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Math Help - bearing problems...please help as soon as possible...

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    bearing 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..
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    Quote Originally Posted by aikenfan View Post
    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..
    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 270^o - x

    we can find x using the sine rule:

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

    Now, continue
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    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..
    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:

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

    \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
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    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..
    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:

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

    \Rightarrow AC = \sqrt{380^2 + 240^2 - 2(380)(240) \cos 100^o} \approx 483.4
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    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..
    here's the diagram for number 29

    A is the position of the point.
    B is the position of the ship (as of noon) moving at 12 mph
    C is the position of the ship (as of noon) moving at 16 mph

    can you continue?

    do you know how i got the side lengths of the triangle?
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