- #1

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I'd appreciate it if anyone could help me out on this question

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- Thread starter david18
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- #1

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I'd appreciate it if anyone could help me out on this question

- #2

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If you want a quick answer upload your image on imageshack...

- #3

HallsofIvy

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Remember that the first vertical line has length [itex]\sqrt{3}L/2[/itex]. We now see that can be divided into a short distance of [itex]\sqrt{3}L/3[/itex] and a longer distance of [itex]\sqrt{3}L/2- \sqrt{3}L/2= \sqrt{3}L/6[/itex]. The length of the longer part is the radius of the larger circle, and the length of the shorter part is the radius of the smaller circle.

[tex]\frac{\frac{\sqrt{3}}{2}L}{\frac{\sqrt{3}}{6}L}= 3[/tex]

not 2.

- #4

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I was working through your solution and knowing the context of the question, thought it seemed a little too complicated.

What I eventually found is that if you draw a line from the centre to a corner of the triangle, and another line from the centre to the tangent, you form a right angled triangle with angles of 30 and 60 degrees. The line from the centre to the tangent is r, so knowing that sin 30 = 1/2, you say that the hypotenuse must be 2r.

Sorry for the poor explanation, I can draw up a solution if you have any trouble going through the lengthy explanation.

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