Recent content by duldin

Aye, I've worked through it all and trying to forget about it now ;). It didn't occur to me originally, but eventually I realized to just leave it as the difference of some inverse trig functions. Messy but it does appear to be the best I can do... substituting theta back in even more messy :P...

I'm sorry I can't get it at all near that form. Using the Asin(x) + Bcos(x), i can solve for theta with a correct answer.

I'm also stuck with this one. I hit the same brick wall by myself with the trig identities, been trying for about 2 hours now to find a way to do this question :( Treating it as a vector addition from a point before the river flow is considered is something I will try.

Ah, dead easy. Simultaneous equations in a vectors question... good stuff :thumbsup: Thanks heaps.

Thanks for the help. I see where you are coming from, but I don't see how you boil that down to x=1/3 ? :(

Thanks. Not sure if that goes where I want it too but I tried this one first: BA/QD = BE/ED Substituting each of those vectors in terms of a and b yields the desired x=1/3... without using y! It's magic. Since it doesn't use y I'm not sure if it constitutes a correct response to the question...

Triangles EDQ and EBA appear to be similar triangles. Does this imply, BE/ED = AE/EQ ?

Ah sorry, I have done that with pen I just forgot to type those lines out. ED = x(BD) ED = x(a+b) Substitute in and I end up with this: ED = ED x(a+b) = y(b-0.5a) + 0.5a I can't work out how to get rid of the y to show x=1/3 :(

Homework Statement Incl. diagrams: http://i.imgur.com/w6Exj.png Homework Equations Just vector addition (?)The Attempt at a Solution a.) ED = x(BD) ED = x(a+b) b) EQ = y(AQ) = y(b-0.5a) From the smaller triangle: ED = EQ + QD ED = [y(b-0.5a)] + (0.5a) I am not sure if this is what part b is...