Is angle in between two angles

Question

I have 3 angles a b c

a=315 b=20 c=45

ok so would like to know giving all three if b is in between a and c

i have the long way of doing this adding a

Answer 1

I personally had the same problem recently and found counterexamples for all the answers posted yet, so I will share my own approach.
Let a be the start angle and b the end angle and we are checking whether c is between them clockwise, that means when you go from a to b you must pass c. The approach of checking whether c is in the range from a to b gives you false positives when b is greater than a numerically. For example: a=80°, b=320° and c=150°: a <= c <= b which would mean that c is between a and b. But it isn't.
The approach that's working is to subtract 360 from b if it is greater than a and also subtract 360 from c if c is greater than a. Then check whether a <= c <= b. In Java:

public static boolean isBetween(double c, double a, double b) {
    if (b > a) b -= 360;
    if (c > a) c -= 360;
    return a <= c && c <= b;
}

This assumes that a, b and c are in range 0 to 360.
Some example:

isBetween(150, 80, 320) => false
isBetween(30, 80, 320) => true
isBetween(340, 80, 320) => true
isBetween(140, 0, 160) => true
isBetween(180, 0, 160) => false

Answer 2

Assuming a > c, you would actually use:

( b < a ) && ( b > c )

This is the same as checking if a value is between a lower and upper bound. Them being angles makes no difference, unless you want to take into account the fact that as you go round a circle, an angle of 405 is the same as an angle of 45. In which case you can just use a % 360 to get the angle betweeen 0 and 360.

Answer 3

1^st off, every angle is between 2 other angles, what you're really asking is:

For given angles: a, b, and g, is g outside the reflex angle between a and b?

You can just go ahead and define a as the leftmost angle and b as the rightmost angle or you can solve for that, for example if either of these statements are true a is your leftmost angle:

1. a ≤ b ∧ b - a ≤ π
2. a > b ∧ a - b ≥ π

For simplicity let's say that your leftmost angle is l and your rightmost angle is r and you're trying to find if g is between them.

The problem here is the seem. There are essentially 3 positive cases that we're looking for:

1. l ≤ g ≤ r
2. l ≤ g ∧ r < l
3. g ≤ r ∧ r < l

If you're just defining a to be leftmost and b to be rightmost you're done here and your condition will look like:

a <= g && g <= b ||
a <= g && b < a ||
g <= b && b < a

If however you calculated the l and r you'll notice there is an optimization opportunity here in doing both processes at once. Your function will look like:

if(a <= b) {
    if(b - a <= PI) {
        return a <= g && g <= b;
    } else {
        return b <= g || g <= a;
    }
} else {
    if(a - b <= PI) {
        return b <= g && g <= a;
    } else {
        return a <= g || g <= b;
    }
}

Or if you need it you could expand into this nightmare condition:

a <= b ?
(b - a <= PI && a <= g && g <= b) || (b - a > PI && (b <= g || g <= a)) :
(a - b <= PI && b <= g && g <= a) || (a - b > PI && (a <= g || g <= b))

Note that all this math presumes that your input is in radians and in the range [0 : 2π].

Live Example

Answer 4

I had a similar problem. I got it. All the calculations are in degrees. I needed to calculate id a gps location is inside a rectangle.

Or, I needed to see if an angle x is between angle check+r and angle check-r.

check-r<x<check+r.

If you need a<x<b, find the angle check in the middle of a and b and then the distance (r) of check from a or b.

The method normalize, changes the angles from -infinity...infinity to -180...180. The method check, takes the arguments x: the angle that we need to see if it is between the angles check-r and check+r. check: the angle to check with. r: the radius around angle check.

private static double normalize(double x) {
        x = x % 360;
        if (x>=180) {
            return x-360;
        }
        if (x<-180) {
            return x+360;
        }
        return x;
}
public static boolean check(double x, double check, double r) {
        x = x - check;
        x = normalize(x);
        return x<r && x>-r;
}