Distance from a point on a circle to another when the angle between them is given
$begingroup$
There are two points on a circle and the angle between them is known, as well as the radius of the circle. What I want to do is find the horizontal and vertical distance between these points. Is there any way to do that?(the distance given by the horizontal&vertical yellow lines)
geometry
asked Apr 15 '14 at 15:54
Suhair ZainSuhair Zain
12
$endgroup$
add a comment |
$begingroup$
There are two points on a circle and the angle between them is known, as well as the radius of the circle. What I want to do is find the horizontal and vertical distance between these points. Is there any way to do that?(the distance given by the horizontal&vertical yellow lines)
geometry
asked Apr 15 '14 at 15:54
Suhair ZainSuhair Zain
12
$endgroup$
1
$begingroup$
use the law of sines : en.wikipedia.org/wiki/Law_of_cosines
$endgroup$
– aflous
Apr 15 '14 at 15:59
add a comment |
$begingroup$
There are two points on a circle and the angle between them is known, as well as the radius of the circle. What I want to do is find the horizontal and vertical distance between these points. Is there any way to do that?(the distance given by the horizontal&vertical yellow lines)
geometry
asked Apr 15 '14 at 15:54
Suhair ZainSuhair Zain
12
$endgroup$
There are two points on a circle and the angle between them is known, as well as the radius of the circle. What I want to do is find the horizontal and vertical distance between these points. Is there any way to do that?(the distance given by the horizontal&vertical yellow lines)
geometry
geometry
asked Apr 15 '14 at 15:54
Suhair ZainSuhair Zain
12
asked Apr 15 '14 at 15:54
Suhair ZainSuhair Zain
12
asked Apr 15 '14 at 15:54
Suhair ZainSuhair Zain
12
asked Apr 15 '14 at 15:54
Suhair ZainSuhair Zain
12
asked Apr 15 '14 at 15:54
Suhair ZainSuhair Zain
12
12
1
$begingroup$
use the law of sines : en.wikipedia.org/wiki/Law_of_cosines
$endgroup$
– aflous
Apr 15 '14 at 15:59
add a comment |
1
$begingroup$
use the law of sines : en.wikipedia.org/wiki/Law_of_cosines
$endgroup$
– aflous
Apr 15 '14 at 15:59
1
1
$begingroup$
use the law of sines : en.wikipedia.org/wiki/Law_of_cosines
$endgroup$
– aflous
Apr 15 '14 at 15:59
$begingroup$
use the law of sines : en.wikipedia.org/wiki/Law_of_cosines
$endgroup$
– aflous
Apr 15 '14 at 15:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Using the law of sines you easily get
$$d= sqrt 2 r sqrt {1 - cos x}$$
where d is the hypotenuse of your yellow triangle.
answered Apr 15 '14 at 16:02
aflousaflous
343113
$endgroup$
add a comment |
$begingroup$
If the radius of the circle is r, then using the law of cosines the length of the chord joining the two points is
$$
begin{align*}
c &= sqrt{r^2 + r^2 - 2r^2cos{x}}
\ &= rsqrt{2(1 - cos{x})}
end{align*}
$$
However, to find the horizontal and vertical components of the distance you need to know the location of the points. This is because if you imagine the same angle in a different position on the circle (such as A and B below), the chord length is the same but the horizontal and vertical components are different. Here is a diagram made with Isosceles:
answered Apr 15 '14 at 16:19
architectpianistarchitectpianist
76148
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add a comment |
$begingroup$
Better to use $t$ or $theta $ for the angle so as not to confuse with $x$ coordinate.
Top Start Point
$$ (x,y)= (0,r) $$
End Point
$$ (x,y)= r [ sin t, cos t] $$
Horizontal and Vertical distances of your yellow lines ( please thicken them ), are $x$ and $y$ distance differences respectively.
Use the distance relation
$$ d^2 = 2 , r^2 (1- cos t), quad d = rsqrt{2 (1- cos t) }. $$
answered Aug 24 '17 at 11:33
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the law of sines you easily get
$$d= sqrt 2 r sqrt {1 - cos x}$$
where d is the hypotenuse of your yellow triangle.
answered Apr 15 '14 at 16:02
aflousaflous
343113
$endgroup$
add a comment |
$begingroup$
Using the law of sines you easily get
$$d= sqrt 2 r sqrt {1 - cos x}$$
where d is the hypotenuse of your yellow triangle.
answered Apr 15 '14 at 16:02
aflousaflous
343113
$endgroup$
add a comment |
$begingroup$
Using the law of sines you easily get
$$d= sqrt 2 r sqrt {1 - cos x}$$
where d is the hypotenuse of your yellow triangle.
answered Apr 15 '14 at 16:02
aflousaflous
343113
$endgroup$
Using the law of sines you easily get
$$d= sqrt 2 r sqrt {1 - cos x}$$
where d is the hypotenuse of your yellow triangle.
answered Apr 15 '14 at 16:02
aflousaflous
343113
edited Apr 15 '14 at 16:13
answered Apr 15 '14 at 16:02
aflousaflous
343113
answered Apr 15 '14 at 16:02
aflousaflous
343113
answered Apr 15 '14 at 16:02
aflousaflous
343113
343113
add a comment |
add a comment |
$begingroup$
If the radius of the circle is r, then using the law of cosines the length of the chord joining the two points is
$$
begin{align*}
c &= sqrt{r^2 + r^2 - 2r^2cos{x}}
\ &= rsqrt{2(1 - cos{x})}
end{align*}
$$
However, to find the horizontal and vertical components of the distance you need to know the location of the points. This is because if you imagine the same angle in a different position on the circle (such as A and B below), the chord length is the same but the horizontal and vertical components are different. Here is a diagram made with Isosceles:
answered Apr 15 '14 at 16:19
architectpianistarchitectpianist
76148
$endgroup$
add a comment |
$begingroup$
If the radius of the circle is r, then using the law of cosines the length of the chord joining the two points is
$$
begin{align*}
c &= sqrt{r^2 + r^2 - 2r^2cos{x}}
\ &= rsqrt{2(1 - cos{x})}
end{align*}
$$
However, to find the horizontal and vertical components of the distance you need to know the location of the points. This is because if you imagine the same angle in a different position on the circle (such as A and B below), the chord length is the same but the horizontal and vertical components are different. Here is a diagram made with Isosceles:
answered Apr 15 '14 at 16:19
architectpianistarchitectpianist
76148
$endgroup$
add a comment |
$begingroup$
If the radius of the circle is r, then using the law of cosines the length of the chord joining the two points is
$$
begin{align*}
c &= sqrt{r^2 + r^2 - 2r^2cos{x}}
\ &= rsqrt{2(1 - cos{x})}
end{align*}
$$
However, to find the horizontal and vertical components of the distance you need to know the location of the points. This is because if you imagine the same angle in a different position on the circle (such as A and B below), the chord length is the same but the horizontal and vertical components are different. Here is a diagram made with Isosceles:
answered Apr 15 '14 at 16:19
architectpianistarchitectpianist
76148
$endgroup$
If the radius of the circle is r, then using the law of cosines the length of the chord joining the two points is
$$
begin{align*}
c &= sqrt{r^2 + r^2 - 2r^2cos{x}}
\ &= rsqrt{2(1 - cos{x})}
end{align*}
$$
However, to find the horizontal and vertical components of the distance you need to know the location of the points. This is because if you imagine the same angle in a different position on the circle (such as A and B below), the chord length is the same but the horizontal and vertical components are different. Here is a diagram made with Isosceles:
answered Apr 15 '14 at 16:19
architectpianistarchitectpianist
76148
answered Apr 15 '14 at 16:19
architectpianistarchitectpianist
76148
answered Apr 15 '14 at 16:19
architectpianistarchitectpianist
76148
answered Apr 15 '14 at 16:19
architectpianistarchitectpianist
76148
76148
add a comment |
add a comment |
$begingroup$
Better to use $t$ or $theta $ for the angle so as not to confuse with $x$ coordinate.
Top Start Point
$$ (x,y)= (0,r) $$
End Point
$$ (x,y)= r [ sin t, cos t] $$
Horizontal and Vertical distances of your yellow lines ( please thicken them ), are $x$ and $y$ distance differences respectively.
Use the distance relation
$$ d^2 = 2 , r^2 (1- cos t), quad d = rsqrt{2 (1- cos t) }. $$
answered Aug 24 '17 at 11:33
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
Better to use $t$ or $theta $ for the angle so as not to confuse with $x$ coordinate.
Top Start Point
$$ (x,y)= (0,r) $$
End Point
$$ (x,y)= r [ sin t, cos t] $$
Horizontal and Vertical distances of your yellow lines ( please thicken them ), are $x$ and $y$ distance differences respectively.
Use the distance relation
$$ d^2 = 2 , r^2 (1- cos t), quad d = rsqrt{2 (1- cos t) }. $$
answered Aug 24 '17 at 11:33
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
Better to use $t$ or $theta $ for the angle so as not to confuse with $x$ coordinate.
Top Start Point
$$ (x,y)= (0,r) $$
End Point
$$ (x,y)= r [ sin t, cos t] $$
Horizontal and Vertical distances of your yellow lines ( please thicken them ), are $x$ and $y$ distance differences respectively.
Use the distance relation
$$ d^2 = 2 , r^2 (1- cos t), quad d = rsqrt{2 (1- cos t) }. $$
answered Aug 24 '17 at 11:33
NarasimhamNarasimham
20.8k52158
$endgroup$
Better to use $t$ or $theta $ for the angle so as not to confuse with $x$ coordinate.
Top Start Point
$$ (x,y)= (0,r) $$
End Point
$$ (x,y)= r [ sin t, cos t] $$
Horizontal and Vertical distances of your yellow lines ( please thicken them ), are $x$ and $y$ distance differences respectively.
Use the distance relation
$$ d^2 = 2 , r^2 (1- cos t), quad d = rsqrt{2 (1- cos t) }. $$
answered Aug 24 '17 at 11:33
NarasimhamNarasimham
20.8k52158
edited Aug 24 '17 at 11:38
answered Aug 24 '17 at 11:33
NarasimhamNarasimham
20.8k52158
answered Aug 24 '17 at 11:33
NarasimhamNarasimham
20.8k52158
answered Aug 24 '17 at 11:33
NarasimhamNarasimham
20.8k52158
20.8k52158
add a comment |
add a comment |
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$begingroup$
use the law of sines : en.wikipedia.org/wiki/Law_of_cosines
$endgroup$
– aflous
Apr 15 '14 at 15:59