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Math Help - Trig help

  1. #1
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    Trig help

    Radio Station A is 100 miles due north of radio station B. Station A receives a distress message froma ship at a bearing of S 50 E, while station B receives the same message at a bearing of N 46 E.

    How would a diagram for this look like?
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  2. #2
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    Quote Originally Posted by sinjid9 View Post
    Radio Station A is 100 miles due north of radio station B. Station A receives a distress message froma ship at a bearing of S 50 E, while station B receives the same message at a bearing of N 46 E.

    How would a diagram for this look like?
    S 50^o E means "point South, then turn to the East 50 degrees".
    Attached Thumbnails Attached Thumbnails Trig help-ship.jpg  
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  3. #3
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    Find the ratio of the sides of a triangle if:
    (a) the angles are in the ratio 1:2:3
    (b) the angles are in the ratio 4:5:6
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  4. #4
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    Quote Originally Posted by sinjid9 View Post
    Find the ratio of the sides of a triangle if:
    (a) the angles are in the ratio 1:2:3
    (b) the angles are in the ratio 4:5:6
    Hi sinjid9,

    if the ratio of angles is 1:2:3 there are 6 parts and the sum of the angles in a triangle is 180 degrees.

    One angle is \frac{180^o}{6}=30^o

    The other angles are twice and three times that, 60^o,\ 90^o

    To calculate the ratios of the sides, you can re-arrange the Sine Rule (Law of Sines).

    \frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}

    to get

    \frac{SinA}{SinB}=\frac{a}{b}

    then a:b=SinA:SinB

    Also \frac{SinB}{b}=\frac{SinC}{c}

    \frac{SinB}{SinC}=\frac{b}{c}

    then a:b:c=SinA:SinB:SinC


    When the ratio of the angles is 4:5:6, there are 15 parts

    \frac{180^o}{15}=12^o

    Then the angles are 4(12^o),\ 5(12^o),\ 6(12^o)

    then solve as above
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  5. #5
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    A satellite travels in a circular orbit 1600km above Earth and takes 2 hours to complete one orbit. It will pass directly over a tracking station at 12:00 noon. If the tracking station can first pick up the satellite at a 30 degrees angle of elevation, at what time will the satellite appear? (The radius of the Earth is about 6400 km)

    Does it look something like this?
    http://img231.imageshack.us/i/mathpic2.png/
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  6. #6
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    Hello, sinjid9!

    You should start another thread for a new problem.


    A satellite has a circular orbit 1600km above Earth and has a 2-hour orbit.
    It will pass directly over a tracking station at 12:00 noon.
    If the tracking station can first pick up the satellite at a 30 degrees angle of elevation,
    at what time will the satellite appear? (The radius of the Earth is about 6400 km.)
    Code:
        S
         o
          \ *
           \  *
       1600 \   *
             \ 30* T
      H - - - \ - - o - - - - -
              *\    :     *
            *   \   :       *
           * 6400\  :6400    *
                  \ : 
          *        \:         *
          * - - - - o - - - - *
                    C

    The center of the earth is C.
    The tracking station is at T\!:\;TC = 6400 km.
    The satellite is at S\!:\;SC = 8000 km.

    The angle of elevation of the satellite is 30.
    . . \angle STH = 30^o,\;\angle STC = 120^o

    In \Delta STC, Law of Sines: . \frac{\sin S}{6400}\:=\:\frac{\sin120^o}{8000} \quad\Rightarrow\quad \sin S \:=\:0.692830323

    . . Hence: . \angle S \:\approx\:43.85^o

    . . And: . \angle C \:=\:180^o - 120^o - 43.85^o \:=\:16.15^o


    The satellite moves 360 in 2 hours.

    To move 16.15, it will take: . \frac{16.15^o}{360^o}\times 2 \;=\;\frac{323}{3600}\text{ hours} \;=\;\frac{323}{60}\text{ minutes}


    The satellite will appear 5 minutes, 23 seconds before 12 noon.

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