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# 3D Geometry

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1. Attached. No idea where to start: I'm not very good at visualising 3D problems.
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2. Maths_3U_1999_HSC6.pdf (5.8 KB, 34 views)
3. (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).
4. (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.
5. (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.
6. (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?
7. (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. . .
8. (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?
9. (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 .
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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