c3 trig

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But waiiiit, if you wanna differentiate lnx isnt the rule

f '(x)/f(x) ??!

Keith Pledger book page 140!!!
lol i know that rule....
ops sorry i made a mistake it should be cos2x/(1+sin2x)... actually i could guide u through the whole process but its hard to write these things here
 
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Manasa I did q8 in the june 13 paper

Okay so you asked for last two parts right?

(c) Just use the sine rule --> you'll have AB/sin90 = 7/sintheta. You found theta in part (a) so use that value. Simple Algebra and voila, you're done :)

(d) Okay this is very simple too and nothing new! Basically they're changing the speed right, so they're changing V

Now, follow the steps:

V = 21 / (24sintheta + 7costheta)
You expressed the denominator in the form Rcos(theta -alpha) so replace it

V = 21/ (Rcos(theta - alpha)) [sidenote: put in your answers for R and alpha, I didnt actually solve the ques lol]
Then sub in 1.68 instead of V

1.68 = 21/ (Rcos(theta - alpha))
Cross Multiply

Rcos(theta - alpha) = 21/1.68 = 12.5

Then just do your normal solving thing!

cos(theta - alpha) = 12.5/R and so on :)
 
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The GCE was hard... Q9 last 2 parts were hell haha, it's definitely harder than older papers, except June 2013, this one is easier but not by much...

i think the A will be 57
 
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