Couple of Questions

g0d · 2007-02-28 11:09:53

Got a couple of problems that I'm stumped on. I got a bit on some of them but I can't get the final answer. X.x
1.

2.

3.

For 1 I can get the length AC and BC from the speed and time taken but then I don't know what to do, maybe cos rule to get some angles?

For 2 & 3 I'm pretty much stumped.

John E. Franklin · 2007-02-28 12:12:22

Okay, for number 2. (also called #7), you need to understand
using cosines and sines to get coordinates when you know
the angle and the hypotenuse.
Define point R as (0,0), the origin.
Now L is therefore (5cos60,5sin60).
And N is simply (1,0)
Now use distance formula (also pythagoreans theorem) to get distance from N to L.
Once you know that, then the sizes are 6 vs. N to L, and therefore the areas are the square of that.

John E. Franklin · 2007-02-28 13:53:28

When you get the answer to #3 (or 8), you might be surprised it appears to be the Sumerian Natural Pi value of 3.14626437 !!!!!!!!!! Pretty cool, whatever it means!!!
Also the reciprocal value of 0.3178372 comes out with 30 degrees, but then the
picture is not really the same.

JaneFairfax · 2007-02-28 14:19:09

For #1:

Let the point C be

. Then

Last edited by JaneFairfax (2007-03-01 02:04:54)

John E. Franklin · 2007-02-28 14:35:37

More on #3 (or 8).
Originally I did it with a computer program, so that's why I can only give you
the answers.
Assign point S as (0,0), the origin.
Point Q ends up being (sqrt{2} + sqrt{3},1)
Point T is (sqrt{3}/4,0.75)
Point R is (sqrt{3},0)
Hope that helps.

John E. Franklin · 2007-02-28 14:37:51

So the vector RQ points 35.26438968 degrees to the upper-right.
And vector TQ points 30 less than that, or 5.264389683 degrees to the upper-right.

JaneFairfax · 2007-02-28 15:11:02

John E. Franklin wrote:

When you get the answer to #3 (or 8), you might be surprised it appears to be the Sumerian Natural Pi value of 3.14626437 !!!!!!!!!! Pretty cool, whatever it means!!!

I worked it out to be √3+√2.

John E. Franklin · 2007-02-28 15:12:46

Cool!!, and without a computer helping you?? Amazing!!

JaneFairfax · 2007-02-28 15:23:53

Actually, Id love to have one of those computer programs to play with.

John E. Franklin · 2007-02-28 15:26:45

I'm doing my calculations for this example with a free language.
It's "Just Basic". You can do atn() for arctangent, and n^0.5 for sqrt.
Very simple stuff. Sometimes I use other languages too, all free ones off the net.

Welp, time for beddy-bye.

Last edited by John E. Franklin (2007-02-28 15:28:46)

Stanley_Marsh · 2007-02-28 18:49:26

No.2 11:8 right?

g0d · 2007-02-28 19:08:21

No.2 answer is 12:7, I got that one now.
Still trying to get no.1 though. ><

g0d · 2007-02-28 19:24:58

Anyone give me some more help with question 1? I found AC = 23.948405 and BC=31.3053 but then to get some angles I tried cos and started getting wierd answers. X.x Someone confirm my lengths are right?

g0d · 2007-02-28 22:55:16

Anyone?

g0d · 2007-02-28 23:36:55

So I'm pretty sure that my AC and BC lengths are right so this is what I put into cos rule to try and get an angle:
AC²=AB²+BC²-2(AB)(BC)cosX

And I get like 2. something so I don't know what to do I've tried it with the other sides and I always get a answer over 1. >< So annoying, got a validation on it tommorrow ><. Any help what so ever is very much appreicaited.

g0d · 2007-03-01 00:27:21

Well, while I wait for someone to reply I got going on question 8, (no.3). I got that side SR is √3 and thus part of PQ is √3 but I can't seem to work out how to get the √2 for the extra bit of PQ. Looks like not much sleep for me tonight. ><

JaneFairfax · 2007-03-01 02:02:03

I already gave you a hint for #1 above. See earlier in the thread. To get each of the distances AC and BC, multiply the speed by the time and divide by 2 (which is what youve already done).

For #2, my answer is also 12:7. The length of each side of ∆LMN should be √21.

For #3, find the legnth PR, which is the sum of PT and RT. Since ∆PST, ∆SRT and ∆PRS are similar (they all have the same interior angles), this should be straightforward to work out. Next, calculate QR. Using the hint that ∆PRQ and ∆QRT are similar, you should get QR:PR = RT:QR. The answer for QR should be √3. Now, if U is the point on PQ such that PSRU is a rectangle, you know that PU = SR and UR = PS. Calculate UR and youre done.

Last edited by JaneFairfax (2007-03-01 06:06:58)

John E. Franklin · 2007-03-01 13:43:30

More tips on # 1, the tumor locator.

By using the law of cosines, shown below,
you can find all the angles of the triangle
between the tumor and the 2 radio sources.

The distance in millimeters to the tumor is one-half 2.99E11 times time-both-ways.

Good Luck. Oh, and if you don't want to use the law of sines, then
write equations of the two different size circles made by the radio waves
going out in every direction. Now find where these 2 circles intersect.

Bye.

Last edited by John E. Franklin (2007-03-01 14:38:32)

John E. Franklin · 2007-03-01 15:03:46

#1.) Distance AC = 23.948405mm
Distance BC = 31.3053mm
Distance AB =

mm

Now you have 3 sides of a triangle.
Let's find

.

Last edited by John E. Franklin (2007-03-01 17:47:48)

John E. Franklin · 2007-03-01 15:15:55

This triangle does not exist!!!!!!!!!
Because squareRoot of 10 is about 3.16 and AC of 23.9 and BC of 31.3 differ by 7.35, which is
more than 3.16, the smallest side of the triangle. Not possible.
The times in the problem are wrong.

John E. Franklin · 2007-03-01 15:26:14

Here's an example of
why the triangle won't work
with different numbers for
the side lengths.

John E. Franklin · 2007-03-01 15:30:14

The only way this might work, is if the machine drawing with the head is
in meters or centimeters. If the sensors are only 3.1622mm apart, it won't work.
They have to be 31.622mm apart (something greater than 7.356mm)
or 316.22mm apart or 3162.2mm apart.

Math Is Fun Forum

#1 2007-02-28 11:09:53

Couple of Questions

#2 2007-02-28 12:12:22

Re: Couple of Questions

#3 2007-02-28 13:53:28

Re: Couple of Questions

#4 2007-02-28 14:19:09

Re: Couple of Questions

#5 2007-02-28 14:35:37

Re: Couple of Questions

#6 2007-02-28 14:37:51

Re: Couple of Questions

#7 2007-02-28 15:11:02

Re: Couple of Questions

#8 2007-02-28 15:12:46

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#9 2007-02-28 15:23:53

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#10 2007-02-28 15:26:45

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#11 2007-02-28 18:49:26

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#12 2007-02-28 19:08:21

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#13 2007-02-28 19:24:58

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#14 2007-02-28 22:55:16

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#15 2007-02-28 23:36:55

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#16 2007-03-01 00:27:21

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#17 2007-03-01 02:02:03

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#18 2007-03-01 13:43:30

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#19 2007-03-01 15:03:46

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#20 2007-03-01 15:15:55

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#22 2007-03-01 15:30:14

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