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# Angles in progression

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.

Note by Rohit Ner
2 years, 6 months ago

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

- 2 years, 6 months ago

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

- 2 years, 6 months ago

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

1. Draw a line passing through centre and all the points

2. Let the radius of circle be $$R$$, distance between centre and first point be $$d$$ and distance between consecutive points be $$x$$

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

4. 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 :)

- 2 years, 6 months ago

Marvelous! Thank you very much for your guiding. I would try to build upon your work. :)

- 2 years, 6 months ago

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.

Staff - 2 years, 6 months ago

Done! :)

- 2 years, 6 months ago