*d*directly without reference to angle

*a*. Without coordinates for P you cannot calculate

*d*without additional information.

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Actually, on rereading the OP it's unclear to me if you have the coordinates of P. I read it as you have coordinates for only A, B, and C, but not for arbitrary point P. If you have the four coordinates then you can calculate *d* directly without reference to angle *a*. Without coordinates for P you cannot calculate *d* without additional information.

Yeah well that's just, like, your opinion, man.

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The coordinates ofJimC wrote: ↑Tue Mar 10, 2020 8:50 pm9B has nailed it. You actually don't need the angle you mentioned in your OP, Brian, simply all 4 coordinates, then use 3 different versions of the formula, one for the perpendicular distance to each of the 3 sides. You could, additionally, calculate the angle if you wanted.

Details on how to do that can be found here.

.

There are two other possibilities: one is paperwork, and the other is nostalgia."

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Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT

P and a give you a line. A and B give a line. Calculate X as the point where the two lines intersect. Then just calculate the distance between P and X.

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Brian Peacock wrote: ↑Tue Mar 10, 2020 9:05 pm... i.e. whenis different the distanceato the boundary of the triangle is different ...d

Details on how to do that can be found here.

.

There are two other possibilities: one is paperwork, and the other is nostalgia."

Frank Zappa

Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT

Brian Peacock wrote: ↑Tue Mar 10, 2020 9:18 pm, and the pointA, B, Cremain as in Figure B in the OP.P

Brian Peacock wrote: ↑Tue Mar 10, 2020 9:05 pm... i.e. whenis different the distanceato the boundary of the triangle is different ...d

The selected attachment doesn't exist

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Ah. What about now?

Details on how to do that can be found here.

.

There are two other possibilities: one is paperwork, and the other is nostalgia."

Frank Zappa

Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT

Yeah, see my earlier post

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Sorry. Not getting it.

Details on how to do that can be found here.

.

There are two other possibilities: one is paperwork, and the other is nostalgia."

Frank Zappa

Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT

- Brian Peacock
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The distance *D* between two points in 2d geometry (say, *P* {x₁, y₁} and *X* {x₂, y₂} as per 9B's post above) is:

D(P,X) = √((x₂ - x₁)² + (y₂ - y₁)²)

...but I still need to know what*d* is in my examples in order to find out where the dotted line intersects the triangle's boundary at 9B's *X* {x₂, y₂}.

What am I missing?

D(P,X) = √((x₂ - x₁)² + (y₂ - y₁)²)

...but I still need to know what

What am I missing?

Details on how to do that can be found here.

.

There are two other possibilities: one is paperwork, and the other is nostalgia."

Frank Zappa

Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT

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Ooh, I love these sorts of problems. I'll have a crack after work today.

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Once you have the coordinates of the 4 points, the issue is fixed, and the angle is not needed as an initial condition; in fact, it can be calculated too, if needed. I'm thinking of making a spreadsheet which, given the 4 initial coordinates, calculates the 3 perpendicular distances, the angles for each, the lengths of the triangle sides, the area of the triangle, and anything else I can think of!Brian Peacock wrote: ↑Tue Mar 10, 2020 9:05 pmThe coordinates ofJimC wrote: ↑Tue Mar 10, 2020 8:50 pm9B has nailed it. You actually don't need the angle you mentioned in your OP, Brian, simply all 4 coordinates, then use 3 different versions of the formula, one for the perpendicular distance to each of the 3 sides. You could, additionally, calculate the angle if you wanted.,A,B, andCare known, as is angleP(here, defined from the horizontal ccw). These are the starting conditions, makingaa function ofd- i.e. whenais different the distanceato the boundary of the triangle is different. See my previous reply to 9B here.d

Nurse, where the fuck's my cardigan?

And my gin!

And my gin!

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You sure know how to have fun!

Again, the problem is not finding the shortest (perpendicular) distance from from any point**P** within the triangle to any of its sides, but calculating the distance from **P** to the boundary of the triangle given any value for **a**.

I think the first Figure B was a visually misleading representation as it looked like a perpendicular intersection, so see the revised Figure B here: viewtopic.php?p=1855598#p1855598

Again, the problem is not finding the shortest (perpendicular) distance from from any point

I think the first Figure B was a visually misleading representation as it looked like a perpendicular intersection, so see the revised Figure B here: viewtopic.php?p=1855598#p1855598

Details on how to do that can be found here.

.

There are two other possibilities: one is paperwork, and the other is nostalgia."

Frank Zappa

Clinton Huxley » 21 Jun 2012 » 14:10:36 GMT

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Well I've been trying for an hour with triangles and I can't work it out. It requires voodoo maths!

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*"The Western world is fucking awesome because of mostly white men"* - DaveDodo007.

*"Socialized medicine is just exactly as morally defensible as gassing and cooking Jews"* - Seth. Yes, he really did say that..

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*"My penis is VERY small"* - Cunt.

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The height is the

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*"The Western world is fucking awesome because of mostly white men"* - DaveDodo007.

*"Socialized medicine is just exactly as morally defensible as gassing and cooking Jews"* - Seth. Yes, he really did say that..

*"Seth you are a boon to this community"* - Cunt.

*"My penis is VERY small"* - Cunt.

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