Help with James Ruse Trial Papers (1 Viewer)

[VampireGirl]

Hi,
I need help with Question 10 a(i) in the James Ruse 1995 2U Maths paper.
Also question 9 part (b) from the 2004 trial paper. If anyone could help me that'd be great. I have attached the questions as a pdf (hopefully it works *fingers crossed*).

View attachment 18846

Thanks.

 

[lyounamu]

1st Q:

call the foot of the perpendicuular from B, y

so in triangle BYC:

cos@ = YC/a
YC = acos@

now look at triangle BYX

angle BXY is 45 degrees because from triangle BAX, BXY is calculated to be 45

so triangle BYX is an isoceles triangle with BY = YX

then sin@ = BY/a
BY = asin@

so YX = a sin@ too

then we know that XC = YC - YX = acos@ - a sin@ = a(cos@-sin@)

2nd Q:

i) since PM is h, OM must be h-8 because PO is 8cm. Using pythagoras' thereom

we can see that (h-8)^2 + OM^2 = 8^2
h^2 - 16h +64 + OM^2 = 64
so OM^2 = 16h - h^2
OM = sqrt(16h-h^2)

then OR = 2OM = 2sqrt(16h-h^2)

Area = 1/2 x a x b = 2sqrt(16h-h^2) x h x 1/2 = h sqrt(16h-h^2)

ii) restriction is that 16h - h^2 > 0
h(16-h) >0
so 0<h<16

iii)

A = h sqrt(16h-h^2)
= sqrt (16h^3 - h^4)

dA/dh = 1/2 (16h^3 - h^4)^-1/2 x (48h^2-4h^3)
so dA/dh = 0 when h = 12 or 16

but h =/= 16 so h = 12 when max area occurs since at h<12, increasing function and at h>12, decreasing function...

sub h=12 into the equation so you get max A = 83.138487...

 

[VampireGirl]

Seriously Namu, you're the best! Thank you sooooo much!!

 

[lyounamu]

Seriously Namu, you're the best! Thank you sooooo much!!

lol it's okay haha
how u going with cosmo?

 

[VampireGirl]

Yeh it's going ok i guess. Just working my way through past papers at the moment. How about you?

 

[lyounamu]

Yeh it's going ok i guess. Just working my way through past papers at the moment. How about you?

ah lol i get sick whenever i start thinking about cosmo :mad1: lol