Complex similarity question

[Stick]

I've attached the question below. I've looked at the solutions and they are awfully involved so I was wondering how likely it was that something of this nature would appear on the VCAA examination. I get the concept, but when it gets so involved I get lost. :S

[ashoni]

Considering the increasing number of students that are capable of doing Methods and/or Specialist doing Further Maths in recent years, I won't be surprised if they put in a few questions similar to yours there in this years VCAA exams. Not sure if you have noticed or not, but the VCAA exam papers from 2006 to 2011 becoming progressively hartderwith every year because of those added trick or mentally draining questions. Given that I'd expect the trend to continue, but I seriously hope not... Last thing I'd want is to panick during the exam, but those types of questions will separate the best out of the best in the end.

Stick, if you don't mind could you post up the solution to that question.. had a serious mind blank attempting to do the question D:

[Stick]

My teacher couldn't even do it and upon looking at the solutions the method they use is actually used in GMA and not really taught in Further. Even though the exams are getting progressively harder, I seriously doubt such a question will find it's way onto the exam - if it's there, I'll be happy to guess.

[Mr Newton]

so what is the answer?

[aishuwa1995]

I thought VCAA exams are supposed to be non-biased, so I don't think these kind of questions will be on them (they better not be)..disadvantaging people 
btw, what is the answer?

[b^3]

While procrastinating from Matlab I've given it a shot, and hope you can read my writing.
Now for the explanation.
For the first two trianges (), we can use similar triangles to get a ratio,
Now if we look at and and we note that and are parallel, then we have a 'Z' angle or 'alternate' angle formed, we know that , so we can tanke the tangent of the two and find an expression for .
Then substituting that back into the first ratio we found we can solve for to get , leaving the answer as .

Anyway, hope thats right


EDIT: Added explanation.

[aishuwa1995]

While procrastinating from Matlab I've given it a shot, and hope you can read my writing.
(Image removed from quote.)
... more to follow.

I tried using similar triangles and simultaneous equations and got 24..but the other variable was 0 so I thought I got it wrong 

[Stick]

Yep, the answer is 24.