bossleymaths
Member
- Joined
- Mar 8, 2008
- Messages
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- HSC
- 2009
i got 101 but i dont seem to have a solution to the circle geometry, could anyone please show me how to do question 6 b) ii
PTQ+TQP+QBP+BPT=360 (angle sum of a quad) (*)i got 101 but i dont seem to have a solution to the circle geometry, could anyone please show me how to do question 6 b) ii
we had this conference with a mx2 marker and he said noone's ever got full marks in an mx2 paper?At least 118/120. Keep in mind, that this is only an educated estimate. It is highly likely that someone (or many people) may get full marks in this year's paper, in which case a raw mark of 118/120 would be lucky to get a state ranking.
Really? I thought there was some guy who now lectures @ Usyd or UNSW? Could be mistaken though (it appears that I am).we had this conference with a mx2 marker and he said noone's ever got full marks in an mx2 paper?
i dunno. i walked in halfway through the conference so i may not have been listening properly but i thought that's what he said?Really? I thought there was some guy who now lectures @ Usyd or UNSW? Could be mistaken though (it appears that I am).
Lol. In my opinion, 110 and 120 imply the same level of skill.i dunno. i walked in halfway through the conference so i may not have been listening properly but i thought that's what he said?
eh.
116+ is close enough to full marks for my liking haha.
NICE thanks man now i get itPTQ+TQP+QBP+BPT=360 (angle sum of a quad) (*)
but from i) TPB+TQB=α+β+2θ and PBQ=180-PBS=180-θ (angles on a straight line)
sub this into (*)
PTQ=180-(α+β+θ) (**)
DAB=2α+θ (exterior angle of triangle equal to the sum of interior, opposite angles)
DAB=BAC=2α+θ (interior angle of cyclic quad is equal to opposite, exterior angle)
CBR=SBP=θ (vert. oppo. angles)
so, now the angle sum of ΔBQR is,
CBR+BRC+RCB=180
subbing in CBR=θ, BRC=2β and DAB=2α+θ
2(α+β+θ)=180
α+β+θ=90 (***)
sub (***) into (**)
PTQ=180-(α+β+θ)=180-90=90
.'. ST is perpendicular to RT
It was 112.The top mark for that paper was 80% at SGS.
Didn't he get 115 for his trials and 119 or so in the HSCAh. The guy who topped MX2 here got 118 for our trials
arg gg.
I think he equalled with some other person. I think there's been more than one person to get 120 for MX2 previous to 1993.Nah, someones got full marks...Anthony Henderson is the name, I believe.
Nah 118 for MX2 trials.Didn't he get 115 for his trials and 119 or so in the HSC
Terence Tao the name you're looking for?Really? I thought there was some guy who now lectures @ Usyd or UNSW? Could be mistaken though (it appears that I am).
correctI believe there has only been one person to get the full 120/120 in the 4U HSC (1993)