Find the type of progression formed by the angle between the tangents to a circle drawn from points which are collinear and at equal distances from each other as depicted.

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TopNewestAll I got is that cosec of half of angles are in AP.

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That's interesting. Will you please let me know about the working part?

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I am not sure that you wanted what I have written above but it was pretty easy to get

Draw a line passing through centre and all the points

Let the radius of circle be \(R\), distance between centre and first point be \(d\) and distance between consecutive points be \(x\)

We will take triangle formed by centre, point of contact and the first point(right angled triangle)

Let the original first angle be \(\theta\) then the angle of triangle be \(\theta/2\)

And we will get \(sin(\theta/2) = \dfrac{R}{d}\).

Similarly for next angle \(sin(\theta_1/2) = \dfrac{R}{d+x}\)

And for next \(sin(\theta_2/2) = \dfrac{R}{d + 2x}\)

Now rest is easy just flip them to make cosec and eliminate R,d,x to get desired result :)

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You should mention what the curve is. It doesn't quite seem like a circle to me, but that might be because of the straight lines.

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Done! :)

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