# 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
3 years, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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

- 3 years, 2 months ago

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

- 3 years, 2 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 :)

- 3 years, 2 months ago

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

- 3 years, 2 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 - 3 years, 2 months ago

Done! :)

- 3 years, 2 months ago