Posted by Bob on Thursday, November 29, 2007 at 3:23pm.
If 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.
there is no protocol that starts with
hottp://...
removing the space and going with the normal http://... gave me a "Not Found" message.
I meant add the 'o' in .c om but it wouldn't let me post a URL.
Sorry for the confusion.
Ok, 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)│
The 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?
OK, 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!!!!
Thank you so so so so so much!