Maths stuff

1. Senior Member
Join Date
Dec 1969
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503

## Maths stuff

Hi there guys, good morning to ya.<BR><BR>I have a slight maths problem which I&#039;m stuck on. <BR><BR>I have to values (OS grid references) obviously X and Y. I need to work out how far the distance is between then (as the crow flies). So basically I need to know how to work out the hypotenuse from the adjacent & opposite sides, while only knowing what that it&#039;s right angled triangle. I&#039;m thinking it&#039;s a two step process right?<BR><BR>Cheers Jonny Tooze

2. Senior Member
Join Date
Dec 1969
Posts
404

## RE: Maths stuff

distance betw 2 pts in x plane = x1-x2<BR>distance betw 2 pts in y plane = y1-y2<BR>Pythagorean theorem A ^ 2 + B ^ 2 = c ^ 2<BR><BR>c = ((x1 - x2) ^ 2 + (y1 - y2) ^ 2) ^ .5

3. Senior Member
Join Date
Dec 1969
Posts
503

## Nice mate, thank you. eop

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Dec 1969
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16,931

## Or...

Put a different way:<BR><BR>c² = (ðx² + ðx²)^½<BR><BR>Craig.

5. Senior Member
Join Date
Dec 1969
Posts
16,931

## Or...

Put a different way:<BR><BR>c = (ðx² + ðx²)^½<BR><BR>Craig.

6. Senior Member
Join Date
Dec 1969
Posts
503

## you can do it by wrking out the angle first too..e

.

7. Senior Member
Join Date
Dec 1969
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16,931

## Of course.... Like...

...as tan(x) = opposite/adjacent.<BR><BR>Therefore:<BR>tan(x) = dx/dy<BR>so x = tan^-1(dx/dy).<BR><BR>however, sin(x) = opposite/hypotenuse, so:<BR>sin(x) = opposite/hypotenuse = sin(tan^-1(dx/dy)) = dx/hypotenuse<BR><BR>Therefore:<BR>hypotenuse = dx/(sin(tan^-1(dx/dy)))<BR><BR>So yeah, you can work out the angles if you want. In fact, if you use the above equation then the angles are not actually used, just the resultant equations are combined. However, you have a trigonometric and an inverse-trigonometric function in there... So Pythatorus&#039; theorum is probably easier in the long-run :)<BR><BR>Craig.

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