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    I just need clarification/help on queston 7. (ii) and 10. (ii).

    If the attachments dont work heres a link to the exam paper:

    http://www.mei.org.uk/files/papers/c206ju_iu78.pdf

    Thanks if you respond!
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    7 ii is just a normal trig equation, just use the acute angle for which the cosine of is 0.5 (you should know it off by heart but you can get it either by considering an equilateral triangle, or from a calculator) then use the symmetry of cos to find the other solutions in the range. Remember that you'll need to find values of 2x between 0 and 720 to find all the values of x.

    On 10 ii, which part are you stuck on? To do the bit involving finding an angle in radians, just use the s = r(theta) formula with r = 5.4, s = 24. For the other bit, you know the angle between LS and north from part i, and converting to degrees will give you the angle "theta". From there you should be able to get the bearing.
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    For Q 10ii)

    The first step is to work out the distance the ship has travelled.

    So this would be 24km/h x 1h(26/60) = 10.4km

    This as a fraction of the entire circumference of the circle of radius r = 5.2km is 10.4 / (5.2 x 2 x pi) = 0.31

    So our angle, theta, is equal to 0.31 x 360 = 114 degrees.

    So, the bearing of the ship from the lighthouse is given by 360 - (114 - 48) = 293 degrees.

    Hope this helps.
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    (Original post by matt2k8)
    7 ii is just a normal trig equation, just use the acute angle for which the cosine of is 0.5 (you should know it off by heart but you can get it either by considering an equilateral triangle, or from a calculator) then use the symmetry of cos to find the other solutions in the range. Remember that you'll need to find values of 2x between 0 and 720 to find all the values of x.

    On 10 ii, which part are you stuck on? To do the bit involving finding an angle in radians, just use the s = r(theta) formula with r = 5.4, s = 24. For the other bit, you know the angle between LS and north from part i, and converting to degrees will give you the angle "theta". From there you should be able to get the bearing.
    Hey do you know what the graph of cos(2x)=0.5 looks like for 0<x<360. I dont know what it looks like.
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    (Original post by tgodkin)
    For Q 10ii)

    The first step is to work out the distance the ship has travelled.

    So this would be 24km/h x 1h(26/60) = 10.4km

    This as a fraction of the entire circumference of the circle of radius r = 5.2km is 10.4 / (5.2 x 2 x pi) = 0.31

    So our angle, theta, is equal to 0.31 x 360 = 114 degrees.

    So, the bearing of the ship from the lighthouse is given by 360 - (114 - 48) = 293 degrees.

    Hope this helps.
    Yea thanks that helps. Your measuring the bearing from the north line to S right?
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    Bearing is always the angle CLOCKWISE to the line you want.
 
 
 
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