Answers for 2002 paper (1 Viewer)

[Lord Ac]

Does anyone one know wwhere i can find the catholic trial ANSWERS for 2002 paper? Cause i did it, but wanna see my results ...

Ac

 

[McLake]

We could mark it for you (ie: post your answers) ...

 

[Lord Ac]

Ok, let me post up a few that had me puzzled:

3) ABC is a triangle and N is a point on AC
<ABn=<CBN=<BCN
BC=2a, CA=b, AB=c
BN=CN=d

i) Given that triangle ABN ||| triangle ACB, show that c^2 = b^2 - 2ac

ii) hence show that (a+c)^2 = a^2 + b^2



6ai) If W= inverse tan A + inverse tan B, show that tan W = A + B / 1 - AB

ii) hence solve the equation inverse tan 3x + inverse tan 2x = pi/4


And if anyone has the paper, question 7 (induction). Its a lot to type out.

Ac

 

[Constip8edSkunk]

as triangle ABN and ACB are similar
(b-d)/c=c/b
c²=b²-bd
but b/c=2a/d => bd=2ac
therefore c²=b²-2ac

c²+2ac=b²
completing the square: (c+a)²-a²=b²
(a+c)²=a²+b²

let tan^-1A = x => tanx = A
let tan^-1B = y => tany = B
tanW = tan(x+y)
=(tanx+tany)/(1-tanxtany)
=(A+B)/(1-AB)

tan(pi/4) = (2x+3x)/(1-2x.3x)
1=5x/(1-6x²)
1-6x²=5x
6x²+5x-1=0
(6x-1)(x+1)=0
x=-1 or 1/6

[edit] 4got minus sign

 

[McLake]

Here's the answer to question 7

EDIT: Here is the answer to the 3U paper (oops!)

 

[McLake]

Originally posted by Constip8edSkunk
tan(pi/4) = (2x+3x)/(1-2x.3x)
1=5x/(1-6x²)
1-6x²=5x
6x²+5x-1=0
(6x-1)(x+1)=0
x=-1 or 1/6

[edit] 4got minus sign

but if x = -1 then: tan^-13x < 0 and tan^-12x < 0.
so x != -1

so x = 1/6