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April 20, 2014

April 20, 2014

Posted by **Bob** on Thursday, November 29, 2007 at 3:23pm.

h ttp://i228.photobucket.cm/albums/ee127/wsu2012/Vector.jpg

If you could take a look at that diagram and help me figure out the displacement relative to point A, after the person walked from point A to point B, that would be incredibly helpful.

I know I have to sort them out into components, but I'm not sure how.

Again, if you can help, thank you very much.

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**Bob**, Thursday, November 29, 2007 at 3:25pmIf you want to look at the drawing, you have to remove the space between h and t in and add the letter 'o'.

Thanks again, it would really help me.

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**Reiny**, Thursday, November 29, 2007 at 3:31pmthere is no protocol that starts with

hottp://...

removing the space and going with the normal http://... gave me a "Not Found" message.

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**Bob**, Thursday, November 29, 2007 at 3:33pmI meant add the 'o' in .c om but it wouldn't let me post a URL.

Sorry for the confusion.

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**Reiny**, Thursday, November 29, 2007 at 4:38pmOk, I got the page, but I am having difficulty reading some of the data.

Especially the second vector running from the end of the 40 m horizontal at a 60º angle.

We have to know how long it is.

Suppose we call the end of that vector P.

Then we could find the magnitude of vector(AB) using the cosine law, and then using the sine law we could find the angle that vector(AB) makes with the 60º angle and the 80 m vectors.

Repeat the process by joining A to the top of the last vector and finding its magnitude and direction.

One more cosine law will allow you to find │vector(AB)│

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**Bob**, Thursday, November 29, 2007 at 5:05pmThe two labels on the right are 50 meters and 60 degrees.

I'm really sorry, but I don't understand what you were trying to say.

Can you please clarify?

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**Reiny**, Thursday, November 29, 2007 at 6:37pmOK, first of all let's label some points.

Let the end of the first vector from A be Q, let the end of the second vector be P and let the end of the third vector be R.

From basic geometry and parallel lines we have

angle AQP = 120º

angle RPQ = 30º and

RP makes a 60º angle with the vertical at R

Look at triangleAPQ, by cosine law

AP^2 = 40^2 + 50^@ - 2(40)(50)cos120º

.

.

AP = 78.1025

by sine law, sin APQ/40 = sin120/78.1025

.

.angle APQ = 26.33, so angle RPA = 3.67º

now look at triangle RPA, by cosine law

RA^2 = 80^2 + 78.1025^2 - 2(80)(78.1025)cos3.67º

..

RA = 5.406

by sine law, sin PRA/78.1025 = sin3.67/5.406

.

.

angle PRA = 67.63º

Finally in triangle RAB

angle BRA = 52.37º,....(180-60-67.63)

so

AB^2 = 30^2 + 5.406^2 - 2 930)(5.406)cos52.37º

AB = 27.04

WOW!!!!

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**Bob**, Thursday, November 29, 2007 at 9:36pmThank you so so so so so much!

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