1988 Hsc (1 Viewer)

[withoutaface]

I was doing the 1988 hsc paper and found that I couldn't do one question and i am unsure of my answer for part ii) of the other. Can someone have a go at them for me?

Thanks in advance,

Justin.

 

[Estel]

I'm sorry I can't do the 2nd as I haven't done 4U...

but here's q1
In triangle GCK
h/CK = tan45
CK = h
In triangle FBK
h/BK = tan30
BK = rt3.h

So in triangle BCK
l^2 + h^2 -2lhcos135 = 3h^2
2h^2 +2lhcos135 -l^2 = 0
2(h/l)^2 -rt2.(h/l) - 1 =0
h/l = [rt2 +- rt10]/4
but h/l >0 as both are positive lengths
hence
h/l = [rt2 + rt10]/4

 

[Xayma]

1st Q

∠KCG = 90°

∴ tan 45 ° =h/CK
h/CK=1
CK=h


∠ KCB =135° ( ∠on a straight line, diagonal of square bisects ∠ )


∠FBK =90 °

∴ tan 30 ° =h/BK
h/BK= 1/ √ 3
BK= h√ 3

Let
∠ CBK = θ
∴ sin θ=1/√ 6 (by application of sine rule and ∠ KCB)

∴ h/l=sin θ/sin (45 - θ ) (sine rule)

=1/&radic 6 (sin 45*cos θ - cos 45*sin θ )

(do a right angled triangle to show that cos θ =√ (5/6)

=1/√ 6 ([√ 5 -1]/ √ 12)
=√ 2 / √ 5 -1
=(√ 10 + √ 2) /4

 

[Estel]

2nd part is.

Part A is simply comparing areas
144 = (16+x)h + (20+x)(4-h)
= -4h +4x
hence x = 16+h
But Volumes is 4U =/

 

[Xayma]

Yeah I realised when the sides were in the same ratio, which means that it isnt just a simple rectangular prism with its top cut off.

 

[Estel]

I can't wait till next term

 

[Xayma]

I can't wait till next term

Neither, I wanna finish these stupid science's, and relearn all 2 pages worth of science present in the entire HSC.

 

[CrashOveride]

Wrong to post here but i didnt feel like creating a new thread.

Find the equation of the two tangents to the circle x^2+y^2-2x-6y+6=0 which pass thru the point P(-1,2). Use lx+my+n=0 (warning: do NOT use y=mx+b)

Thanks in advance. (Also why the warning?)

 

[Estel]

Eq through (-1,2)
(x+1) + k(y-2) = 0

2x + 2y.dy/dx - 2 -6.dy/dx = 0
dy/dx = (x-1)/(3-y)
ie k = (x-1)/(y-3)

so (x+1) + (x-1)(y-2)/(y-3) = 0
x + 1 + [xy + 2 - 2x - y]/y-3 = 0
xy - 3x + y - 3 + xy + 2 - 2x - y = 0
2xy - 5x -1 =0

I could, of course, have this horribly wrong.
=P

Warning is to avert algebra bash?

EDIT: yup it's wrong.
i'll think abt this...

 

[withoutaface]

Eq through (-1,2)
(x+1) + k(y-2) = 0

2x + 2y.dy/dx - 2 -6.dy/dx = 0
dy/dx = (x-1)/(3-y)
ie k = (x-1)/(y-3)

so (x+1) + (x-1)(y-2)/(y-3) = 0
x + 1 + [xy + 2 - 2x - y]/y-3 = 0
xy - 3x + y - 3 + xy + 2 - 2x - y = 0
2xy - 5x -1 =0

I could, of course, have this horribly wrong.
=P

Warning is to avert algebra bash?

EDIT: yup it's wrong.
i'll think abt this...

I said get the circle with y in terms of x, then equate mx+m+2=y to the circle equation, find the discriminant, and equate that to zero, but apparently the disc. was linear (and hence had only one solution), just a warning for those who were going to try that.

 

[Estel]

Yay...

lx+my+n=0,
where - l + 2m + n = 0 [passes (-1, 2)]
Also,
|l + 3m+n|/rt(l^2+m^2)=2 (perp distance of line to (1,3) is 2)
i.e. (m+2l)^2 = 4m^2 + 4l^2
m^2 + 4ml + 4l^2 = 4m^2 + 4l^2
4ml=3m^2 i.e. m = 0 or m = 4l/3
So one tangent is x= - 1
The other:
x+4y/3+n=0
but n = l-2m
= -5/3
So x+4y/3-5/3=0

 

[CrashOveride]

Ok actually i found one of them through inspection. The point (-1,3) is the far most left and (-1,2) lies below it so x=-1 must be one of the equations.

Now for the 2nd..

 

[CrashOveride]

Thnx Estel...just one thing how did u get l=1

 

[Estel]

k and l are essentially a ratio
m = 4l/3
so if we go for l = 1 [easiest scenario], then m=4/3

EDIT: an example for you:
suppose we take l = 2 then m=8/3
eq is 2x+8y/3-10/3=0
which simplifies as required.

 

[CrashOveride]

Cheers mate