- #1

SolCon

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I've got 2 simple questions relating to trigonometry in circular measure. I've got answers to both but they are not satisfactory. So here goes.

1) This question requires that we show that A (which is the area of a circle) is a maximum and not a minimum, when r=3/2.

A=3r-r^2

I've got the answer coming as 9/4 after substituting the value of 'r' in the equation. However, I do not know how this proves the question's requirement.

2) This is slightly bigger. I have the following figure:

http://usera.ImageCave.com/biosyn/q1.PNG

Q: . We are required to find the perimeter of the shaded region.

The conditions:

. OPQ is an equilateral triangle.

. OS=PS=QS

. arc PXQ has centre O with radius 12 cm

. Answer must be in terms of pi and under-root 3

First of all, since OS,PS and QS are equal, I believe that the angles PSQ, PSO and QSO are equal and =360. This means each is 120 degrees or 2pi/3. I've also established the radius PS as 6/under-root 3/4. I've used s=r(theta) to find PXQ as 12pi/3[under-root(3/4)].

However, adding them all together is not giving me the correct answer.

What am I missing here?

Thanks for any help.