For the math-heads
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I am assuming that ptC is just a little to the right of straight down from B. Just to check my vision of what you want is correct. ptC.x = ptB.x - |BC| cos (Theta); ptC.y = ptB.y - |BC| sin (Theta); |BC| = |AB| = 90 Theta = 100o = PI/4 + a bit. So, C = (115.63, -78.63). How's that? Iain.
Remind me - what does Theta represent?
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C.x = B.x+sine(ABC-90)*BC C.y = B.y-cosine(ABC-90)*BC = C.x = 100+sine(10)*100 C.y = 10-cosine(10)*100
I think it's cool that Shog's coding johnson is longer than everyone elses -- JoeSox 10/8/03
Got it all except that C.y thing. Can't figure out how you got that :-D Smitha Every person, all the events of your life, are there because you have drawn them there. What you choose to do with them is up to you. -- Richard Bach
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Got it all except that C.y thing. Can't figure out how you got that :-D Smitha Every person, all the events of your life, are there because you have drawn them there. What you choose to do with them is up to you. -- Richard Bach
I'm assuming coordinate system where negative Y is down. Invert (10+ not 10-) for typical screen coordinates.
I think it's cool that Shog's coding johnson is longer than everyone elses -- JoeSox 10/8/03
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I'm assuming coordinate system where negative Y is down. Invert (10+ not 10-) for typical screen coordinates.
I think it's cool that Shog's coding johnson is longer than everyone elses -- JoeSox 10/8/03
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Thanks *wanders off to find some coffee*
I think it's cool that Shog's coding johnson is longer than everyone elses -- JoeSox 10/8/03
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Remind me - what does Theta represent?
Theta was the angle ABC. In this case, 100 degrees. Iain.
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I am assuming that ptC is just a little to the right of straight down from B. Just to check my vision of what you want is correct. ptC.x = ptB.x - |BC| cos (Theta); ptC.y = ptB.y - |BC| sin (Theta); |BC| = |AB| = 90 Theta = 100o = PI/4 + a bit. So, C = (115.63, -78.63). How's that? Iain.
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I'm playing around with graphing, and I'm trying to figure out the following: Draw a line from point A (0,10) to point B (100,10). Now draw a line from point B to point C, in such a way that C is below B, and angle ABC is 100 degrees. AB and BC have the same length. How do you calculate the coordinates of point C?
Draw a line from point A (0,10) to point B (100,10). Now draw a line from point B to point C, in such a way that C is below B, and angle ABC is 100 degrees. AB and BC have the same length. OK, Now my solution is as follows: (NB: I haven't done any trig. maths for quite a few years, so my memory of most trig equations are a little sketchy. Forgive my pathetic excuse of a text graph :~
^
|
|A(0,10) B(100,10) x D *'''''''''''''''''''''''''''''''*''''''''''''+ -
| 100 \ 90'
| \ '
| \ '
| \ '
| \ ' y
| \ '
| \ '
| \ '
| \ '
| \ '
+-------------------------------+--------> \ '
|(0,0) 100 \'
| * CAB = BC = 100
A<B>C = 100 deg.
~ C<B>D = 180 - 100 = 80 deg.
~ B<C>D = 180 - (90+80) = 10 deg.since BC = 100:
y/BC = COS(B<C>D)
y/100 = COS(10);
:= y = 100 * COS(10) = 98.48X/BC = SIN(10)
X = 100 * Sin(10) = 17.365~ C = ( 100 + X, 10 - Y )
= (117.365, -88.48).QED.
(i'm sure some of you will prove me wrong) **I Dream of Absolute Zero
**
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AB and BC are the same length, so we can assume they are radii of a circle with the origin at (100,10). Using polar coordinates, x=radius*(Cosine(Theta)) and y=radius*(Sine(Theta)): x=(100*cos(280)+100) = 117.365 y=(100*sin(280)+10) = -88.481 Roughly. If you apply the distance formula: distance=sqrt((x2-x1)^2 + (y2-y1)^2)) distance=sqrt(17^2 + (-98)^2)=sqrt(289+9604)=99.464 Pretty close to 100.
"It is better to remain silent and be thought a fool than to open one's mouth and remove all doubt." - Abraham Lincoln
Well, you got it right...but my question is why didn't you just use cos(80) and sin(80)
We're given this:
100
A--------------------B
θ=100° \
\
100\
\
\
\
CNow, if we draw an imaginary triange the other way, we can say that φ=80° because of supplementary angles.
Then we can say that Cx = 100 + 100 cos(80°) and Cy = 100 sin(80°):100A--------------------B======|
θ=100° \φ=80°|
\ |
100\ |
\ |
\ |
\|
CWhat's with cos and sin of 280? Where does that come from? [EDIT] Hey, if you highlight everything in the pre tags, it looks like a guy with a really long nose and a little cap on. :-D [/EDIT]
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
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Draw a line from point A (0,10) to point B (100,10). Now draw a line from point B to point C, in such a way that C is below B, and angle ABC is 100 degrees. AB and BC have the same length. OK, Now my solution is as follows: (NB: I haven't done any trig. maths for quite a few years, so my memory of most trig equations are a little sketchy. Forgive my pathetic excuse of a text graph :~
^
|
|A(0,10) B(100,10) x D *'''''''''''''''''''''''''''''''*''''''''''''+ -
| 100 \ 90'
| \ '
| \ '
| \ '
| \ ' y
| \ '
| \ '
| \ '
| \ '
| \ '
+-------------------------------+--------> \ '
|(0,0) 100 \'
| * CAB = BC = 100
A<B>C = 100 deg.
~ C<B>D = 180 - 100 = 80 deg.
~ B<C>D = 180 - (90+80) = 10 deg.since BC = 100:
y/BC = COS(B<C>D)
y/100 = COS(10);
:= y = 100 * COS(10) = 98.48X/BC = SIN(10)
X = 100 * Sin(10) = 17.365~ C = ( 100 + X, 10 - Y )
= (117.365, -88.48).QED.
(i'm sure some of you will prove me wrong) **I Dream of Absolute Zero
**
You got it right. You just went through a very roundabout way of doing it. :)
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
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Well, you got it right...but my question is why didn't you just use cos(80) and sin(80)
We're given this:
100
A--------------------B
θ=100° \
\
100\
\
\
\
CNow, if we draw an imaginary triange the other way, we can say that φ=80° because of supplementary angles.
Then we can say that Cx = 100 + 100 cos(80°) and Cy = 100 sin(80°):100A--------------------B======|
θ=100° \φ=80°|
\ |
100\ |
\ |
\ |
\|
CWhat's with cos and sin of 280? Where does that come from? [EDIT] Hey, if you highlight everything in the pre tags, it looks like a guy with a really long nose and a little cap on. :-D [/EDIT]
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
David Stone wrote: What's with cos and sin of 280? Where does that come from? Think of it as a circle with the origin at 100,10 with a radius of 100. AB is 180 from BX (BX being the line from (100,10)-(200,10). When we add another 100 to get the ABC angle, that equals 280 on our circle. I haven't had trig for a while, but the polar coordinate system is pretty simple for me. For problems like this, I always try to think in terms of a circle.
"It is better to remain silent and be thought a fool than to open one's mouth and remove all doubt." - Abraham Lincoln
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David Stone wrote: What's with cos and sin of 280? Where does that come from? Think of it as a circle with the origin at 100,10 with a radius of 100. AB is 180 from BX (BX being the line from (100,10)-(200,10). When we add another 100 to get the ABC angle, that equals 280 on our circle. I haven't had trig for a while, but the polar coordinate system is pretty simple for me. For problems like this, I always try to think in terms of a circle.
"It is better to remain silent and be thought a fool than to open one's mouth and remove all doubt." - Abraham Lincoln
Jason Henderson wrote: Think of it as a circle with the origin at 100,10 with a radius of 100. AB is 180 from BX (BX being the line from (100,10)-(200,10). When we add another 100 to get the ABC angle, that equals 280 on our circle. Ah, that makes sense...okay. I see your point. Jason Henderson wrote: I haven't had trig for a while, but the polar coordinate system is pretty simple for me. I use trig a lot in Physics and 3rd semester calculus. So that's why I immediately go for trig. I learned to hate polar coordinates last semester. Ever try integrating polar equations? X|
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
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You got it right. You just went through a very roundabout way of doing it. :)
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
I haven't done these sort of maths in YEARS! I've still got it.... :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: yeah! **I Dream of Absolute Zero
**
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I haven't done these sort of maths in YEARS! I've still got it.... :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: :jig: yeah! **I Dream of Absolute Zero
**
Isn't math fun? :)
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
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Remind me - what does Theta represent?
-
Jason Henderson wrote: Think of it as a circle with the origin at 100,10 with a radius of 100. AB is 180 from BX (BX being the line from (100,10)-(200,10). When we add another 100 to get the ABC angle, that equals 280 on our circle. Ah, that makes sense...okay. I see your point. Jason Henderson wrote: I haven't had trig for a while, but the polar coordinate system is pretty simple for me. I use trig a lot in Physics and 3rd semester calculus. So that's why I immediately go for trig. I learned to hate polar coordinates last semester. Ever try integrating polar equations? X|
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
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Jason Henderson wrote: Trig is all about circles. Trig is all about triangles. At least that's what I was always told...and that's what I've seen. Jason Henderson wrote: Not that I recall. Lucky...it's not fun at all.
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
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Isn't math fun? :)
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
Some of my students are not enjoying it much.:( Last week's exam results were so bad that I'm retesting in a week. Seriously, you'd think that people who do perfectly on homework would pay attention on an exam, rather than racing through it and failing to read the questions completely.:doh: "Your village called -
They're missing their idiot." -
Jason Henderson wrote: Trig is all about circles. Trig is all about triangles. At least that's what I was always told...and that's what I've seen. Jason Henderson wrote: Not that I recall. Lucky...it's not fun at all.
When I can talk about 64 bit processors and attract girls with my computer not my car, I'll come out of the closet. Until that time...I'm like "What's the ENTER key?" -Hockey on being a geek
The basics of trig were taught to me using a circle of radius 1. Almost anything can be broken into right triangles and the hypotenouses of those triangles can be seen as the radius of a circle.
"It is better to remain silent and be thought a fool than to open one's mouth and remove all doubt." - Abraham Lincoln
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I'm playing around with graphing, and I'm trying to figure out the following: Draw a line from point A (0,10) to point B (100,10). Now draw a line from point B to point C, in such a way that C is below B, and angle ABC is 100 degrees. AB and BC have the same length. How do you calculate the coordinates of point C?
The answer is a set of coordinates, isn't it? anything on the ray starting at B through some point C which satisfies your criteria, is actually a valid response. I'm I way off my thinking here?
"The beat goes on.. da-da-dum dadum dum"
BW