I have no idea how to work these problems...there are four of them...I have the answer for each of them (for the back of the book)...i would really appreciate some guidance..

http://i103.photobucket.com/albums/m...19/22001-1.jpg

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- Nov 8th 2007, 05:51 PMaikenfanbearing problems...please help as soon as possible...
I have no idea how to work these problems...there are four of them...I have the answer for each of them (for the back of the book)...i would really appreciate some guidance..

http://i103.photobucket.com/albums/m...19/22001-1.jpg - Nov 10th 2007, 08:10 PMJhevon
for the first (see the diagram below).

we want the angle indicated by the red line (not x). to get it, we need to find x.

the bearing we seek is given by $\displaystyle 270^o - x$

we can find x using the sine rule:

$\displaystyle \frac {\sin x}{500} = \frac {\sin 46}{720}$

Now, continue - Nov 10th 2007, 08:37 PMJhevon
for the second, see the diagram below.

i used F for Franklin, C for Centerville, and R for Rosemont

Note that angle FCR = 32 + 90 + 15 = 137 degrees

we want the distance FR

By the cosine rule:

$\displaystyle FR^2 = 810^2 + 648^2 - 2(810)(648) \cos 137^o$

$\displaystyle \Rightarrow FR = \sqrt{810^2 + 648^2 - 2(810)(648) \cos 137^o} \approx 1357.8$ km

you can use the sine rule to find angle CRF, and then finding the bearing is easy from there - Nov 10th 2007, 09:05 PMJhevon
to number 27

see the diagram below

since angle DBC and angle ABC are angles on a straight line, they are supplementary (add up to 180 degrees), thus, angle ABC = 100 degrees.

thus, by the cosine rule:

$\displaystyle AC^2 = 380^2 + 240^2 - 2(380)(240) \cos 100^o$

$\displaystyle \Rightarrow AC = \sqrt{380^2 + 240^2 - 2(380)(240) \cos 100^o} \approx 483.4$ - Nov 10th 2007, 09:22 PMJhevon