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MATH AS QUESTIONS>>>

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Eamyzz:
Personally I solved all da past papers starting 4rm 2000 till 2009
so i kind of gt bored...so i thought about studying in groups ie we will choose da hardest questions n post da years n all of us will start solving dem then we can compare answers n methods this will encouage us to work, study and revise
i hope u lykd ma idea  ;) ;)



If any1 has any doubts on any question u can just post it regardless hw easy or hard it is n gd luck 2 all of us  :)

Eamyzz:
lets start wid
O/N05 - 3
O/ N05 -5
M/J 06-7
O/N06-7
O/N08-7

thats all lets start solving dem n post da answers n check them....plz if any1 has any suggested questions just post them
Good luck  ;)

astarmathsandphysics:
Shall i stay out of this?

slvri:

--- Quote from: Eamyzz on October 14, 2009, 01:18:18 am ---lets start wid
O/N05 - 3
O/ N05 -5
M/J 06-7
O/N06-7
O/N08-7

thats all lets start solving dem n post da answers n check them....plz if any1 has any suggested questions just post them
Good luck  ;)

--- End quote ---
is this for p1?
ok then.......ono5 q3
(i)in triangle CBE
sin 60=CE/(2sqrt(3)d)
CE=(sqrt3/2)*(2sqrt(3)d)=3d
let X be a point on AD such that BX=ED and BE=XD
sin 30=BX/2d
BX=0.5*2d=d
thus ED=d
so CD=3d+d=4d
part ii will be answered in next post

slvri:
ok...part ii
cos 30=AX/2d
AX=2d*(sqrt3/2)=(sqrt3)(d)
cos 60=BE/2sqrt3(d)
BE=2sqrt3*0.5=sqrt3(d)
XD=sqrt3(d)
AD=AX+XD=sqrt3(d)+sqrt3(d)=2sqrt3(d)
tan CAD=CD/AD
tan CAD=4d/2sqrt3d
tan CAD=2/sqrt3
CAD=tan-1(2/sqrt3)
hope this helped
more answers coming soon ;)

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