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Hmm seems to be working for me. Here's a direct link, does it work? http://i.imgur.com/RGuzr.jpg

I've answered part a) of this question which was to find the exact length of AC. Now I need to answer part b). I'm not sure what to do. http://i.imgur.com/RGuzr.jpg I won't embed image, as it is rather large. Thank you.

Yes that as well as Economics. :'( I'm screwed.

Yeah, maths is something I really need to dedicate more time to, because it'll play a big part in my ATAR.

Thanks. That makes sense. I just find it difficult to approach these questions and I never know if I am actually getting anywhere or not. Thanks for all the help everyone.

Thanks Carrotsticks. One more thing the following lines I'm not 100% sure on: I'm not sure how you go from one line to the next.

I understand all that, except for the second last line on the RHS. You go from \frac{sin\beta}{sin^{2}\beta+cos^{2}\beta} to sin\beta. EDIT: sin^{2}\beta+cos^{2}\beta = 1, right? :)

Thank you :)

I'm not sure how to manipulate it, that's the problem. :)

Yes but how do you do so?

Hey guys, trig identities are probably my biggest downfall when it comes to trigonometry and I am up to the final question for trig identities in my textbook, but I don't know how to approach it. So I was hoping I could get some help and also some advice on how you would usually approach a trig...

Nevermind the last question, figured it out. Now I need to solve cosx=1

Hmm, so given the last question you'd approach it the same way then what? \text{Acute angle = 45} or as the first question actually is: cos\theta = 0.187

Yeah, I understood that after I wrote it, my bad. I got the answer. Now I'm answering questions for -180 \le \theta \le 180 :) What's the difference for that?

That makes sense, but can I just solve the positive value?

Solve. I'm not sure what to do when it's squared, that is all. If I square root both sides, I get \sqrt{1} on the right, but that isn't an exact value.

Hey guys, I was wondering if I could get some help for the following question. { tan }^{ 2 }3\theta=1 \quad \quad 0 \le \theta \le 360 Thanks :)

Why are they so hard? I never did anything like this in year 9 or 10, nothing even close.

I seriously haven't known how to do 1 question in this thread and I'm 2U year 11. Is this some sort of accelerated maths? :/

I get it now. :)