Mixing
What is the angle between two intersecting tangents to a circle?
$begingroup$
A circle of radius $r$ with centre $C$ is located at distance $d$ from a point $P$.
There are two tangents to the circle which pass through point $P$ - one on each side. They intersect the circle at points $A$ and $B$.
What is the angle through $P$ between these two tangents? In other words, angle $APB$?
I know that angle $APB$ + angle $ACB$ add up to 180.
(Not homework, for graphics programming)
Thanks,
Louise
geometry
asked Jan 15 at 16:54
LouiseLouise
1031
$endgroup$
add a comment |
$begingroup$
A circle of radius $r$ with centre $C$ is located at distance $d$ from a point $P$.
There are two tangents to the circle which pass through point $P$ - one on each side. They intersect the circle at points $A$ and $B$.
What is the angle through $P$ between these two tangents? In other words, angle $APB$?
I know that angle $APB$ + angle $ACB$ add up to 180.
(Not homework, for graphics programming)
Thanks,
Louise
geometry
asked Jan 15 at 16:54
LouiseLouise
1031
$endgroup$
$begingroup$
What kind of graphics programming problem?
$endgroup$
– lightxbulb
Jan 15 at 16:58
add a comment |
$begingroup$
A circle of radius $r$ with centre $C$ is located at distance $d$ from a point $P$.
There are two tangents to the circle which pass through point $P$ - one on each side. They intersect the circle at points $A$ and $B$.
What is the angle through $P$ between these two tangents? In other words, angle $APB$?
I know that angle $APB$ + angle $ACB$ add up to 180.
(Not homework, for graphics programming)
Thanks,
Louise
geometry
asked Jan 15 at 16:54
LouiseLouise
1031
$endgroup$
A circle of radius $r$ with centre $C$ is located at distance $d$ from a point $P$.
There are two tangents to the circle which pass through point $P$ - one on each side. They intersect the circle at points $A$ and $B$.
What is the angle through $P$ between these two tangents? In other words, angle $APB$?
I know that angle $APB$ + angle $ACB$ add up to 180.
(Not homework, for graphics programming)
Thanks,
Louise
geometry
geometry
asked Jan 15 at 16:54
LouiseLouise
1031
asked Jan 15 at 16:54
LouiseLouise
1031
asked Jan 15 at 16:54
LouiseLouise
1031
asked Jan 15 at 16:54
LouiseLouise
1031
asked Jan 15 at 16:54
LouiseLouise
1031
1031
$begingroup$
What kind of graphics programming problem?
$endgroup$
– lightxbulb
Jan 15 at 16:58
add a comment |
$begingroup$
What kind of graphics programming problem?
$endgroup$
– lightxbulb
Jan 15 at 16:58
$begingroup$
What kind of graphics programming problem?
$endgroup$
– lightxbulb
Jan 15 at 16:58
$begingroup$
What kind of graphics programming problem?
$endgroup$
– lightxbulb
Jan 15 at 16:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is a picture:
$overline {CP} = d$
$angle CAP$ and $angle BAP$ are right angles, and $triangle APB$ is isosceles.
$mangle APC = arcsin frac rd\
mangle APB = 2arcsin frac rd\
mangle BAP = arccos frac rd$
answered Jan 15 at 17:58
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
Thank you! How did you generate the picture?
$endgroup$
– Louise
Jan 16 at 17:14
$begingroup$
I use MS Paint.
$endgroup$
– Doug M
Jan 16 at 18:57
add a comment |
$begingroup$
I presume "at distance $d$" means that $|CP|=r+d$.
The triangle $ACP$ is right angled, with $|AC|=r$. Then
$$sinangle APC=frac{r}{r+d}.$$
Then
$$angle ABP=2angle APC=2sin^{-1}frac r{r+d}.$$
answered Jan 15 at 16:59
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a picture:
$overline {CP} = d$
$angle CAP$ and $angle BAP$ are right angles, and $triangle APB$ is isosceles.
$mangle APC = arcsin frac rd\
mangle APB = 2arcsin frac rd\
mangle BAP = arccos frac rd$
answered Jan 15 at 17:58
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
Thank you! How did you generate the picture?
$endgroup$
– Louise
Jan 16 at 17:14
$begingroup$
I use MS Paint.
$endgroup$
– Doug M
Jan 16 at 18:57
add a comment |
$begingroup$
Here is a picture:
$overline {CP} = d$
$angle CAP$ and $angle BAP$ are right angles, and $triangle APB$ is isosceles.
$mangle APC = arcsin frac rd\
mangle APB = 2arcsin frac rd\
mangle BAP = arccos frac rd$
answered Jan 15 at 17:58
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
Thank you! How did you generate the picture?
$endgroup$
– Louise
Jan 16 at 17:14
$begingroup$
I use MS Paint.
$endgroup$
– Doug M
Jan 16 at 18:57
add a comment |
$begingroup$
Here is a picture:
$overline {CP} = d$
$angle CAP$ and $angle BAP$ are right angles, and $triangle APB$ is isosceles.
$mangle APC = arcsin frac rd\
mangle APB = 2arcsin frac rd\
mangle BAP = arccos frac rd$
answered Jan 15 at 17:58
Doug MDoug M
45.2k31854
$endgroup$
Here is a picture:
$overline {CP} = d$
$angle CAP$ and $angle BAP$ are right angles, and $triangle APB$ is isosceles.
$mangle APC = arcsin frac rd\
mangle APB = 2arcsin frac rd\
mangle BAP = arccos frac rd$
answered Jan 15 at 17:58
Doug MDoug M
45.2k31854
answered Jan 15 at 17:58
Doug MDoug M
45.2k31854
answered Jan 15 at 17:58
Doug MDoug M
45.2k31854
answered Jan 15 at 17:58
Doug MDoug M
45.2k31854
45.2k31854
$begingroup$
Thank you! How did you generate the picture?
$endgroup$
– Louise
Jan 16 at 17:14
$begingroup$
I use MS Paint.
$endgroup$
– Doug M
Jan 16 at 18:57
add a comment |
$begingroup$
Thank you! How did you generate the picture?
$endgroup$
– Louise
Jan 16 at 17:14
$begingroup$
I use MS Paint.
$endgroup$
– Doug M
Jan 16 at 18:57
$begingroup$
Thank you! How did you generate the picture?
$endgroup$
– Louise
Jan 16 at 17:14
$begingroup$
Thank you! How did you generate the picture?
$endgroup$
– Louise
Jan 16 at 17:14
$begingroup$
I use MS Paint.
$endgroup$
– Doug M
Jan 16 at 18:57
$begingroup$
I use MS Paint.
$endgroup$
– Doug M
Jan 16 at 18:57
add a comment |
$begingroup$
I presume "at distance $d$" means that $|CP|=r+d$.
The triangle $ACP$ is right angled, with $|AC|=r$. Then
$$sinangle APC=frac{r}{r+d}.$$
Then
$$angle ABP=2angle APC=2sin^{-1}frac r{r+d}.$$
answered Jan 15 at 16:59
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
I presume "at distance $d$" means that $|CP|=r+d$.
The triangle $ACP$ is right angled, with $|AC|=r$. Then
$$sinangle APC=frac{r}{r+d}.$$
Then
$$angle ABP=2angle APC=2sin^{-1}frac r{r+d}.$$
answered Jan 15 at 16:59
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
I presume "at distance $d$" means that $|CP|=r+d$.
The triangle $ACP$ is right angled, with $|AC|=r$. Then
$$sinangle APC=frac{r}{r+d}.$$
Then
$$angle ABP=2angle APC=2sin^{-1}frac r{r+d}.$$
answered Jan 15 at 16:59
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
I presume "at distance $d$" means that $|CP|=r+d$.
The triangle $ACP$ is right angled, with $|AC|=r$. Then
$$sinangle APC=frac{r}{r+d}.$$
Then
$$angle ABP=2angle APC=2sin^{-1}frac r{r+d}.$$
answered Jan 15 at 16:59
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 15 at 16:59
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 15 at 16:59
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 15 at 16:59
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
add a comment |
add a comment |
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$begingroup$
What kind of graphics programming problem?
$endgroup$
– lightxbulb
Jan 15 at 16:58