# Calculating apparent separation

1. Jan 18, 2014

### bobo1455

I am trying to calculate the apparent separation between the 2 stars Mizar and Alcor. In the question I am given the total distance from me to the 2 stars which is 83 light years and the angle of separation which is 12 arcminutes.

The question is multiple choice: (A) 0.1 ly, (B) 0.3 ly, (C) 1 ly, (D) 3 ly, (e) 10 ly

What I have tried is using the formula from Wikipedia on Angular Distance to get a value that matches one from the choices A to E

\theta ≈ \frac {a}{D}

The problem is, I am having a hard time understanding what to exactly do with 83 light years and 12 arcminutes. I have read online that the 2 stars Mizar and Alcor have distances of 81 light years and 78 light years, so the distance between them is 3 light years, so I'm assuming the answer is (D), but I don't know at all how to get there.

I am also given these formulas:

distance = actual size / angular size

angular size is proportional to actual size / distance

Any help is appreciated.

2. Jan 18, 2014

### BvU

The angle of separation is the angle between the lines from you to the stars. Theta in the formula.
Don't forget to convert to radians.

3. Jan 18, 2014

### bobo1455

Okay, I convert 12 arcminutes to degrees (0.2) to radians (0.0034906585) and plugged in for theta:

then I have:

0.0034906585 = x / 83 ly (not sure if I am supposed to convert ly to AU or some other unit of measurement)

then I get x = 0.2897246555, which is approx 0.3 ly, and (B) is 0.3 ly, but I am still not 100% sure it is correct or not.

4. Jan 18, 2014

### HallsofIvy

The original distances were given in light years so the answer should be given in light years. Yes, 12 minutes is 12/60= 1/5 degrees and so (1/5)(3.14159/180)= 0.0034906585 so the distance between them, along a circular arc, is (0.0034906585)(83)= 0.2897 light years. I don't think I would put it to as many significant figures as you do. Since "12 arcminutes" and "83 light years" both have two significant figures, the most reasonable answer is 0.29 light years.

You could also do this as a "trig" problem. Dropping a vertical from the vertex (the earth) to the line between the two stars, we have two congruent right triangles with angle 6 arcminutes and hypotenuse 83 ly. The opposite side of that right triangle (half the distance between the two stars), x, satisfies $$sin(6')= x/83$$ so that x= 83sin(6')=0.1449 so that the entire distance, 2x, is 0.2898 light years, still 0.29 ly as before.