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There is a line inside a circle that starts from a point on the diameter and reaches the circumference of the circle. It thus subtends an angle between itself and the diameter. It is as the following image :

enter link description here

With the radius, r, the offset from center, a, and the angle subtended by the line with the diameter, b, known, how can we find :

  1. The length of the line, x
  2. The length of the sector, y
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  • $\begingroup$ what's been tried ? also I think stackechange has it's own image site. $\endgroup$ – user451844 Jul 14 '17 at 19:05
  • $\begingroup$ $r^2=a^2+x^2-2ax \cos b$ $\endgroup$ – N74 Jul 14 '17 at 19:12
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For $x$, join the center of the circle to the intersection of the "inside" line (marked with length $x$) and the circle. A triangle thus formed has side lengths $a,r$ and $x$ and an internal angle $b$. Now apply cosine rule to solve for $x$. $$\cos b = \frac{a^2+x^2-r^2}{2ax}$$

After this construction and finding $x$ it will be easy to get $y$. Can you take it from here?

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  • $\begingroup$ Thanks for the equation. I solved it to an extent, but am stuck with a final form of the equation : x^2 - 6x = 21.9. My math is pretty much rustic at the moment, and I'm trying to figure out how this can be solved further. $\endgroup$ – Ashley Jul 14 '17 at 19:21
  • $\begingroup$ So you are solving a quadratic equation of the form $ax^2+bx+c=0$. The solutions to that are given by $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. In your case $a=1,b=-6$ and $c=-21.9$. $\endgroup$ – Anurag A Jul 14 '17 at 19:22
  • $\begingroup$ Thanks for pointing out the way. Moreover, can you say what could the equation be to similarly find out the length of the arc between the diameter and the X line. $\endgroup$ – Ashley Jul 14 '17 at 20:02
  • $\begingroup$ Anurag, going by your equation, I was getting the value for x as 3 when the angle B, was 90, for a 5cm radius circle and a as 3cm, when it is actually 4, according to the right angle theorem, and also my own measurements. I have double checked my calculations, as per your formula, but still getting the value of 3. Could you clarify this anomaly. $\endgroup$ – Ashley Jul 15 '17 at 11:42
  • $\begingroup$ @Ashley When angle $b=90^{\circ}$, then $\cos b=0$. So the equation reduces to $3^2+x^2-5^2=0$, which gives $x=4$. I think you might have plugged in the incorrect value somewhere. $\endgroup$ – Anurag A Jul 15 '17 at 13:04

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