Calculate Angle between Two Intersecting Line Segments
Clash Royale CLAN TAG#URR8PPP
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Need some help/direction, haven't had trig in several decades.
On a 2 dimensional grid, I have two line segments.
The first line Segment always starts at the origin (0,0), and extends to (1,0) along the X axis.
The second line Segment intersects the first at the origin, and can extend to potentially anywhere within (-1,-1) to (1,1).
I need to always calculate the angle to the right of the first segment...
If this is already addressed in another post, please comment me the link.
UPDATE
I will have a single input of (x,y) for the end of the 2nd segment...
so segment A would be (0,0) -> (1,0) and segment B would be (0,0) -> (x,y) where (x,y) can be anywhere inside (-1,-1) and (1,1) assuming that the scale is 0.1.
Let me know If I can provide any additional information that will help.
UPDATE
OK... assuming that the first segment is running along the Y axis... $A(0,0)$ and $B(0,1)$
And the second segment is running from $A(0,0)$ to $C(.4,.4)$ with a scale of .2....
$$theta= tan^-1dfrac.4.4= 45$$
If I change C to $C(.4,-.4)$ I get.
$$theta= tan^-1dfrac.4-.4= -45$$
Do I have to manually compensate for the quadrant because this seems to calculate based specifically on the axis... I would expect the 2nd one to come up as 135 degrees from the positive Y Axis...
What am I missing?
Just for posterity...
If I had $C(-0.4,-0.1)$ I would expect the result for the angle from the positive Y axis to this line segment to be roughly 255 degrees...
$$theta= tan^-1dfrac.4-.1= 75.9637$$
Plus 180 from starting at the positive Y axis....
trigonometry
 |Â
show 2 more comments
up vote
4
down vote
favorite
Need some help/direction, haven't had trig in several decades.
On a 2 dimensional grid, I have two line segments.
The first line Segment always starts at the origin (0,0), and extends to (1,0) along the X axis.
The second line Segment intersects the first at the origin, and can extend to potentially anywhere within (-1,-1) to (1,1).
I need to always calculate the angle to the right of the first segment...
If this is already addressed in another post, please comment me the link.
UPDATE
I will have a single input of (x,y) for the end of the 2nd segment...
so segment A would be (0,0) -> (1,0) and segment B would be (0,0) -> (x,y) where (x,y) can be anywhere inside (-1,-1) and (1,1) assuming that the scale is 0.1.
Let me know If I can provide any additional information that will help.
UPDATE
OK... assuming that the first segment is running along the Y axis... $A(0,0)$ and $B(0,1)$
And the second segment is running from $A(0,0)$ to $C(.4,.4)$ with a scale of .2....
$$theta= tan^-1dfrac.4.4= 45$$
If I change C to $C(.4,-.4)$ I get.
$$theta= tan^-1dfrac.4-.4= -45$$
Do I have to manually compensate for the quadrant because this seems to calculate based specifically on the axis... I would expect the 2nd one to come up as 135 degrees from the positive Y Axis...
What am I missing?
Just for posterity...
If I had $C(-0.4,-0.1)$ I would expect the result for the angle from the positive Y axis to this line segment to be roughly 255 degrees...
$$theta= tan^-1dfrac.4-.1= 75.9637$$
Plus 180 from starting at the positive Y axis....
trigonometry
Are you taking as input a pair of coordinates? Like, (1,5), for example?
â turkeyhundt
Jan 9 '15 at 15:22
2
Does angel = angle? So as it extends along X axis, you are using $(y, x)$ rather than $(x, y) $ coordinates?
â servabat
Jan 9 '15 at 15:23
Thanks for pointing it out... I've updated it.
â Patrick
Jan 9 '15 at 15:28
You should try and learn Latex if you are going to use this site. It is the formatting standard everyone goes by.
â Wintermute
Jan 9 '15 at 15:40
@mtiano yeah.. I tried using it for Pi.. and it doesn't want to play nice.
â Patrick
Jan 9 '15 at 16:00
 |Â
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Need some help/direction, haven't had trig in several decades.
On a 2 dimensional grid, I have two line segments.
The first line Segment always starts at the origin (0,0), and extends to (1,0) along the X axis.
The second line Segment intersects the first at the origin, and can extend to potentially anywhere within (-1,-1) to (1,1).
I need to always calculate the angle to the right of the first segment...
If this is already addressed in another post, please comment me the link.
UPDATE
I will have a single input of (x,y) for the end of the 2nd segment...
so segment A would be (0,0) -> (1,0) and segment B would be (0,0) -> (x,y) where (x,y) can be anywhere inside (-1,-1) and (1,1) assuming that the scale is 0.1.
Let me know If I can provide any additional information that will help.
UPDATE
OK... assuming that the first segment is running along the Y axis... $A(0,0)$ and $B(0,1)$
And the second segment is running from $A(0,0)$ to $C(.4,.4)$ with a scale of .2....
$$theta= tan^-1dfrac.4.4= 45$$
If I change C to $C(.4,-.4)$ I get.
$$theta= tan^-1dfrac.4-.4= -45$$
Do I have to manually compensate for the quadrant because this seems to calculate based specifically on the axis... I would expect the 2nd one to come up as 135 degrees from the positive Y Axis...
What am I missing?
Just for posterity...
If I had $C(-0.4,-0.1)$ I would expect the result for the angle from the positive Y axis to this line segment to be roughly 255 degrees...
$$theta= tan^-1dfrac.4-.1= 75.9637$$
Plus 180 from starting at the positive Y axis....
trigonometry
Need some help/direction, haven't had trig in several decades.
On a 2 dimensional grid, I have two line segments.
The first line Segment always starts at the origin (0,0), and extends to (1,0) along the X axis.
The second line Segment intersects the first at the origin, and can extend to potentially anywhere within (-1,-1) to (1,1).
I need to always calculate the angle to the right of the first segment...
If this is already addressed in another post, please comment me the link.
UPDATE
I will have a single input of (x,y) for the end of the 2nd segment...
so segment A would be (0,0) -> (1,0) and segment B would be (0,0) -> (x,y) where (x,y) can be anywhere inside (-1,-1) and (1,1) assuming that the scale is 0.1.
Let me know If I can provide any additional information that will help.
UPDATE
OK... assuming that the first segment is running along the Y axis... $A(0,0)$ and $B(0,1)$
And the second segment is running from $A(0,0)$ to $C(.4,.4)$ with a scale of .2....
$$theta= tan^-1dfrac.4.4= 45$$
If I change C to $C(.4,-.4)$ I get.
$$theta= tan^-1dfrac.4-.4= -45$$
Do I have to manually compensate for the quadrant because this seems to calculate based specifically on the axis... I would expect the 2nd one to come up as 135 degrees from the positive Y Axis...
What am I missing?
Just for posterity...
If I had $C(-0.4,-0.1)$ I would expect the result for the angle from the positive Y axis to this line segment to be roughly 255 degrees...
$$theta= tan^-1dfrac.4-.1= 75.9637$$
Plus 180 from starting at the positive Y axis....
trigonometry
edited Jan 9 '15 at 16:40
asked Jan 9 '15 at 15:19
Patrick
1213
1213
Are you taking as input a pair of coordinates? Like, (1,5), for example?
â turkeyhundt
Jan 9 '15 at 15:22
2
Does angel = angle? So as it extends along X axis, you are using $(y, x)$ rather than $(x, y) $ coordinates?
â servabat
Jan 9 '15 at 15:23
Thanks for pointing it out... I've updated it.
â Patrick
Jan 9 '15 at 15:28
You should try and learn Latex if you are going to use this site. It is the formatting standard everyone goes by.
â Wintermute
Jan 9 '15 at 15:40
@mtiano yeah.. I tried using it for Pi.. and it doesn't want to play nice.
â Patrick
Jan 9 '15 at 16:00
 |Â
show 2 more comments
Are you taking as input a pair of coordinates? Like, (1,5), for example?
â turkeyhundt
Jan 9 '15 at 15:22
2
Does angel = angle? So as it extends along X axis, you are using $(y, x)$ rather than $(x, y) $ coordinates?
â servabat
Jan 9 '15 at 15:23
Thanks for pointing it out... I've updated it.
â Patrick
Jan 9 '15 at 15:28
You should try and learn Latex if you are going to use this site. It is the formatting standard everyone goes by.
â Wintermute
Jan 9 '15 at 15:40
@mtiano yeah.. I tried using it for Pi.. and it doesn't want to play nice.
â Patrick
Jan 9 '15 at 16:00
Are you taking as input a pair of coordinates? Like, (1,5), for example?
â turkeyhundt
Jan 9 '15 at 15:22
Are you taking as input a pair of coordinates? Like, (1,5), for example?
â turkeyhundt
Jan 9 '15 at 15:22
2
2
Does angel = angle? So as it extends along X axis, you are using $(y, x)$ rather than $(x, y) $ coordinates?
â servabat
Jan 9 '15 at 15:23
Does angel = angle? So as it extends along X axis, you are using $(y, x)$ rather than $(x, y) $ coordinates?
â servabat
Jan 9 '15 at 15:23
Thanks for pointing it out... I've updated it.
â Patrick
Jan 9 '15 at 15:28
Thanks for pointing it out... I've updated it.
â Patrick
Jan 9 '15 at 15:28
You should try and learn Latex if you are going to use this site. It is the formatting standard everyone goes by.
â Wintermute
Jan 9 '15 at 15:40
You should try and learn Latex if you are going to use this site. It is the formatting standard everyone goes by.
â Wintermute
Jan 9 '15 at 15:40
@mtiano yeah.. I tried using it for Pi.. and it doesn't want to play nice.
â Patrick
Jan 9 '15 at 16:00
@mtiano yeah.. I tried using it for Pi.. and it doesn't want to play nice.
â Patrick
Jan 9 '15 at 16:00
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
In fact, you want to know the angle between the $x$ axis and any line going through the origin $(0,0)$ and a point $A (x_a,y_a)$, with $-1 leq x_a,y_a leq 1?$
This is fairly easy. Since there are two angles between two lines, you said you wanted the "right angle", that is, the one to the right hand side of the $x$ axis.
If $x_a times y_a<0$, $theta= tan^-1dfracy_ax_a$
If $x_a=0$, $theta=dfracpi2$
If $x_atimes y_a>0$, $theta= tan^-1dfracy_ax_a-pi$
Yes... you appear to know what I want more so than I do... Let me verify everything and I'll select answer.
â Patrick
Jan 9 '15 at 15:33
You can also use $theta = cos^-1left(fracxsqrtx^2+y^2 right)$ for example, as it will also work when $x = 0$
â servabat
Jan 9 '15 at 15:42
I ran (-.4, -.4) through the bottom formula and got -64.9 Degrees... that doesn't seem right to me... if the first part is running along the X axis.... I would expect a positive answer between 90 and 180 degrees... what am I missing?
â Patrick
Jan 9 '15 at 15:52
@Patrick First, angle are negative when counting them down... (that is, clockwise). Second, did you substract $pi$ or $180°$?
â Martigan
Jan 9 '15 at 15:54
OK... I can handle the negative... I subtracted by Pi
â Patrick
Jan 9 '15 at 15:56
 |Â
show 4 more comments
up vote
0
down vote
Not sure EXACTLY what you are asking, but I will answer this to the best of my ability. If you could include a visual that would greatly help me.
When we have two intersecting line segments like this
finding any single value (a, c, b, d) will reveal all other values. For example, if we have the value of a, then c = a, and we have (in degrees) b = d = 180-a.
Therefore, I can equip with the tools to find the angle between two vectors, as you have given. For example, we can treat the first vector as you have said as $[1, 0]$ and the second vector as $[1, 1]$. We take the dot product between them, which just means that we multiply the corresponding values and sum them up, or $<a, b> = sumlimits_i = 1^na_ib_i$ where $n$ is the number of elements in the vector. We use the geometric fact that $<a, b> = |a||b|cos(theta_a, b)$ where $|a|$ means its norm or magnitude (I only deal with standard inner product and norm here). This gives us that the dot product, which is $1$, is equal to $sqrt2cos(theta)$, which means $cos(theta) = frac1sqrt2$
This in turn gives us that the angle between these two is 45 degrees, and we can figure out the adjacent angle as 135, and the vertical angles all share the same degrees.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
In fact, you want to know the angle between the $x$ axis and any line going through the origin $(0,0)$ and a point $A (x_a,y_a)$, with $-1 leq x_a,y_a leq 1?$
This is fairly easy. Since there are two angles between two lines, you said you wanted the "right angle", that is, the one to the right hand side of the $x$ axis.
If $x_a times y_a<0$, $theta= tan^-1dfracy_ax_a$
If $x_a=0$, $theta=dfracpi2$
If $x_atimes y_a>0$, $theta= tan^-1dfracy_ax_a-pi$
Yes... you appear to know what I want more so than I do... Let me verify everything and I'll select answer.
â Patrick
Jan 9 '15 at 15:33
You can also use $theta = cos^-1left(fracxsqrtx^2+y^2 right)$ for example, as it will also work when $x = 0$
â servabat
Jan 9 '15 at 15:42
I ran (-.4, -.4) through the bottom formula and got -64.9 Degrees... that doesn't seem right to me... if the first part is running along the X axis.... I would expect a positive answer between 90 and 180 degrees... what am I missing?
â Patrick
Jan 9 '15 at 15:52
@Patrick First, angle are negative when counting them down... (that is, clockwise). Second, did you substract $pi$ or $180°$?
â Martigan
Jan 9 '15 at 15:54
OK... I can handle the negative... I subtracted by Pi
â Patrick
Jan 9 '15 at 15:56
 |Â
show 4 more comments
up vote
0
down vote
In fact, you want to know the angle between the $x$ axis and any line going through the origin $(0,0)$ and a point $A (x_a,y_a)$, with $-1 leq x_a,y_a leq 1?$
This is fairly easy. Since there are two angles between two lines, you said you wanted the "right angle", that is, the one to the right hand side of the $x$ axis.
If $x_a times y_a<0$, $theta= tan^-1dfracy_ax_a$
If $x_a=0$, $theta=dfracpi2$
If $x_atimes y_a>0$, $theta= tan^-1dfracy_ax_a-pi$
Yes... you appear to know what I want more so than I do... Let me verify everything and I'll select answer.
â Patrick
Jan 9 '15 at 15:33
You can also use $theta = cos^-1left(fracxsqrtx^2+y^2 right)$ for example, as it will also work when $x = 0$
â servabat
Jan 9 '15 at 15:42
I ran (-.4, -.4) through the bottom formula and got -64.9 Degrees... that doesn't seem right to me... if the first part is running along the X axis.... I would expect a positive answer between 90 and 180 degrees... what am I missing?
â Patrick
Jan 9 '15 at 15:52
@Patrick First, angle are negative when counting them down... (that is, clockwise). Second, did you substract $pi$ or $180°$?
â Martigan
Jan 9 '15 at 15:54
OK... I can handle the negative... I subtracted by Pi
â Patrick
Jan 9 '15 at 15:56
 |Â
show 4 more comments
up vote
0
down vote
up vote
0
down vote
In fact, you want to know the angle between the $x$ axis and any line going through the origin $(0,0)$ and a point $A (x_a,y_a)$, with $-1 leq x_a,y_a leq 1?$
This is fairly easy. Since there are two angles between two lines, you said you wanted the "right angle", that is, the one to the right hand side of the $x$ axis.
If $x_a times y_a<0$, $theta= tan^-1dfracy_ax_a$
If $x_a=0$, $theta=dfracpi2$
If $x_atimes y_a>0$, $theta= tan^-1dfracy_ax_a-pi$
In fact, you want to know the angle between the $x$ axis and any line going through the origin $(0,0)$ and a point $A (x_a,y_a)$, with $-1 leq x_a,y_a leq 1?$
This is fairly easy. Since there are two angles between two lines, you said you wanted the "right angle", that is, the one to the right hand side of the $x$ axis.
If $x_a times y_a<0$, $theta= tan^-1dfracy_ax_a$
If $x_a=0$, $theta=dfracpi2$
If $x_atimes y_a>0$, $theta= tan^-1dfracy_ax_a-pi$
answered Jan 9 '15 at 15:31
Martigan
4,384714
4,384714
Yes... you appear to know what I want more so than I do... Let me verify everything and I'll select answer.
â Patrick
Jan 9 '15 at 15:33
You can also use $theta = cos^-1left(fracxsqrtx^2+y^2 right)$ for example, as it will also work when $x = 0$
â servabat
Jan 9 '15 at 15:42
I ran (-.4, -.4) through the bottom formula and got -64.9 Degrees... that doesn't seem right to me... if the first part is running along the X axis.... I would expect a positive answer between 90 and 180 degrees... what am I missing?
â Patrick
Jan 9 '15 at 15:52
@Patrick First, angle are negative when counting them down... (that is, clockwise). Second, did you substract $pi$ or $180°$?
â Martigan
Jan 9 '15 at 15:54
OK... I can handle the negative... I subtracted by Pi
â Patrick
Jan 9 '15 at 15:56
 |Â
show 4 more comments
Yes... you appear to know what I want more so than I do... Let me verify everything and I'll select answer.
â Patrick
Jan 9 '15 at 15:33
You can also use $theta = cos^-1left(fracxsqrtx^2+y^2 right)$ for example, as it will also work when $x = 0$
â servabat
Jan 9 '15 at 15:42
I ran (-.4, -.4) through the bottom formula and got -64.9 Degrees... that doesn't seem right to me... if the first part is running along the X axis.... I would expect a positive answer between 90 and 180 degrees... what am I missing?
â Patrick
Jan 9 '15 at 15:52
@Patrick First, angle are negative when counting them down... (that is, clockwise). Second, did you substract $pi$ or $180°$?
â Martigan
Jan 9 '15 at 15:54
OK... I can handle the negative... I subtracted by Pi
â Patrick
Jan 9 '15 at 15:56
Yes... you appear to know what I want more so than I do... Let me verify everything and I'll select answer.
â Patrick
Jan 9 '15 at 15:33
Yes... you appear to know what I want more so than I do... Let me verify everything and I'll select answer.
â Patrick
Jan 9 '15 at 15:33
You can also use $theta = cos^-1left(fracxsqrtx^2+y^2 right)$ for example, as it will also work when $x = 0$
â servabat
Jan 9 '15 at 15:42
You can also use $theta = cos^-1left(fracxsqrtx^2+y^2 right)$ for example, as it will also work when $x = 0$
â servabat
Jan 9 '15 at 15:42
I ran (-.4, -.4) through the bottom formula and got -64.9 Degrees... that doesn't seem right to me... if the first part is running along the X axis.... I would expect a positive answer between 90 and 180 degrees... what am I missing?
â Patrick
Jan 9 '15 at 15:52
I ran (-.4, -.4) through the bottom formula and got -64.9 Degrees... that doesn't seem right to me... if the first part is running along the X axis.... I would expect a positive answer between 90 and 180 degrees... what am I missing?
â Patrick
Jan 9 '15 at 15:52
@Patrick First, angle are negative when counting them down... (that is, clockwise). Second, did you substract $pi$ or $180°$?
â Martigan
Jan 9 '15 at 15:54
@Patrick First, angle are negative when counting them down... (that is, clockwise). Second, did you substract $pi$ or $180°$?
â Martigan
Jan 9 '15 at 15:54
OK... I can handle the negative... I subtracted by Pi
â Patrick
Jan 9 '15 at 15:56
OK... I can handle the negative... I subtracted by Pi
â Patrick
Jan 9 '15 at 15:56
 |Â
show 4 more comments
up vote
0
down vote
Not sure EXACTLY what you are asking, but I will answer this to the best of my ability. If you could include a visual that would greatly help me.
When we have two intersecting line segments like this
finding any single value (a, c, b, d) will reveal all other values. For example, if we have the value of a, then c = a, and we have (in degrees) b = d = 180-a.
Therefore, I can equip with the tools to find the angle between two vectors, as you have given. For example, we can treat the first vector as you have said as $[1, 0]$ and the second vector as $[1, 1]$. We take the dot product between them, which just means that we multiply the corresponding values and sum them up, or $<a, b> = sumlimits_i = 1^na_ib_i$ where $n$ is the number of elements in the vector. We use the geometric fact that $<a, b> = |a||b|cos(theta_a, b)$ where $|a|$ means its norm or magnitude (I only deal with standard inner product and norm here). This gives us that the dot product, which is $1$, is equal to $sqrt2cos(theta)$, which means $cos(theta) = frac1sqrt2$
This in turn gives us that the angle between these two is 45 degrees, and we can figure out the adjacent angle as 135, and the vertical angles all share the same degrees.
add a comment |Â
up vote
0
down vote
Not sure EXACTLY what you are asking, but I will answer this to the best of my ability. If you could include a visual that would greatly help me.
When we have two intersecting line segments like this
finding any single value (a, c, b, d) will reveal all other values. For example, if we have the value of a, then c = a, and we have (in degrees) b = d = 180-a.
Therefore, I can equip with the tools to find the angle between two vectors, as you have given. For example, we can treat the first vector as you have said as $[1, 0]$ and the second vector as $[1, 1]$. We take the dot product between them, which just means that we multiply the corresponding values and sum them up, or $<a, b> = sumlimits_i = 1^na_ib_i$ where $n$ is the number of elements in the vector. We use the geometric fact that $<a, b> = |a||b|cos(theta_a, b)$ where $|a|$ means its norm or magnitude (I only deal with standard inner product and norm here). This gives us that the dot product, which is $1$, is equal to $sqrt2cos(theta)$, which means $cos(theta) = frac1sqrt2$
This in turn gives us that the angle between these two is 45 degrees, and we can figure out the adjacent angle as 135, and the vertical angles all share the same degrees.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Not sure EXACTLY what you are asking, but I will answer this to the best of my ability. If you could include a visual that would greatly help me.
When we have two intersecting line segments like this
finding any single value (a, c, b, d) will reveal all other values. For example, if we have the value of a, then c = a, and we have (in degrees) b = d = 180-a.
Therefore, I can equip with the tools to find the angle between two vectors, as you have given. For example, we can treat the first vector as you have said as $[1, 0]$ and the second vector as $[1, 1]$. We take the dot product between them, which just means that we multiply the corresponding values and sum them up, or $<a, b> = sumlimits_i = 1^na_ib_i$ where $n$ is the number of elements in the vector. We use the geometric fact that $<a, b> = |a||b|cos(theta_a, b)$ where $|a|$ means its norm or magnitude (I only deal with standard inner product and norm here). This gives us that the dot product, which is $1$, is equal to $sqrt2cos(theta)$, which means $cos(theta) = frac1sqrt2$
This in turn gives us that the angle between these two is 45 degrees, and we can figure out the adjacent angle as 135, and the vertical angles all share the same degrees.
Not sure EXACTLY what you are asking, but I will answer this to the best of my ability. If you could include a visual that would greatly help me.
When we have two intersecting line segments like this
finding any single value (a, c, b, d) will reveal all other values. For example, if we have the value of a, then c = a, and we have (in degrees) b = d = 180-a.
Therefore, I can equip with the tools to find the angle between two vectors, as you have given. For example, we can treat the first vector as you have said as $[1, 0]$ and the second vector as $[1, 1]$. We take the dot product between them, which just means that we multiply the corresponding values and sum them up, or $<a, b> = sumlimits_i = 1^na_ib_i$ where $n$ is the number of elements in the vector. We use the geometric fact that $<a, b> = |a||b|cos(theta_a, b)$ where $|a|$ means its norm or magnitude (I only deal with standard inner product and norm here). This gives us that the dot product, which is $1$, is equal to $sqrt2cos(theta)$, which means $cos(theta) = frac1sqrt2$
This in turn gives us that the angle between these two is 45 degrees, and we can figure out the adjacent angle as 135, and the vertical angles all share the same degrees.
answered Jul 6 '17 at 16:09
Charlie Tian
9110
9110
add a comment |Â
add a comment |Â
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Are you taking as input a pair of coordinates? Like, (1,5), for example?
â turkeyhundt
Jan 9 '15 at 15:22
2
Does angel = angle? So as it extends along X axis, you are using $(y, x)$ rather than $(x, y) $ coordinates?
â servabat
Jan 9 '15 at 15:23
Thanks for pointing it out... I've updated it.
â Patrick
Jan 9 '15 at 15:28
You should try and learn Latex if you are going to use this site. It is the formatting standard everyone goes by.
â Wintermute
Jan 9 '15 at 15:40
@mtiano yeah.. I tried using it for Pi.. and it doesn't want to play nice.
â Patrick
Jan 9 '15 at 16:00