Can someone tell me how you can find the radius of the circle in the figure given below:
Given: $$RS = WS$$
I have tried all I know but I just dont understand how can you use only information related to angles to find the length of the radius.
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Could you solve the problem if the angle were 90 degrees instead? (This is meant to be a hint.) – kimchi lover Apr 03 '19 at 18:57
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You can not. You need the length of $RS$ for example. – Michael Rozenberg Apr 03 '19 at 18:57
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I guess you're expected to find it in terms of the lengths of those chords. Like the radius is $c \cdot |RS|$ for whatever the length of $RS$ is. @MichaelRozenberg – PrincessEev Apr 03 '19 at 18:58
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If so it's just $RS$. – Michael Rozenberg Apr 03 '19 at 18:59
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1So there is no way - no fricking way- that I can find the length of the radius without the length of the chords. RIGHT???? – ng.newbie Apr 03 '19 at 19:00
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1No. You can scale the figure by any factor (say 2), and the angle will be the same. The only thing you can say is the radius in terms of $|RS|$ – Andrei Apr 03 '19 at 19:02
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1Clearly the picture is scale invariant...you can get any length you want for the radius. It seems clear that you are meant to show that the radius equals the length $RS$. – lulu Apr 03 '19 at 19:02
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But, you can use the Inscribed angle theorem to find the radius in terms of $\overline{RS}$. – avs Apr 03 '19 at 19:03
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Maybe you could apply the radius-of-a-sector if you shift the angle to be on the center as in:https://math.stackexchange.com/questions/1274036/how-to-the-find-the-radius-of-a-sector – NoChance Apr 03 '19 at 19:08
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Set $S = (0,0)$, and W.L.O.G. assume that the vertical coordinate of W and R is the same. Thus $W = (\sqrt{3}L/2,L/2)$ and $R = (-\sqrt{3}L/2,L/2)$, with $L = RS=SW$.
Because $R,S$ and $W$ are points on a circle $(x-a)^2+(y-b)^2 = r^2$, the following equations should be satisfied
$$ (0-a)^2+(0-b)^2 = r^2$$ $$ (\sqrt{3}L/2-a)^2+(L/2-b)^2 = r^2$$ $$ (-\sqrt{3}L/2-a)^2+(L/2-b)^2 = r^2$$
From the two last equations, we have $a=0.$ Thus, we will have the system
$$ b^2 = r^2 $$
$$ 3L^2/4 +(L/2-b)^2 = r^2,$$
which gives $r = L.$
Alex Silva
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Connect points $R,S,W$ to the center of the circle $C$. Note that the triangles $RCS$ and $SCW$ are congruent and isosceles (as two out of three sides are equal to the radius). Thus, angles $RSC$ and $WSC$ are equal and must be of $60^0$. Hence, all angles in both triangles are the same $(60^0)$, and the radius = $RS = WS$.
dnqxt
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Yes I already know that but if RS and WS are not given can you find the radius? – ng.newbie Apr 09 '19 at 12:13
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Thinking out loud: Using the information from Find the radius from a sector, and knowing that a central angel is 2 times the angle given, I reached this solution.
I am not sure it is correct though, but I don't know what is wrong with it if it is incorrect.
NoChance
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I suggest you verify this by actually drawing it. Unfort. I don't have the tools! – NoChance Apr 09 '19 at 12:29