Preliminary mathematics marathon (2 Viewers)

[edmundsung]

Re: another question

lol I read "distance between the centres" as "distance between the circles", haha, I thought AB was 17+10+2=29. lol.

So yeah, PQ is 15 cm.

haha anyway thank you very much again

 

[fullonoob]

Re: another question

i did it diff way but same answer
Let X be on AP such that AX = 2cm

PQ^2 = 17^2 - 8^2
PQ = 15

 

[ohexploitable]

Re: another question

ehh i got
let point X be on AP such that AX = 2cm
PQ^2 = 8^2 + 17^2
PQ = 18.79

Umm... you mean 17² - 8²?

 

[ohexploitable]

Re: another question

New Question:



Lemniscate of Bernoulli - Wikipedia, the free encyclopedia

 

[fullonoob]

Re: another question

i'll let 011er's answer this
if not done by 6 then ill do it, then merlin time

 

[mirakon]

Re: another question

New Question:



Lemniscate of Bernoulli - Wikipedia, the free encyclopedia

Hint for those attempting it: Use implicit differentiation. Although I think this is in HSC course, not prelim anyway.

 

[hungwell1337]

 

[edmundsung]

another question

AB, CD are perpendicular chords of a circle, centre O. Prove that angle DAB = angle OAC


thxx

 

[hungwell1337]

let angle DAB = X
so angle CDA = 90-X (right angled triangle)
let angle OAB = a
let angle OCD = ODC = b isosceles
let angle OAC = OCA = Y "
let the intersction of the two chords be E

now we can see triangle OAD is isosceles
angle OAD = angle ODA

a + X = 90 - X + b

rearranges into 2X = 90 - a + b

now look at triangle AEC a right angled triangle
angle ACE + angle CAE = 90

a + Y + (Y-b) = 90

rearranges into 2Y = 90 - a + b

therefore Y = X

 

[findsome1]

If sin x = 2/3 and cos y= 3/4 find the exact value of

sin (x+y)

 

[hscishard]

If sin x = 2/3 and cos y= 3/4 find the exact value of

sin (x+y)

(6+root7)/12

Expand sinxcosy + sinycosx

sinx = 2/3
Therefore cosx=1/3 (pythag for adjacent side)
cosy=3/4
Therefore siny = root7/4

Sub into the forumla and simplify.

 

[findsome1]

(6+root7)/12

Expand sinxcosy + sinycosx

sinx = 2/3
Therefore cosx=1/3 (pythag for adjacent side)
cosy=3/4
Therefore siny = root7/4

Sub into the forumla and simplify.

Really?

I got this:

sin x= 2/3 (given) cos x= root5/3 (use pythagoras theorem) sin y= root 7/4 (pythagoras theorem) cos y = 3/4 (given)

sub into formula sinxcosy+cosxsiny

= 2/3 x3/4 + root35/ 12

(multiply)

= 6/12+ root 35/12

= (6+root 35)/12<!-- google_ad_section_end --></SPAN> <!-- / message --><!-- sig -->

 

[fullonoob]

sinx = 2/3 cosx = root 5/3
cosy = 3/4 siny = root 7 /4
sin(x+y) = sinxcosy + sinycosx

 

[hungwell1337]

(6+root7)/12

Expand sinxcosy + sinycosx

sinx = 2/3
Therefore cosx=1/3 (pythag for adjacent side)
cosy=3/4
Therefore siny = root7/4

Sub into the forumla and simplify.

you cant do that

 

[findsome1]

Ok lets try another one:

Prove that sinycotysecy= 1

 

[hscishard]

Yea..whoops I went 1 2 and 3 are pythag triads.
cos x = root5

 

[hscishard]

siny x 1/cosy x cosy/siny secy = 1/cosy

= siny/cosy x cosy/siny
=1

 

[nikkifc]

If , then find the value of dy/dx when .

 

[hscishard]

-2

 

[fullonoob]

split cos^2x 2x into cos 2x . cos2x
product rule
did anyone do it another way?