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Math Help - How do you solve this bearing trig problem?

  1. #1
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    How do you solve this bearing trig problem?

    A ship leaves base and sails 30 km on a bearing of 034. It is then ordered to take up position due North of the base in exactly 1 hour's time. If the ship has a speed of 20km/h, find the two possible bearings it could take.
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  2. #2
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    Hello womatama

    Welcome to Math Help Forum!
    Quote Originally Posted by womatama View Post
    A ship leaves base and sails 30 km on a bearing of 034. It is then ordered to take up position due North of the base in exactly 1 hour's time. If the ship has a speed of 20km/h, find the two possible bearings it could take.
    In the attached diagram, the ship's two possible courses are given by BC_1 and BC_2. Then use the Sine Rule:

     \frac{\sin\theta}{30}=\frac{\sin34^o}{20}

    \Rightarrow \sin\theta =0.8388

    There are then two possible values of \theta: \theta_2 = 57.0^o and \theta_1 = 180 - 57.0 = 123.0^o

    So the two possible courses to steer are 360 - \theta_2 = 303^o and 360-\theta_1 = 237^o.

    Grandad
    Attached Thumbnails Attached Thumbnails How do you solve this bearing trig problem?-untitled.jpg  
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello womatama

    Welcome to Math Help Forum!
    In the attached diagram, the ship's two possible courses are given by BC_1 and BC_2. Then use the Sine Rule:

     \frac{\sin\theta}{30}=\frac{\sin34^o}{20}

    \Rightarrow \sin\theta =0.8388

    There are then two possible values of \theta: \theta_2 = 57.0^o and \theta_1 = 180 - 57.0 = 123.0^o

    So the two possible courses to steer are 360 - \theta_2 = 303^o and 360-\theta_1 = 237^o.

    Grandad
    Oh thanks alot !! i drew the thing but didnt know what to do lol. Thanks.
    One question: How come you took 57 away from 180 to get the other value? i dont understand that
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  4. #4
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    Hello womatama
    Quote Originally Posted by womatama View Post
    Oh thanks alot !! i drew the thing but didnt know what to do lol. Thanks.
    One question: How come you took 57 away from 180 to get the other value? i dont understand that
    I don't know how much you know about sines and cosines, but amng the rules are, for any value of \theta:

    • \sin\theta = \sin(180-\theta)


    • \cos\theta = \cos(180-\theta)

    which tells you how to find the sine and cosine of obtuse angles. So, for example, \sin57^o = \sin(180-57)^o=\sin123^o (and \cos 123^o = -\cos57^o). Check these out on your calculator if you like.

    Grandad
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  5. #5
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    Oh i get it now
    i looked at the sine graph and cosine graph and then i understood it. THX
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