularity

Looking for Recommendations on Math Textbooks.

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Hi all,

 

Kind of gone quiet recently, been filling my time with other things seeing as the forums aren't moving too quickly and there hasn't been a whole lot I've been interested in.

 

Now on to the questions at hand. First off I was wondering if any of the rather intelligent fellows here could recommend math textbooks from basic arithmetic all the way through to the end of high school and then even into university level. As I'm currently looking at buying some and working through them and reading them but finding out what ones are good has been difficult as most don't include basic arithmetic and high school level, all the lists I've read generally skip to university level.

 

Now the other question was regarding Right Angled Trigonometry, basically that is what I'm learning at school, it's basic sure but I need more practice questions then what I've got in the textbook I'm using to make sure I fully understand what I'm doing and so far there hasn't been any luck finding anything. I thought I had something with ck-12.org but it skims over Right Angled Trigonometry like all the other sites do, same as PatrickJMT and Khan Academy.

 

To the last problem at hand which is an actual math problem a Right Angled Trigonometry problem of course. So here is the problem: A balloon travels horizontally at a distance of h kilometres above the ground between two points A and B, which are two kilometres apart. From a point C on the ground, the angle of elevation of the balloon at A is 40 degrees and at B is 25 degrees. Assume that A, B and C are in the same plane and that A and B are on the same side of the observation point. Find the height h of the balloon.

 

See the main problem with the above question is I can't visualise it and even after a friend tried drawing it on a collaborative drawing site I still got the answer wrong, and then I thought I had it again and nope no dice but the answer was much closer which really means nothing though as it still wasn't right.

 

Anyway any help with the above is appreciated.

 

And if anyone wants to know yes I have posted this on Reddit and still no dice. So...

Edited by TheSingularity

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If this is high school level I don't think you could go past the popular HSC study guides.

 

the 4 unit levels (or whatever the most advanced HS levels are these days) will cover most of what you learn in first year uni - unless you're doing a maths degree, maybe.

 

As to your question, isn't it just a triangle with the longest side at 2km, and one angle (touching that side) of 40 degrees and the other angle at 25 degrees?

 

Then, you have to cut it in 2 using a vertical line, which is perpendicular to the longest side, and goes up to meet the corner made by the 2 shorter lines. Then find the length of this line.

 

You don't have to know the speed of the balloon - that's just misdirection.

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With math books I've found they are all reasonably similar, your best bet is to find a field you like and do a search on websites like amazon and then read the reviews attached.

 

With your question, it's a 2D problem yeah?, If I interpreted the question correctly I got an answer of h = 0.5995, if that ended up being correct I can help you got with some working if you need it.

 

EDIT: Ignore my answer I assumed that A and B were also on the ground, not that the points were at the same height as the balloon.

Edited by GoFaster

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Have a look at the books and resources here:

 

http://www.haesemathematics.com.au/

http://adelaidetuition.com.au/

http://www.masa.on.net/

 

To get a good handle on algebra, I recommend this book:

 

http://www.amazon.com/dp/0471530123/

 

The Trig for Dummies book and Workbook are OK and would be a good supplement through to Specialist Maths.

 

As for uni, I found the prescribed textbooks good enough for first year.

 

For the question, here's one way: Note A and B are to one side of C. Draw that and drop a vertical line from B to a point on the ground; let's call that point D and the line is the height, h.

 

By Pythagoras, CB^2 = CD^2 + h^2. You can find CB and CD in terms of h. Plug those in and solve for h.

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Just for those that are interested this is the textbook I'm using http://www.pussycat-ox.com/share/yr11_geom_trig_noPW.pdf (don't ask about the url only place I've found it uploaded to the net as the full textbook). Bottom of page 30 contains the problem.

 

And for those that wish to know the answer that I can't seem to solve page 412, Exercise 1E Q18 answer 2.099km.

 

http://i.imgur.com/vCAdvQn.jpg I'm not sure but is this anything like you describe sponger. Or well anyone...I'm thinking not...

 

Edit: Oh and basically where I'm up to in that textbook is everything I know on Right Angled Trigonometry.

 

Kothos were you referring to these books: http://www.fivesenseseducation.com.au/cata...aths-study-aids

Edited by TheSingularity

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Either I'm getting very rusty or this question is a little trickier than I expected. (Probably both.)

 

Try this:

 

http://img443.imageshack.us/img443/9428/trig.png

 

It should be a correct partial answer.

 

A, B and C are all on the ground, but you want the height, h, which is not on the ground, yet you haven't included it in your drawing.

 

I had to use simultaneous equations as well as basic trig to get the answer. Let me know if it's correct (I'm at work and am not going to work it out to the end).

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Either I'm getting very rusty or this question is a little trickier than I expected. (Probably both.)

 

Try this:

 

http://img443.imageshack.us/img443/9428/trig.png

 

It should be a correct partial answer.

 

A, B and C are all on the ground, but you want the height, h, which is not on the ground, yet you haven't included it in your drawing.

 

I had to use simultaneous equations as well as basic trig to get the answer. Let me know if it's correct (I'm at work and am not going to work it out to the end).

A and B are both h above the ground, they are the points the balloon travels between, C is on the ground. If your still having trouble getting the right answer one hint is that B is further from C than A and A is an arbitrary distance from C.

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A and B are both h above the ground, they are the points the balloon travels between, C is on the ground.

Ah, I see now. Still, that results in exactly the same triangle as the one I drew, except up-side-down and back-to-front. But that makes no difference to the answer. The partial answer I've worked out should hopefully be correct.

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A and B are both h above the ground, they are the points the balloon travels between, C is on the ground.

Ah, I see now. Still, that results in exactly the same triangle as the one I drew, except up-side-down and back-to-front. But that makes no difference to the answer. The partial answer I've worked out should hopefully be correct.

 

Your picture also has C as between A and B rather than to one side, which will also affect the answer.

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Tried another way going by Kothos picture and boy that didn't work (not using Kothos working out just tried some obscure way to make a larger triange using the 2km AB and the 25 degrees angle of B).

 

Been thinking about the language used and I've noticed they mention angle of elevation of A and B which doesn't make sense as that would mean there is a triangle above A and B instead of below like they describe...Started to think they wrote the question wrong...Or I'm just really confused and not seeing it at all.

Edited by TheSingularity

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Your picture also has C as between A and B rather than to one side, which will also affect the answer.

I interpret the requirement for A and B to be "on the same side" of C, to mean that C is closer to the reader/observer, but nevertheless between A and B.

 

That is, C is along the Z plane.

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I do think the question is a bit ambiguous somewhere - A, B and C are always going to be on the same plane, it's impossible for them not to be as they are only 3 points. So what is the questioner referring to when saying we can "assume" they are in the same plane?

 

 

Well, you've got the right answer GoFaster, so you obviously have the intended interpretation!

Edited by Kothos

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I do think the question is a bit ambiguous somewhere - A, B and C are always going to be on the same plane, it's impossible for them not to be as they are only 3 points. So what is the questioner referring to when saying we can "assume" they are in the same plane?

It's reducing a potentially 3D problem to a 2D one, I imagine the context of the question (in the sense of other questions from the same chapter) would make this clearer.

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I imagine you're right there as well. Perhaps some simpler problems led up to this one and so the ambiguity would have been much reduced, through precedence.

 

Good going working it out though.

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Well, you've got the right answer GoFaster, so you obviously have the intended interpretation!

Yeah, it was a poorly worded question the first time tried it as a 3D problem and the 2nd time I did it the same as you.

 

EDIT: that's what I get for trying to use to autoappend :P.

 

Good going working it out though.

Thanks :) Edited by GoFaster

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Okay now that looks really simple, thanks.

 

Although now I look at your working out I'm quite certain they fucked up the wording of it. As angle A is 40 degrees not angle Ca same as B was 25 degrees not angle Cb. Which without knowing they mean Ca and Cb then the angle of elevation makes no sense at all as that would result in another triangle above A and B.

 

Edit: Also seems way over what has been taught so far.

Edited by TheSingularity

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Okay now that looks really simple, thanks.

 

Although now I look at your working out I'm quite certain they fucked up the wording of it. As angle A is 40 degrees not angle Ca same as B was 25 degrees not angle Cb. Which without knowing they mean Ca and Cb then the angle of elevation makes no sense at all as that would result in another triangle above A and B.

 

Edit: Also seems way over what has been taught so far.

I think their wording is fine, my short hand Ca and Cb is equivalent to the what they say.

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Okay well you got it right after all.

 

All I'm trying to say is if it was angle ACM is 40 degrees and BCN is 25 degrees then that would make sense basically what you call Ca and Cb but what is taught as angles here. Unless though it is stated angle A or B then that just means to me the angle inside the triangle or shape at the point A. The same goes for B. Thus this ACM and BCN angles to me come from no where but make sense as then the angle of elevation wording makes sense.

 

It just clicked in my mind how you did h and that I could have gotten the same from what I had worked out with a friend but the angle problem I mentioned above would have still thrown me off.

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