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Math Help - Finding degree and bearing

  1. #1
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    Finding degree and bearing

    Town C is 5 miles due east of town D.
    Town E is 12 miles from town C at a bearing (from C) of N52E.

    Two parts of the problem, I'm not sure about the second part:

    1: How far apart are towns DE?

    Using the law of cosines, I found that DE is 16 miles.

    2: Find the bearing of two E from town D, round to nearest one-half mile.

    Using the law of sines, I found that angle D is 11 degrees. If I use the NS line, and assume that D is 11 degrees of a full 90 degree angle, I can subtract 11 from 90, resulting in 79 degrees.

    Am I calculating part two correctly?

    Thank you!
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  2. #2
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    Quote Originally Posted by Snowcrash View Post
    Town C is 5 miles due east of town D.
    Town E is 12 miles from town C at a bearing (from C) of N52E.

    Two parts of the problem, I'm not sure about the second part:

    1: How far apart are towns DE?

    Using the law of cosines, I found that DE is 16 miles.

    2: Find the bearing of two E from town D, round to nearest one-half mile.

    Using the law of sines, I found that angle D is 11 degrees. If I use the NS line, and assume that D is 11 degrees of a full 90 degree angle, I can subtract 11 from 90, resulting in 79 degrees.

    Am I calculating part two correctly?

    Thank you!
    agree with #1 ... DE = 16.2 miles

    using the law of sines, I get m\angle{D} = \arcsin\left(\frac{12\sin(142^{\circ})}{DE}\right) \approx 27^{\circ}
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  3. #3
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    Thanks - I do see that angle E = 27 degrees - would the bearing from D to E then be S27W?

    I thought when the problem asked the bearing from town D to E, I'd need to find the angle of Town D because it's "From D to E" and the result would be N79E for bearing. Setting up word problems is always difficult for me!
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  4. #4
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    Quote Originally Posted by Snowcrash View Post
    Thanks - I do see that angle E = 27 degrees - would the bearing from D to E then be S27W?

    I thought when the problem asked the bearing from town D to E, I'd need to find the angle of Town D because it's "From D to E" and the result would be N79E for bearing. Setting up word problems is always difficult for me!
    m\angle{E} = 11^{\circ}

    bearing of town E from town D would be N 63^{\circ} E

    look at the rough layout ...
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  5. #5
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    Eek - I'm still not seeing it - sorry - I do have my drawing as yours is set up too -

    So, to find angle E we use the law of sines I did this:

    sin E / 5 = sin 142 / 16
    16 sin E = 5 sin142
    sin E = 5 sin142 / 16

    As I'm typing this I think I see what I did - I flipped the values, using the side length 5 instead of 12. I came up with the same answers, just reversed.

    Thanks Skeeter!
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  6. #6
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    Quote Originally Posted by Snowcrash View Post
    Eek - I'm still not seeing it - sorry - I do have my drawing as yours is set up too -

    So, to find angle E we use the law of sines I did this:

    sin E / 5 = sin 142 / 16
    16 sin E = 5 sin142
    sin E = 5 sin142 / 16

    \textcolor{red}{m\angle{E} = \arcsin\left(\frac{5\sin(142)}{16.2}\right) = 11} degrees
    ...
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