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# OCR June 2001 Pure 3 Help watch

1. hi

question 2)ii) on OCR June 2001 Pure 3 (MEI) i cant seem to see how to get to the answer, please can someone give me an explanation of this im bit baffeled, to much drink last night me thinks n the exam is on thursday!

thanks
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2. Use the sine rule:
1/sin(pi/6) = PQ/sin[pi - (pi/6+a)]
PQ = sin[pi - (pi/6+a)]/sin(pi/6)
= [sin(pi)cos(pi/6+a) - sin(pi/6+a)cos(pi)]/sin(pi/6)
= sin(pi/6+a)/sin(pi/6)
= sin(pi/6+a)/0.5
= 2sin(pi/6+a)

And:
RS = sina

So:
A = 0.5 * PQ * RS = sin(a+pi/6)sina, as required.

To find the value of a for which A is maximum, solve dA/da=0.
3. Beat me to it, but I think me solution is slightly clearer.
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