Mixing
calculating angle in circle
How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is $(x_1,x_2)$. I have a point on circumstance of circle that is $(x_2,y_2)$. Also I know the radius of circle that is R.
How to calculate the angle between these two lines
-- line starting from origin and ending on circumstance at 0 degree
-- line starting from origin and ending on $(x_2,y_2)$.
geometry trigonometry circle angle
edited Jul 20 '17 at 17:30
Martin Sleziak
44.7k7115270
asked Dec 27 '11 at 7:57
coure2011
200249
add a comment |
How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is $(x_1,x_2)$. I have a point on circumstance of circle that is $(x_2,y_2)$. Also I know the radius of circle that is R.
How to calculate the angle between these two lines
-- line starting from origin and ending on circumstance at 0 degree
-- line starting from origin and ending on $(x_2,y_2)$.
geometry trigonometry circle angle
edited Jul 20 '17 at 17:30
Martin Sleziak
44.7k7115270
asked Dec 27 '11 at 7:57
coure2011
200249
1
Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
– Ramana Venkata
Dec 27 '11 at 8:01
Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
– J. M. is not a mathematician
Dec 27 '11 at 9:29
add a comment |
How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is $(x_1,x_2)$. I have a point on circumstance of circle that is $(x_2,y_2)$. Also I know the radius of circle that is R.
How to calculate the angle between these two lines
-- line starting from origin and ending on circumstance at 0 degree
-- line starting from origin and ending on $(x_2,y_2)$.
geometry trigonometry circle angle
edited Jul 20 '17 at 17:30
Martin Sleziak
44.7k7115270
asked Dec 27 '11 at 7:57
coure2011
200249
How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is $(x_1,x_2)$. I have a point on circumstance of circle that is $(x_2,y_2)$. Also I know the radius of circle that is R.
How to calculate the angle between these two lines
-- line starting from origin and ending on circumstance at 0 degree
-- line starting from origin and ending on $(x_2,y_2)$.
geometry trigonometry circle angle
geometry trigonometry circle angle
edited Jul 20 '17 at 17:30
Martin Sleziak
44.7k7115270
asked Dec 27 '11 at 7:57
coure2011
200249
edited Jul 20 '17 at 17:30
Martin Sleziak
44.7k7115270
asked Dec 27 '11 at 7:57
coure2011
200249
edited Jul 20 '17 at 17:30
Martin Sleziak
44.7k7115270
edited Jul 20 '17 at 17:30
Martin Sleziak
44.7k7115270
edited Jul 20 '17 at 17:30
Martin Sleziak
44.7k7115270
44.7k7115270
asked Dec 27 '11 at 7:57
coure2011
200249
asked Dec 27 '11 at 7:57
coure2011
200249
asked Dec 27 '11 at 7:57
coure2011
200249
200249
1
Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
– Ramana Venkata
Dec 27 '11 at 8:01
Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
– J. M. is not a mathematician
Dec 27 '11 at 9:29
add a comment |
1
Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
– Ramana Venkata
Dec 27 '11 at 8:01
Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
– J. M. is not a mathematician
Dec 27 '11 at 9:29
1
1
Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
– Ramana Venkata
Dec 27 '11 at 8:01
Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
– Ramana Venkata
Dec 27 '11 at 8:01
Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
– J. M. is not a mathematician
Dec 27 '11 at 9:29
Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
– J. M. is not a mathematician
Dec 27 '11 at 9:29
add a comment |
6 Answers
6
active
oldest
votes
I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:
angle = atan2(y2-y1, x2-x1)
(which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.
The result comes out in radians, of course.
answered Dec 28 '11 at 2:34
Henning Makholm
238k16303538
1
thanks for searching about me and giving me exactly what I was looking for.
– coure2011
Dec 28 '11 at 7:37
add a comment |
HINT:
Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.
answered Dec 27 '11 at 8:07
davidlowryduda♦
74.4k7117251
add a comment |
The angle can be unambiguously computed as
$$
theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
$$
This can be seen in the following diagram:
answered Dec 28 '11 at 23:20
robjohn♦
265k27303624
This is how to computeatan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
– robjohn♦
Dec 28 '11 at 23:27
add a comment |
Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.
Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.
The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.
Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$
By solving for $cos A$ we get:
$$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Which simplifies to:
$$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Therefore the angle is:
$$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$
I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.
Note: Please feel free to comment if I have made any careless mistakes.
answered Dec 27 '11 at 11:19
E.O.
5,01452550
add a comment |
You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
At least end point of another line should also known to find the angle between those two lines in circle.
answered Dec 27 '11 at 9:18
mevada.yogesh
1013
2
It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
– Henning Makholm
Dec 28 '11 at 2:37
add a comment |
Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.
We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.
edited Dec 28 '11 at 22:02
Brandon Carter
7,21322538
answered Dec 27 '11 at 12:15
Tomarinator
1,32111023
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f94379%2fcalculating-angle-in-circle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:
angle = atan2(y2-y1, x2-x1)
(which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.
The result comes out in radians, of course.
answered Dec 28 '11 at 2:34
Henning Makholm
238k16303538
1
thanks for searching about me and giving me exactly what I was looking for.
– coure2011
Dec 28 '11 at 7:37
add a comment |
I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:
angle = atan2(y2-y1, x2-x1)
(which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.
The result comes out in radians, of course.
answered Dec 28 '11 at 2:34
Henning Makholm
238k16303538
1
thanks for searching about me and giving me exactly what I was looking for.
– coure2011
Dec 28 '11 at 7:37
add a comment |
I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:
angle = atan2(y2-y1, x2-x1)
(which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.
The result comes out in radians, of course.
answered Dec 28 '11 at 2:34
Henning Makholm
238k16303538
I see you have a bunch of rep at SO, so in case you're looking for a practical, coding-oriented answer:
angle = atan2(y2-y1, x2-x1)
(which may need to be Math.atan2 or something such, depending on your language of choice). This usually maps down to a hardware operation specifically constructed for solving exactly this problem, and is therefore more efficient than the formulas involving standard trigonometric functions you find in the other answers.
The result comes out in radians, of course.
answered Dec 28 '11 at 2:34
Henning Makholm
238k16303538
answered Dec 28 '11 at 2:34
Henning Makholm
238k16303538
answered Dec 28 '11 at 2:34
Henning Makholm
238k16303538
answered Dec 28 '11 at 2:34
Henning Makholm
238k16303538
238k16303538
1
thanks for searching about me and giving me exactly what I was looking for.
– coure2011
Dec 28 '11 at 7:37
add a comment |
1
thanks for searching about me and giving me exactly what I was looking for.
– coure2011
Dec 28 '11 at 7:37
1
1
thanks for searching about me and giving me exactly what I was looking for.
– coure2011
Dec 28 '11 at 7:37
thanks for searching about me and giving me exactly what I was looking for.
– coure2011
Dec 28 '11 at 7:37
add a comment |
HINT:
Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.
answered Dec 27 '11 at 8:07
davidlowryduda♦
74.4k7117251
add a comment |
HINT:
Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.
answered Dec 27 '11 at 8:07
davidlowryduda♦
74.4k7117251
add a comment |
HINT:
Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.
answered Dec 27 '11 at 8:07
davidlowryduda♦
74.4k7117251
HINT:
Don't think of it as a circle. Instead, think of it as a triangle with three points $(x_1, y_1), ; (x_1 + R, y_1), ; (x_2, y_2)$. Then you can use any of the standard techniques for finding angles in a triangle, like the law of sines or cosines, etc.
answered Dec 27 '11 at 8:07
davidlowryduda♦
74.4k7117251
answered Dec 27 '11 at 8:07
davidlowryduda♦
74.4k7117251
answered Dec 27 '11 at 8:07
davidlowryduda♦
74.4k7117251
answered Dec 27 '11 at 8:07
davidlowryduda♦
74.4k7117251
74.4k7117251
add a comment |
add a comment |
The angle can be unambiguously computed as
$$
theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
$$
This can be seen in the following diagram:
answered Dec 28 '11 at 23:20
robjohn♦
265k27303624
This is how to computeatan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
– robjohn♦
Dec 28 '11 at 23:27
add a comment |
The angle can be unambiguously computed as
$$
theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
$$
This can be seen in the following diagram:
answered Dec 28 '11 at 23:20
robjohn♦
265k27303624
This is how to computeatan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
– robjohn♦
Dec 28 '11 at 23:27
add a comment |
The angle can be unambiguously computed as
$$
theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
$$
This can be seen in the following diagram:
answered Dec 28 '11 at 23:20
robjohn♦
265k27303624
The angle can be unambiguously computed as
$$
theta=2;tan^{-1}left(dfrac{y_2-y_1}{x_2-x_1+sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right)
$$
This can be seen in the following diagram:
answered Dec 28 '11 at 23:20
robjohn♦
265k27303624
answered Dec 28 '11 at 23:20
robjohn♦
265k27303624
answered Dec 28 '11 at 23:20
robjohn♦
265k27303624
answered Dec 28 '11 at 23:20
robjohn♦
265k27303624
265k27303624
This is how to computeatan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
– robjohn♦
Dec 28 '11 at 23:27
add a comment |
This is how to computeatan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.
– robjohn♦
Dec 28 '11 at 23:27
This is how to compute
atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.– robjohn♦
Dec 28 '11 at 23:27
This is how to compute
atan2, as in @HenningMakholm's answer, when all you have is $tan^{-1}$.– robjohn♦
Dec 28 '11 at 23:27
add a comment |
Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.
Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.
The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.
Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$
By solving for $cos A$ we get:
$$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Which simplifies to:
$$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Therefore the angle is:
$$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$
I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.
Note: Please feel free to comment if I have made any careless mistakes.
answered Dec 27 '11 at 11:19
E.O.
5,01452550
add a comment |
Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.
Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.
The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.
Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$
By solving for $cos A$ we get:
$$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Which simplifies to:
$$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Therefore the angle is:
$$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$
I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.
Note: Please feel free to comment if I have made any careless mistakes.
answered Dec 27 '11 at 11:19
E.O.
5,01452550
add a comment |
Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.
Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.
The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.
Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$
By solving for $cos A$ we get:
$$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Which simplifies to:
$$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Therefore the angle is:
$$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$
I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.
Note: Please feel free to comment if I have made any careless mistakes.
answered Dec 27 '11 at 11:19
E.O.
5,01452550
Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.
Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.
The distance between $A$ and $B$ is given by $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.
Now, using the law of cosines which is $a^2=b^2+c^2-2bccos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}cos A$
By solving for $cos A$ we get:
$$cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2over -2cdot rcdot sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Which simplifies to:
$$cos A={x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$
Therefore the angle is:
$$A=cos ^{-1} left( {x_2-x_1over sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}right )$$
I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.
Note: Please feel free to comment if I have made any careless mistakes.
answered Dec 27 '11 at 11:19
E.O.
5,01452550
edited Dec 27 '11 at 11:47
answered Dec 27 '11 at 11:19
E.O.
5,01452550
answered Dec 27 '11 at 11:19
E.O.
5,01452550
answered Dec 27 '11 at 11:19
E.O.
5,01452550
5,01452550
add a comment |
add a comment |
You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
At least end point of another line should also known to find the angle between those two lines in circle.
answered Dec 27 '11 at 9:18
mevada.yogesh
1013
2
It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
– Henning Makholm
Dec 28 '11 at 2:37
add a comment |
You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
At least end point of another line should also known to find the angle between those two lines in circle.
answered Dec 27 '11 at 9:18
mevada.yogesh
1013
2
It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
– Henning Makholm
Dec 28 '11 at 2:37
add a comment |
You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
At least end point of another line should also known to find the angle between those two lines in circle.
answered Dec 27 '11 at 9:18
mevada.yogesh
1013
You have only provided the radius and end point co-ordinates of a radius/line only. So, the information provided by you is insufficient for calculating the angle between those two lines.
At least end point of another line should also known to find the angle between those two lines in circle.
answered Dec 27 '11 at 9:18
mevada.yogesh
1013
answered Dec 27 '11 at 9:18
mevada.yogesh
1013
answered Dec 27 '11 at 9:18
mevada.yogesh
1013
answered Dec 27 '11 at 9:18
mevada.yogesh
1013
1013
2
It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
– Henning Makholm
Dec 28 '11 at 2:37
add a comment |
2
It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
– Henning Makholm
Dec 28 '11 at 2:37
2
2
It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
– Henning Makholm
Dec 28 '11 at 2:37
It seems clear from the diagram that the other line goes straight to the right, in the same direction as the $x$ axis.
– Henning Makholm
Dec 28 '11 at 2:37
add a comment |
Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.
We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.
edited Dec 28 '11 at 22:02
Brandon Carter
7,21322538
answered Dec 27 '11 at 12:15
Tomarinator
1,32111023
add a comment |
Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.
We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.
edited Dec 28 '11 at 22:02
Brandon Carter
7,21322538
answered Dec 27 '11 at 12:15
Tomarinator
1,32111023
add a comment |
Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.
We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.
edited Dec 28 '11 at 22:02
Brandon Carter
7,21322538
answered Dec 27 '11 at 12:15
Tomarinator
1,32111023
Let $theta$ denote the unknown angle. Then we can easily find $tan(180 - theta)$ by completing a triangle with vertices $(x_1, y_1)$ and $(x_2,y_2)$.
We have $tan (180-theta) = frac{y_2-y_1}{x_2-x_1} = - tan theta$. Hence we know $theta = arctan (frac{y1-y2}{x2-x1})$.
edited Dec 28 '11 at 22:02
Brandon Carter
7,21322538
answered Dec 27 '11 at 12:15
Tomarinator
1,32111023
edited Dec 28 '11 at 22:02
Brandon Carter
7,21322538
edited Dec 28 '11 at 22:02
Brandon Carter
7,21322538
edited Dec 28 '11 at 22:02
Brandon Carter
7,21322538
7,21322538
answered Dec 27 '11 at 12:15
Tomarinator
1,32111023
answered Dec 27 '11 at 12:15
Tomarinator
1,32111023
answered Dec 27 '11 at 12:15
Tomarinator
1,32111023
1,32111023
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f94379%2fcalculating-angle-in-circle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Is the origin is at $(x_1,x_2)$ or $(x_1,y_1)$??
– Ramana Venkata
Dec 27 '11 at 8:01
Two-argument arctangent, $arctan(x_2-x_1,y_2-y_1)$, works quite well here.
– J. M. is not a mathematician
Dec 27 '11 at 9:29