## Basic Question about angle of inclination of a circle

I've seen some measurements of supernovae, and they often mention a ring that's expanding. I can follow that. What is often stated is that they assume the ring is a circle, but of course it's seen at an angle from Earth.

One such example was a ring that looks like it's a certain size in the sky (x arcseconds, by y arcseconds). Assuming it was a circular ring, the auther just jumps to a conclusion, something like: "using geometry, we can see that the ring is inclined at an angle of e.g. i=29 degrees to face-on"

This should be simple geometry, and yet I'm stumped as to how to get the angle of inclination of the ring, assuming it's circular and looking at it's size in the sky in arcseconds.

Does anyone have any ideas?
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 Quote by kungfuscious … One such example was a ring that looks like it's a certain size in the sky (x arcseconds, by y arcseconds). Assuming it was a circular ring, the auther just jumps to a conclusion, something like: "using geometry, we can see that the ring is inclined at an angle of e.g. i=29 degrees to face-on" This should be simple geometry, and yet I'm stumped as to how to get the angle of inclination of the ring, assuming it's circular and looking at it's size in the sky in arcseconds.
Hi kungfuscious ! Welcome to PF!

Yes, it's simple geometry …

a circle seen at a angle is an ellipse …

take a vertical circle of angular diameter x, and turn it round the vertical diameter, through an angle θ …

the vertical diameter will have the same angular size, x, but the horizontal diameter will be reduced to angular size y = x cosθ …

so just measure the longest and shortest diameters, x and y, and the angle is cos-1 y/x

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