For the math-heads
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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?
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
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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?
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?
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
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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
-
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
-
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?
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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