2013 HSC Question 13c (1 Viewer)

[porcupinetree]

https://imgur.com/lLxdv2l

I have been stumped with one simple part in part iii: both of the solutions that I have looked at state: "similarly, CE = 2rcosBsinA" but I can't seem to work out how to actually show this.
It's probably something heaps simple that I've overlooked but I'm getting nowhere so I thought I'd post.
Thanks

 

[InteGrand]

https://imgur.com/lLxdv2l

I have been stumped with one simple part in part iii: both of the solutions that I have looked at state: "similarly, CE = 2rcosBsinA" but I can't seem to work out how to actually show this.
It's probably something heaps simple that I've overlooked but I'm getting nowhere so I thought I'd post.
Thanks

Try using a similar method to what you did in part (ii), but maybe with a different triangle.

Another way to show it without the 'similarly' is to note that .

 

[Carrotsticks]

Imagine flipping the diagram around so beta is where alpha is now. So you're essentially interchanging beta with alpha and consequently CE with AE.

That's why the "similarly" result is exactly the same, just with alpha and beta swapped.

 

[porcupinetree]

I still can't get it. I've tried two ways but I can't find how to do it:

cosB = DB/2r, then to show our desired result we'd have to show that sinA = EC/DB, which I can't see how to do. or:
cosB = EC/DC, then we'd have to prove that sinA = DC/2r, which I can't see how to do.

Can somebody spell it out to me plain and clear? (otherwise I doubt I'll get it)