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    Attached. No idea where to start: I'm not very good at visualising 3D problems.
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    (Original post by HapaxOromenon2)
    Attached. No idea where to start: I'm not very good at visualising 3D problems.
    For (i) Try using Pythagoras, remember that D is on the surface of the hemisphere - so OD is. . .?
    You'll probably want to use their hint regarding the angles for (ii).
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    (Original post by joostan)
    For (i) Try using Pythagoras, remember that D is on the surface of the hemisphere - so OD is. . .?
    You'll probably want to use their hint regarding the angles for (ii).
    OD is 10, so by Pythag triple (6,8,10), FD=8. Got it.
    Then for (ii), <DFE = <BOC, and by the arc length formula 10 * <BOC = 4 -> <BOC = 2/5. Is that right?
    Still unsure for (iii) though.
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    (Original post by HapaxOromenon2)
    OD is 10, so by Pythag triple (6,8,10), FD=8. Got it.
    Then for (ii), <DFE = <BOC, and by the arc length formula 10 * <BOC = 4 -> <BOC = 2/5. Is that right?
    Still unsure for (iii) though.
    That's what I wrote for (ii) yeah.
    From there I guess you can get the length of the straight line DE, and from there you should be able to get the angle DOE.

    There might be a slicker way than this, my Geometry isn't great.
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    (Original post by joostan)
    That's what I wrote for (ii) yeah.
    Well you can then work out the arc length DE.
    From there I guess you can get the length of the straight line DE, from there you should be able to get the angle DOE.

    There might be a slicker way than this, my Geometry isn't great.
    So arc DE = 8*2/5 = 16/5, but how do I get from there to the length of the straight line DE?
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    (Original post by HapaxOromenon2)
    So arc DE = 8*2/5 = 16/5, but how do I get from there to the length of the straight line DE?
    Yeah sorry I don't know why I put the arc length in, I did calculate that, but then realised I didn't really need it.

    The triangle FED is isoceles, and you have the angle DFE. . .
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    (Original post by joostan)
    Yeah sorry I don't know why I put the arc length in, I did calculate that, but then realised I didn't really need it.

    The triangle FED is isoceles, and you have the angle DFE. . .
    So FE=FD=8 and <DFE=2/5, so using the Cosine Rule, DE=3.1787...
    Then OD=OE=10, so by the Cosine Rule, <DOE=0.319 (3.d.p.)
    Is that right?
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    (Original post by HapaxOromenon2)
    So FE=FD=8 and <DFE=2/5, so using the Cosine Rule, DE=3.1787...
    Then OD=OE=10, so by the Cosine Rule, <DOE=0.319 (3.d.p.)
    Is that right?
    Yup, I got an exact value of 2\arcsin\left(\dfrac{4\sin(\frac  {1}{5})}{5} \right) \approx 0.319.
    I did this using the half angle formula on my expression for DE, and then simple trig to get the angle.
 
 
 
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