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Mathematics: Post your doubts here!

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This is May June 10 P43. I kind of did this but can some one explain Q7 i) and ii)
Screenshot_2013-05-11-23-26-20.png

Here's the marking scheme:
Screenshot_2013-05-11-23-32-04.png
 

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If u r not clear by the diagram only:
  • I have used tangent theorem
  • first i found out the angle uOX
  • then multiplied by two!
  • Well i did it dis way cz u c both OM n OX r tangent to d circle n so Ou bisects the angle MOX.
One question on ma part: If min arg ws to be found which angle would it hv been?wud it hv bn angle uOX?


n guyzz cn u help me wid the sum here i nid help here desperately!:
https://www.xtremepapers.com/community/threads/pure-mathematics-3-help.25895/

Thank u!N hope i was helpful!:)

Erm thanks I'll get to argand diagrams when my mechanics paper is over :)
 
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For particle P:
u=20, a=-10,
s=ut+0.5at^2
s=20t+(0.5 x -10 x t^2)
s=20t - 0.5t^2

For Particle Q:
u=25, a=-10, time=t-4
s=25(t-4) - 0.5(t-4)^2
After simplifying you get
s=29t + 10.8 - 5t^2

Now equate both the distances :
20t-0.5t^2 = 29t + 10.8 -5t^2
After solving this you'll get t=1.2

For Particle P:
v=u+at
v=20+(-10 x 1.2) = 8m/s

For Particle Q:
v=25-10(t-0.4)
v=25-10(1.2-0.4) = 17m/s

I hope you got it :)
 
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The one in the first part, really sorry I forgot to specify.
Theres nthin to be sorry!
Well 1 lame thing but it works out: take out ur O level bks n go through some geometry n trigonometry basic stuffs or simply surf the net! n practice d simple sums!It worked out 4 me wen i realised i gt stuck in some kind of sums!

1 mr thng read d question thoroughly ... many hints are there already!uderline em!so that u cn mentally work out solutions like puzzle!Use all info in Q.
I gs deres no othr way to clear such general doubts!

As 4 dis sum: D solution is:
u c AMN forms a segment n we knw arc length,s=radius*theta
so here MN=r*x
nw perimeter of the segment=AN+AM+MN=2r+rx
Nw considering, the trianfle ADN
ang AND=x
[Cz ang AND n ang NAM r alternate angs]
hence, using trig for triangle AND
we cn write: sinx=a/r
=> a=rsinx
Nw using the info that perimeter of segment is half of perimeter of rectangle:
Perimeter of rectangle= 2*(3a+a)=8a
Nw perimeter of segment hence= 4a
We had calculated b4 that perimeter of segment=2r+rx
So we gt 2 prsn 4 d perimeter so lets equate both equns n so we get:
r(2+x)=4(rsinx)
so sinx=(2+x)/4 ....
Hope i cud help!
so I thnk practising mr sums lyk dis or goin' through mensuration,trigonometry n geometry sums frm O levels may clear ur doubt! :)
 
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Theres nthin to be sorry!
Well 1 lame thing but it works out: take out ur O level bks n go through some geometry n trigonometry basic stuffs or simply surf the net! n practice d simple sums!It worked out 4 me wen i realised i gt stuck in some kind of sums!

1 mr thng read d question thoroughly ... many hints are there already!uderline em!so that u cn mentally work out solutions like puzzle!Use all info in Q.
I gs deres no othr way to clear such general doubts!

As 4 dis sum: D solution is:
u c AMN forms a segment n we knw arc length,s=radius*theta
so here MN=r*x
nw perimeter of the segment=AN+AM+MN=2r+rx
Nw considering, the trianfle ADN
ang AND=x
[Cz ang AND n ang NAM r alternate angs]
hence, using trig for triangle AND
we cn write: sinx=a/r
=> a=rsinx
Nw using the info that perimeter of segment is half of perimeter of rectangle:
Perimeter of rectangle= 2*(3a+a)=8a
Nw perimeter of segment hence= 4a
We had calculated b4 that perimeter of segment=2r+rx
So we gt 2 prsn 4 d perimeter so lets equate both equns n so we get:
r(2+x)=4(rsinx)
so sinx=(2+x)/4 ....
Hope i cud help!
so I thnk practising mr sums lyk dis or goin' through mensuration,trigonometry n geometry sums frm O levels may clear ur doubt! :)
Many thanks for the answer! <3

And I will go through the basics real quick whenever I stumble across such types of questions. I have the habit of freaking out and leaving this question to the end and sometimes I get it wrong :/. Ty again ^^
 
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